Displaying 1-10 of 81 results found.
p-INVERT of the positive integers, where p(S) = 1 - S^2.
+10
65
0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0. Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027: p(S) t(1,2,3,4,5,...)
FORMULA
G.f.: x/(1 - 4 x + 5 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).
MATHEMATICA
z = 60; s = x/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)
p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.
+10
57
1, 3, 9, 26, 74, 210, 596, 1692, 4804, 13640, 38728, 109960, 312208, 886448, 2516880, 7146144, 20289952, 57608992, 163568448, 464417728, 1318615104, 3743926400, 10630080640, 30181847168, 85694918912, 243312448256, 690833811712, 1961475291648, 5569190816256
COMMENTS
Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). In the following guide to p-INVERT sequences using s = (1,1,1,1,1,...) = A000012, in some cases t(1,1,1,1,1,...) is a shifted version of the cited sequence: p(S) t(1,1,1,1,1,...)
(1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S) A2910031 + S - S^2 A000045 (Fibonacci numbers starting with -1) 1 - S - S^2 - S^3 - S^4 - S^5 A2910071 - S - S^2 - S^3 - S^4 + S^5 A2910201 - S - S^2 - S^3 + S^4 + S^5 A291021
FORMULA
G.f.: (-1 + x - x^2)/(-1 + 4 x - 4 x^2 + 2 x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) for n >= 4.
MATHEMATICA
z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291000 *)
Expansion of 1/(x + sqrt(1-4x)).
+10
18
1, 1, 3, 9, 29, 97, 333, 1165, 4135, 14845, 53791, 196417, 721887, 2667941, 9907851, 36950465, 138320021, 519515209, 1957091277, 7392602917, 27992976565, 106236268337, 404005515873, 1539293204549, 5875059106769, 22459721336977
COMMENTS
Number of irreducible ordered pairs of compositions of n. A pair of compositions of n into the same number of (positive) parts, say n = a1 + ... + ak and n = b1 + ... + bk, is irreducible if for all j < k, a1 + ... + aj is not equal to b1 + ... + bj. E.g., a(3)=3 because the irreducible pairs are (1+2,2+1), (2+1,1+2), (3,3). - Herbert S. Wilf, May 22 2004
Hankel transform is 2^n. - Paul Barry, Nov 22 2007 (1 + 3x + 9x^2 + 29x^3 + ...) * 1/(1 + x + 3x^2 + 9x^3 + 29x^4 + ...) = (1 + 2x + 4x^2 + 10x^3 + 28x^4 + ...); where A068875 = (1, 2, 4, 10, 28, ...). - Gary W. Adamson, Nov 21 2011 Apparently the number of grand Motzkin paths of length n with two types of flat step (F,f) and avoiding F at level 0. - David Scambler, Jul 04 2013
LINKS
Ron M. Adin, Arkady Berenstein, Jacob Greenstein, Jian-Rong Li, Avichai Marmor, and Yuval Roichman, Transitive and Gallai colorings, arXiv:2309.11203 [math.CO], 2023. See p. 25.
FORMULA
G.f.: 1/(x + sqrt(1-4*x)).
D-finite with recurrence: n*a(n) + 2*(-4*n+3)*a(n-1) + 3*(5*n-8)*a(n-2) + 2*(2*n-3)*a(n-3) = 0.
a(n) = Sum_{k=0..n} binomial(2n-k,n+k)*(3k+1)/(n+k+1). - Emanuele Munarini, Apr 02 2011 A Catalan transform of the Fibonacci numbers F(n+1) under the mapping G(x) -> G(xc(x)), c(x) the g.f. of A000108. The inverse mapping is H(x) -> H(x(1-x)). a(n) = Sum{k=0..n} (k/(2n-k))binomial(2n-k, n-k)F(k+1). (End)
We have (per Wouter Meeussen): a(n) = (Sum_{k=1..n} k*Fibonacci(k+1)*(-1)^(n+k)*binomial(-n,n-k))/n = (Sum_{k=1..n} k*Fibonacci(k+1)*binomial(2*n-k-1,n-1))/n. If we introduce an alternating sign, defining b(n) = (Sum_{k=1..n} k*Fibonacci(k+1)*binomial(-n,n-k))/n = (Sum_{k=1..n} k*Fibonacci(k+1)*(-1)^(n+k)*binomial(2*n-k-1,n-1))/n, then b(n) = 1 for all n. (Not obvious--I proved it satisfies b(n+2) = ((17*n^2 + 37*n + 18)*b(n+1) - 4*(2*n+1)*(2*n+3)*b(n))/((n+2)*(n+3)).) (End)
G.f.: 1/(1-x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009 a(n) = the upper left term in M^n, M = an infinite square production matrix in which a column of (1,2,2,2,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros:
1, 1, 0, 0, 0, 0, ...
