Search: a090079 -id:a090079
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A000975
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a(2n) = 2*a(2n-1), a(2n+1) = 2*a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).
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+10
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0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845, 43690, 87381, 174762, 349525, 699050, 1398101, 2796202, 5592405, 11184810, 22369621, 44739242, 89478485, 178956970, 357913941, 715827882, 1431655765, 2863311530, 5726623061, 11453246122
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OFFSET
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0,3
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COMMENTS
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Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)
Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.
Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000
a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002
Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022
a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005
Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006
In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008
For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009
This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).
2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)
Represented in binary form each term after the second one contains every previous term as a substring.
The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.
Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n.
(End)
Also the number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013
a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013
a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015
a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015
Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016
Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows: 2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017
Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) = x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017
a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018
Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019
Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.
Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)
With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:
..1 3 7 15...
..1 0 1 1 1 0 1 0 1 0 1 1 1 0 1...
..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)
.....1...........2.......................5...; as to zeros.
..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:
..0, 1, 2, 0, 1, 2, ... whereas even numbered disks move in the pattern:
..0, 2, 1, 0, 2, 1, ... (End)
Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021
Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:
{1} {1} {1} {1}
{2} {2} {2}
{3} {3}
{1,3} {4}
{1,2,3} {1,3}
{2,4}
{1,2,3}
{1,2,4}
{1,3,4}
{2,3,4}
The complement is counted by A005578.
For mean instead of median we have A051293, counting empty sets A327475.
(End)
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REFERENCES
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Thomas Fink and Yong Mao, The 85 Ways to Tie a Tie, Broadway Books, New York (1999), p. 138.
Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009.
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LINKS
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FORMULA
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a(n) = ceiling(2*(2^n-1)/3).
Alternating sum transform (PSumSIGN) of {2^n - 1} (A000225).
a(n) = a(n-1) + 2*a(n-2) + 1.
a(n) = 2*2^n/3 - 1/2 - (-1)^n/6.