2, 1, 1, 0, 0, 0, ...
2, 1, 1, 1, 0, 0, ...
2, 1, 1, 1, 1, 0, ...
2, 1, 1, 1, 1, 1, ...
... (End)
MATHEMATICA
y[n_] := y[n] = (2*(4*n - 3)*y[n - 1] - (15*n - 24)*y[n - 2] - (4*n - 6)*y[n - 3])/n; y[0] = 1; y[1] = 1; y[2] = 3; (* corrected by Wouter Meeussen, Apr 30 2011 *) CoefficientList[Series[1/(x+Sqrt[1-4x] ), {x, 0, 30}], x] (* Harvey P. Dale, May 05 2021 *)
PROG
(Maxima) makelist(sum(binomial(2*n-k, n+k)*(3*k+1)/(n+k+1), k, 0, n), n, 0, 12); /* Emanuele Munarini, Apr 02 2011 */ (PARI) x='x+O('x^66); Vec(1/(x+sqrt(1-4*x))) \\ Joerg Arndt, Jul 06 2013
EXTENSIONS
Wouter credited with first sums in Gosper's FORMULA Comment, which were mistyped by NJAS (caught by Julian Ziegler Hunts), May 14 2011
Row sums of convolution triangle A030523.
+10
13
1, 4, 15, 55, 200, 725, 2625, 9500, 34375, 124375, 450000, 1628125, 5890625, 21312500, 77109375, 278984375, 1009375000, 3651953125, 13212890625, 47804687500, 172958984375, 625771484375, 2264062500000, 8191455078125
COMMENTS
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 5.
The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x*(1-x)/(1 + (t-1)*x*(1-x)). See A091867 for more info on this family. Here t = -4 (mod signs in the results). Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t). O.g.f.: G(x) = x*(1-x)/(1 - 5x*(1-x)) = P(Cinv(x),-5).
Inverse O.g.f.: Ginv(x) = (1 - sqrt(1 - 4*x/(1+5x)))/2 = C(P(x,5)) (signed A026378). Cf. A030528. (End)
FORMULA
G.f.: x*(1-x)/(1-5*x+5*x^2) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523). Binomial transform of Fibonacci(2n+2).
a(n) = (sqrt(5)/2 + 5/2)^n*(3*sqrt(5)/10 + 1/2) - (5/2 - sqrt(5)/2)^n*(3*sqrt(5)/10 - 1/2). (End)
a(n) = (1/5)*Sum_{r=1..9} sin(3*r*Pi/10)*sin(r*Pi/2)*(2*cos(r*Pi/10))^(2n)).
a(n) = 5*a(n-1) - 5*a(n-2).
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(n, i)*binomial(k+i+1, 2k+1). - Paul Barry, Jun 22 2004 Limit_{k->infinity} a(n+k)/a(k) = ( A020876(n) + A093131(n)*sqrt(5))/2. (End)
Limit_{k->infinity} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.
a(n) = a(n-1) * 3 + A081567(n-2) (not proved). (End)
MATHEMATICA
CoefficientList[Series[(1 - x) / (1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
PROG
(PARI) Vec(x*(1-x)/(1-5*x+5*x^2) + O(x^40)) \\ Altug Alkan, Nov 20 2015
p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^8.
+10
10
0, 0, 0, 0, 0, 0, 0, 1, 8, 36, 120, 330, 792, 1716, 3432, 6436, 11456, 19584, 32640, 54264, 93024, 170544, 341088, 735472, 1653632, 3749760, 8386560, 18289440, 38724480, 79594560, 159189120, 311058496, 597137408, 1133991936, 2147450880, 4089171840
COMMENTS
Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). See A291000 for a guide to related sequences.
FORMULA
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) for n >= 9.