a(n) = n + 2*Sum_{k = 1..n-2} a(k).
G.f.: x/((1+x)*(1-x)*(1-2*x)) = x/(1-2*x-x^2+2*x^3). - Paul Barry, Feb 11 2003
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3). - Paul Barry, Feb 11 2003
a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2*k-1). - Paul Barry, Nov 11 2003
a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2*k); a(n+1) = Sum_{k = 0..n} Sum_{j = 0..k} (-1)^(j+k)*2^j. - Paul Barry, Nov 12 2003
(-1)^(n+1)*a(n) = Sum_{i = 0..n} Sum_{k = 1..i} k!*k* Stirling2(i, k)*(-1)^(k-1) = (1/3)*(-2)^(n+1)-(1/6)(3*(-1)^(n+1)-1). - Mario Catalani (mario.catalani(AT)unito.it), Dec 22 2003
a(n+1) = (n!/3)*Sum_{i - (-1)^i + j = n, i = 0..n, j = 0..n} 1/(i - (-1)^i)!/j!. - Benoit Cloitre, May 24 2004
a(n) = Sum_{k = 0..n} 2^(n-k-1)*(1-(-1)^k), row sums of A130125. - Paul Barry, Jul 28 2004
a(n) = Sum_{k = 0..n} binomial(k, n-k+1)2^(n-k); a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k, k+1)2^k. - Paul Barry, Oct 07 2004
a(n) = floor(2^(n+1)/3) = ceiling(2^(n+1)/3) - 1 = A005578(n+1) - 1. - Paul Barry, Oct 08 2005
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n." (A059570) with "1-n" (A024000), treating the result as if offset was 0. - Graeme McRae, Jul 12 2006
Starting (1, 2, 5, 10, 21, 42, ...), these are the row sums of triangle A135228. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), a(n-1)]. - Gary W. Adamson, Dec 25 2007
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = round((2^(n+2)-3)/6) = floor((2^(n+1)-1)/3) = round((2^(n+1)-2)/3); a(n) = a(n-2) + 2^(n-1), n > 1. - Mircea Merca, Dec 27 2010
a(n) = binary XOR of 2^k-1 for k=0..n. - Paul D. Hanna, Nov 05 2011
E.g.f.: 2/3*exp(2*x) - 1/2*exp(x) - 1/6*exp(-x) = 2/3*U(0); U(k) = 1 - 3/(4*(2^k) - 4*(2^k)/(1+3*(-1)^k - 24*x*(2^k)/(8*x*(2^k)*(-1)^k - (k+1)/U(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
a(n) = 2*(2^n - 1)/3, for even n; a(n) = (2^(n+1) - 1)/3 = (1/3)*(2^((n+1)/2) - 1)*(2^((n+1)/2) + 1), for odd n. - Hieronymus Fischer, Nov 22 2012
a(n) + a(n+1) = 2^(n+1) - 1. - Arie Bos, Apr 03 2013
G.f.: Q(0)/(3*(1-x)), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
a(n) = (2^(n+1) - 2 + (n mod 2))/3. - Paul Toms, Mar 18 2015
a(0) = 0, a(n) = 2*(a(n-1)) + (n mod 2). - Paul Toms, Mar 18 2015
Binary: a(n) = (a(n-1) shift left 1) + (a(n-1)) NOR (...11110). - Paul Toms, Mar 18 2015
a(n) = -(2^n)*a(-n) for even n; a(n) = -(2^(n+1))*a(-n) + 1 for odd n.
0 = +a(n)*(+2 +4*a(n) -4*a(n+1)) + a(n+1)*(-1 +a(n+1)) for all n in Z. (End)
a(4*k+d) = 2^(d+1)*a(4*k-1) + a(d), a(n+4) = a(n) + 2^n*10, a(0..3) = [0,1,2,5]. So the last digit is always 0,1,2,5 repeated. - Yuchun Ji, May 22 2023
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EXAMPLE
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a(4)=10 since 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111 are the 10 binary strings switching 0000 to 1111.
a(3) = 1 because "lrc" is the only way to tie a tie with 3 half turns, namely, pass the business end of the tie behind the standing part to the left, bring across the front to the right, then behind to the center. The final motion of tucking the loose end down the front behind the "lr" piece is not considered a "step".
a(4) = 2 because "lrlc" is the only way to tie a tie with 4 half turns. Note that since the number of moves is even, the first step is to go to the left in front of the tie, not behind it. This knot is the standard "four in hand", the most commonly known men's tie knot. By contrast, the second most well known tie knot, the Windsor, is represented by "lcrlcrlc".
a(n) = (2^0 - 1) XOR (2^1 - 1) XOR (2^2 - 1) XOR (2^3 - 1) XOR ... XOR (2^n - 1). - Paul D. Hanna, Nov 05 2011
G.f. = x + 2*x^2 + 5*x^3 + 10*x^4 + 21*x^5 + 42*x^6 + 85*x^7 + 170*x^8 + ...
a(9) = 341 = 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 4^4 + 4^3 + 4^2 + 4^1 + 4^0 = A002450(5). a(10) = 682 = 2*a(9) = 2*A002450(5). - Gregory L. Simay, Sep 27 2017
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MAPLE
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A000975 := proc(n) option remember; if n <= 1 then n else if n mod 2 = 0 then 2*A000975(n-1) else 2*A000975(n-1)+1 fi; fi; end;
f:=n-> if n mod 2 = 0 then (2^n-1)/3 else (2^n-2)/3; fi; [seq(f(n), n=0..40)]; # N. J. A. Sloane, Mar 21 2017
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MATHEMATICA
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Array[Ceiling[2(2^# - 1)/3] &, 41, 0]
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == a[n - 1] + 2a[n - 2] + 1}, a, {n, 40}] (* or *)
LinearRecurrence[{2, 1, -2}, {0, 1, 2}, 40] (* Harvey P. Dale, Aug 10 2013 *)
f[n_] := Block[{exp = n - 2}, Sum[2^i, {i, exp, 0, -2}]]; Array[f, 33] (* Robert G. Wilson v, Oct 30 2015 *)
f[s_List] := Block[{a = s[[-1]]}, Append[s, If[OddQ@ Length@ s, 2a + 1, 2a]]]; Nest[f, {0}, 32] (* Robert G. Wilson v, Jul 20 2017 *)
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PROG
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(PARI) {a(n) = if( n<0, 0, 2 * 2^n \ 3)}; /* Michael Somos, Sep 04 2006 */
(PARI) a(n)=if(n<=0, 0, bitxor(a(n-1), 2^n-1)) \\ Paul D. Hanna, Nov 05 2011
(PARI) concat(0, Vec(x/(1-2*x-x^2+2*x^3) + O(x^100))) \\ Altug Alkan, Oct 30 2015
(PARI) {a(n) = (4*2^n - 3 - (-1)^n) / 6}; /* Michael Somos, Jul 23 2017 */
(Haskell)
a000975 n = a000975_list !! n
a000975_list = 0 : 1 : map (+ 1)
(zipWith (+) (tail a000975_list) (map (* 2) a000975_list))
(Magma) [(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // Vincenzo Librandi, Mar 18 2015
(GAP) List([0..35], n->(2^(n+1)-2+(n mod 2))/3); # Muniru A Asiru, Nov 01 2018
(Python)
def a(n): return (2**(n+1) - 2 + (n%2))//3
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CROSSREFS
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Cf. A000295, A005578, A015441, A043291, A053404, A059260, A077854, A119440, A127824, A130125, A135228, A155051, A179970, A264784.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A005811
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Number of runs in binary expansion of n (n>0); number of 1's in Gray code for n.
(Formerly M0110)
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+10
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0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 5, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 2, 3, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 5, 4, 3, 4, 5, 6, 5, 6, 7, 6, 5, 4, 5, 6, 5, 4, 5
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,3
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COMMENTS
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Starting with a(1) = 0 mirror all initial 2^k segments and increase by one.
a(n) gives the net rotation (measured in right angles) after taking n steps along a dragon curve. - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
This sequence generates A082410: (0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) and A014577; identical to the latter except starting 1, 1, 0, ...; by writing a "1" if a(n+1) > a(n); if not, write "0". E.g., A014577(2) = 0, since a(3) < a(2), or 1 < 2. - Gary W. Adamson, Sep 20 2003
Starting with 1 = partial sums of A034947: (1, 1, -1, 1, 1, -1, -1, 1, 1, 1, ...). - Gary W. Adamson, Jul 23 2008
The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.
Can be used as a binomial transform operator: Let a(n) = the n-th term in any S(n); then extract 2^k strings, adding the terms. This results in the binomial transform of S(n). Say S(n) = 1, 3, 5, ...; then we obtain the strings: (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), ...; = the binomial transform of (1, 3, 5, ...) = (1, 4, 12, 32, 80, ...). Example: the 8-bit string has a sum of 32 with a distribution of (1, 3, 3, 1) or one 1, three 3's, three 5's, and one 7; as expected. - Gary W. Adamson, Jun 21 2012
Considers all positive odd numbers as nodes of a graph. Two nodes are connected if and only if the sum of the two corresponding odd numbers is a power of 2. Then a(n) is the distance between 2n + 1 and 1. - Jianing Song, Apr 20 2019
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted with addendum in Donald E. Knuth, Selected Papers on Fun and Games, 2010, pages 571-614. Equation 3.2 g(n) = a(n-1).
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FORMULA
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a(2^k + i) = a(2^k - i + 1) + 1 for k >= 0 and 0 < i <= 2^k. - Reinhard Zumkeller, Aug 14 2001
a(2n+1) = 2a(n) - a(2n) + 1, a(4n) = a(2n), a(4n+2) = 1 + a(2n+1).
a(j+1) = a(j) + (-1)^A014707(j). - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^2^k/(1+x^2^(k+1)). - Ralf Stephan, May 02 2003
Delete the 0, make subsets of 2^n terms; and reverse the terms in each subset to generate A088696. - Gary W. Adamson, Oct 19 2003
a(0) = 0, a(2n) = a(n) + [n odd], a(2n+1) = a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(0) = 0 then a(n) = a(floor(n/2)) + (a(floor(n/2)) + n) mod 2. - Benoit Cloitre, Jan 20 2014
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EXAMPLE
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Considered as a triangle with 2^k terms per row, the first few rows are:
1
2, 1
2, 3, 2, 1
2, 3, 4, 3, 2, 3, 2, 1
...
The n-th row becomes right half of next row; left half is mirrored terms of n-th row increased by one. - Gary W. Adamson, Jun 20 2012
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MAPLE
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local i, b, ans;
if n = 0 then
return 0 ;
end if;
ans := 1;
b := convert(n, base, 2);
for i from nops(b)-1 to 1 by -1 do
if b[ i+1 ]<>b[ i ] then
ans := ans+1
fi
od;
return ans ;
end proc:
# second Maple program:
a:= n-> add(i, i=Bits[Split](Bits[Xor](n, iquo(n, 2)))):
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MATHEMATICA
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Table[ Length[ Length/@Split[ IntegerDigits[ n, 2 ] ] ], {n, 1, 255} ]
a[n_] := DigitCount[BitXor[n, Floor[n/2]]]; Array[a, 100, 0] (* Amiram Eldar, Jul 11 2024 *)
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PROG
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(PARI) a(n)=sum(k=1, n, (-1)^((k/2^valuation(k, 2)-1)/2))
(PARI) a(n)=if(n<1, 0, a(n\2)+(a(n\2)+n)%2) \\ Benoit Cloitre, Jan 20 2014
(Haskell)
import Data.List (group)
a005811 0 = 0
a005811 n = length $ group $ a030308_row n
a005811_list = 0 : f [1] where
f (x:xs) = x : f (xs ++ [x + x `mod` 2, x + 1 - x `mod` 2])
(Python)
def a(n): return bin(n^(n>>1))[2:].count("1") # Indranil Ghosh, Apr 29 2017
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CROSSREFS
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Cf. A056539, A014707, A014577, A082410, A030308, A090079, A044813, A165413, A226227, A226228, A226229.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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1, 2, 4, 2, 4, 8, 6, 2, 4, 12, 16, 8, 6, 12, 6, 2, 4, 12, 36, 12, 16, 32, 24, 8, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 16, 48, 64, 32, 24, 72, 24, 8, 6, 30, 60, 30, 24, 48, 60, 12, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 36, 180, 144, 72, 60, 180, 60, 12, 16, 48, 144, 48, 64, 128, 96, 32, 24, 120, 216, 72, 24, 72
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OFFSET
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0,2
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LINKS
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FORMULA
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MATHEMATICA
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f[n_, i_, x_] := Which[n == 0, x, EvenQ@ n, f[n/2, i + 1, x], True, f[(n - 1)/2, i, x Prime@ i]]; g[n_] := If[n == 1, 1, Times @@ MapIndexed[ Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]]];
Table[g@ f[BitXor[n, Floor[n/2]], 1, 1], {n, 0, 93}] (* Michael De Vlieger, May 09 2017 *)
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PROG
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(Python)
from sympy import prime, factorint
import math
def A(n): return n - 2**int(math.floor(math.log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a005940(n): return b(n - 1)
def P(n):
f = factorint(n)
return sorted([f[i] for i in f])
def a046523(n):
x=1
while True:
if P(n) == P(x): return x
else: x+=1
def a003188(n): return n^int(n/2)
def a243353(n): return a005940(1 + a003188(n))
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CROSSREFS
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Sequences that (seem to) partition N into same or coarser equivalence classes are at least these: A005811, A136004, A033264, A037800, A069010, A087116, A090079 and many others like A105500, A106826, A166242, A246960, A277561, A037834, A225081 although these have not been fully checked yet.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A090077
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In binary expansion of n: reduce contiguous blocks of 1's to 1.
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+10
5
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0, 1, 2, 1, 4, 5, 2, 1, 8, 9, 10, 5, 4, 5, 2, 1, 16, 17, 18, 9, 20, 21, 10, 5, 8, 9, 10, 5, 4, 5, 2, 1, 32, 33, 34, 17, 36, 37, 18, 9, 40, 41, 42, 21, 20, 21, 10, 5, 16, 17, 18, 9, 20, 21, 10, 5, 8, 9, 10, 5, 4, 5, 2, 1, 64, 65, 66, 33, 68, 69, 34, 17, 72, 73, 74, 37, 36, 37, 18, 9
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OFFSET
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0,3
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LINKS
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FORMULA
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EXAMPLE
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100 -> '1100100' -> [11]00[1]00 -> [1]00[1]00 -> '100100' -> 36=a(100).
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MATHEMATICA
|
Array[FromDigits[Flatten[Split@ IntegerDigits[#, 2] /. w_List /; First[w] == 1 -> {1}], 2] &, 80, 0] (* Michael De Vlieger, Jul 28 2022 *)
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PROG
|
(Python)
def a(n):
b = bin(n)[2:]
while "11" in b: b = b.replace("11", "1")
return int(b, 2)
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CROSSREFS
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KEYWORD
|
nonn,base
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AUTHOR
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STATUS
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approved
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A090078
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In binary expansion of n, reduce contiguous blocks of 0's to 0.
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+10
3
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0, 1, 2, 3, 2, 5, 6, 7, 2, 5, 10, 11, 6, 13, 14, 15, 2, 5, 10, 11, 10, 21, 22, 23, 6, 13, 26, 27, 14, 29, 30, 31, 2, 5, 10, 11, 10, 21, 22, 23, 10, 21, 42, 43, 22, 45, 46, 47, 6, 13, 26, 27, 26, 53, 54, 55, 14, 29, 58, 59, 30, 61, 62, 63, 2, 5, 10, 11, 10, 21, 22, 23, 10, 21
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OFFSET
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0,3
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LINKS
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FORMULA
|
a(a(n)) = a(n).
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EXAMPLE
|
100 -> '1100100' -> 11[00]1[00] -> 11[0]1[0] -> '11010' -> 26=a(100).
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PROG
|
(PARI) a(n)=my(v=binary(n), t); for(i=1, #v, if(v[i], t+=t+1, t%2, t+=t)); t \\ Charles R Greathouse IV, Aug 17 2016
(Python)
def a(n):
b = bin(n)[2:]
while "00" in b: b = b.replace("00", "0")
return int(b, 2)
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CROSSREFS
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KEYWORD
|
nonn,base,easy
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AUTHOR
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STATUS
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approved
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A337864
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a(n) is the number formed by removing from n each digit if it is a duplicate of the previous digit, from left to right.
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+10
3
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 2, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 3, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 4, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 5, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 6, 67, 68, 69, 70, 71, 72, 73, 74
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OFFSET
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0,3
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COMMENTS
|
Similar to A137564 from which first differs at a(101) = 101 here, there a(101) = 10.
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LINKS
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EXAMPLE
|
a(100) = 10. Note that the second zero from the index n = 100 has been removed.
a(101) = 101.
a(1211323171) = 121323171. Note that the third "1" from the index n has been removed).
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PROG
|
(Perl) sub a {my($n)=@_; $n =~ s/(.)\1+/$1/g; $n} # Kevin Ryde, Oct 04 2020
(Python)
from itertools import groupby
def a(n): return int("".join(k for k, g in groupby(str(n))))
(PARI) a(n) = if(n < 10, return(n)); if(n%10 == (n\10)%10, return(a(n\10)), return(a(n\10)*10+n%10)) \\ David A. Corneth, Jul 23 2022
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CROSSREFS
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KEYWORD
|
base,nonn
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AUTHOR
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STATUS
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approved
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A356014
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Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).
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+10
3
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1, 2, 3, 4, 3, 2, 3, 8, 9, 10, 3, 12, 3, 10, 3, 16, 3, 18, 3, 20, 21, 10, 3, 24, 9, 10, 27, 20, 3, 2, 3, 32, 21, 10, 3, 4, 3, 10, 21, 40, 3, 10, 3, 20, 45, 10, 3, 48, 9, 50, 21, 20, 3, 54, 21, 40, 21, 10, 3, 12, 3, 10, 63, 64, 21, 10, 3, 20, 21, 10, 3, 72, 3
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OFFSET
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1,2
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COMMENTS
|
We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.
This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.
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LINKS
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FORMULA
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a(a(n)) = a(n).
a(n^k) = a(n)^k for any k >= 0.
a(n) = 3 iff n belongs to A294674 \ {1}.
a(n) = 4 iff n belongs to A061742 \ {1}.
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EXAMPLE
|
For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- compressing consecutive values, we obtain: 1 0 2 0,
- so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.
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PROG
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(PARI) a(n) = { my (v=1, e=-1, k=0); forprime (p=2, oo, if (n==1, return (v), if (e!=e=valuation(n, p), v*=prime(k++)^e); n/=p^e)) }
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CROSSREFS
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KEYWORD
|
nonn
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AUTHOR
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STATUS
|
approved
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A090080
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|
In binary expansion of n-th prime: reduce contiguous blocks of 0's to 0 and contiguous blocks of 1's to 1.
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|
+10
1
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2, 1, 5, 1, 5, 5, 5, 5, 5, 5, 1, 21, 21, 21, 5, 21, 5, 5, 5, 5, 21, 5, 21, 21, 5, 21, 5, 21, 21, 5, 1, 5, 21, 21, 85, 21, 21, 21, 21, 85, 21, 85, 5, 5, 21, 5, 21, 5, 5, 21, 21, 5, 5, 5, 5, 5, 21, 5, 85, 21, 21, 85, 21, 21, 21, 21, 85, 85, 85, 85, 21, 21, 21, 85, 21, 5, 21, 21, 21
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OFFSET
|
1,1
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COMMENTS
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LINKS
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EXAMPLE
|
A000040(25)=97 -> '1100001' -> [11][0000][1] -> [1][0][1] ->
'101' -> 5=a(25).
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CROSSREFS
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KEYWORD
|
nonn,base
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AUTHOR
|
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STATUS
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approved
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A339674
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Irregular triangle T(n, k), n, k >= 0, read by rows; for any number m with runs in binary expansion (r_1, ..., r_j), let R(m) = {r_1 + ... + r_j, r_2 + ... + r_j, ..., r_j}; row n corresponds to the numbers k such that R(k) is included in R(n), in ascending order.
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|
+10
1
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0, 0, 1, 0, 1, 2, 3, 0, 3, 0, 3, 4, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 6, 7, 0, 7, 0, 7, 8, 15, 0, 1, 6, 7, 8, 9, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 3, 4, 7, 8, 11, 12, 15, 0, 3, 12, 15, 0, 1, 2, 3, 12, 13, 14, 15, 0, 1, 14, 15, 0
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OFFSET
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0,6
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COMMENTS
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For any m > 0, R(m) contains the partial sums of the m-th row of A227736; by convention, R(0) = {}.
The underlying idea is to take some or all of the rightmost runs of a number, and possibly merge some of them.
For any n >= 0, the n-th row:
- has first term 0 and last term A003817(n),
- corresponds to the distinct terms in n-th row of table A341840.
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LINKS
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FORMULA
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T(n, 0) = 0.
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EXAMPLE
|
The triangle starts:
0;
0, 1;
0, 1, 2, 3;
0, 3;
0, 3, 4, 7;
0, 1, 2, 3, 4, 5, 6, 7;
0, 1, 6, 7;
0, 7;
0, 7, 8, 15;
0, 1, 6, 7, 8, 9, 14, 15;
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15;
0, 3, 4, 7, 8, 11, 12, 15;
0, 3, 12, 15;
0, 1, 2, 3, 12, 13, 14, 15;
0, 1, 14, 15;
0, 15;
...
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PROG
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(PARI) See Links section.
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CROSSREFS
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|
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KEYWORD
|
nonn,base,tabf
|
|
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AUTHOR
|
|
|
|
STATUS
|
approved
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