G.f.: x^7 / ((1 - 2*x)*(1 - 2*x + 2*x^2)*(1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4)). - Colin Barker, Aug 22 2017
MATHEMATICA
z = 60; s = x/(1 - x); p = 1 - s^8;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290995 *)
PROG
(PARI) concat(vector(7), Vec(x^7 / ((1 - 2*x)*(1 - 2*x + 2*x^2)*(1 - 4*x + 6*x^2 - 4*x^3 + 2*x^4)) + O(x^50))) \\ Colin Barker, Aug 22 2017 (Magma) R<x>:=PowerSeriesRing(Integers(), 60); [0, 0, 0, 0, 0, 0, 0] cat Coefficients(R!( x^7/((1-x)^8 - x^8) )); // G. C. Greubel, Apr 11 2023 (SageMath)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( x^7/((1-x)^8 - x^8) ).list()
p-INVERT of the nonnegative integers ( A000027), where p(S) = 1 - S - S^2.
+10
6
0, 1, 2, 5, 12, 28, 64, 145, 328, 743, 1686, 3830, 8704, 19781, 44950, 102133, 232048, 527208, 1197808, 2721421, 6183108, 14048151, 31917714, 72517738, 164761792, 374342057, 850512458, 1932380869, 4390407092, 9975090996, 22663602720, 51492150953
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). See A290890 for a guide to related sequences.
FORMULA
a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3) + a(n-4).
G.f.: x*(1 - 2*x + 2*x^2) / (1 - 4*x + 5*x^2 - 2*x^3 - x^4). - Colin Barker, Aug 24 2017
MATHEMATICA
z = 60; s = x^2/(1-x)^2; p = 1 -s -s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290990 *) LinearRecurrence[{4, -5, 2, 1}, {0, 1, 2, 5}, 50] (* G. C. Greubel, Apr 12 2023 *)
PROG
(PARI) concat(0, Vec(x*(1-2*x+2*x^2)/(1-4*x+5*x^2-2*x^3-x^4) + O(x^50))) \\ Colin Barker, Aug 24 2017 (Magma) I:=[0, 1, 2, 5]; [n le 4 select I[n] else 4*Self(n-1) -5*Self(n-2) +2*Self(n-3) +Self(n-4): n in [1..50]]; // G. C. Greubel, Apr 12 2023 (SageMath)
@CachedFunction
if (n<4): return (0, 1, 2, 5)[n]
else: return 4*a(n-1) -5*a(n-2) +2*a(n-3) +a(n-4)
p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^4.
+10
6
0, 0, 1, 4, 10, 21, 43, 92, 205, 462, 1035, 2301, 5099, 11303, 25088, 55728, 123800, 274969, 610628, 1355970, 3011157, 6686979, 14850196, 32978725, 73237462, 162641499, 361184653, 802098203, 1781254927, 3955712256, 8784625824, 19508406192, 43323176177
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). See A291000 for a guide to related sequences. For n >= 1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 3 blocks (i.e., subintervals) from each interval. For example, for n=9, a(8)=205 since the number of ways to split [9] into intervals and then select 3 blocks from each interval is C(9,3) + C(6,3)*C(3,3) + C(5,3)*C(4,3) + C(4,3)*C(5,3) + C(3,3)*C(6,3) + C(3,3)*C(3,3)*C(3,3) for a total of 205 ways. - Enrique Navarrete, Dec 23 2023 a(n-1) is also the number of compositions of n using parts of size at least 3 where there are binomial(i,3) types of i, n>=1, i>=3 (see example). - Enrique Navarrete, Dec 25 2023
FORMULA
a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - a(n-4) for n >= 5.
G.f.: x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4). - Colin Barker, Aug 22 2017
EXAMPLE
Since there are binomial(3,3) = 1 type of 3, binomial(4,3) = 4 types of 4, binomial(5,3) = 10 types of 5, binomial(6,3) = 20 types of 6, and binomial(9,3) = 84 types of 9, we can write 9 in the following ways:
9 in 84 ways;
6+3 in 20 ways;
5+4 in 40 ways;
4+5 in 40 ways;
3+6 in 20 ways;
3+3+3 in 1 way, for a total of 205 ways. (End)
MATHEMATICA
z = 60; s = x/(1 - x); p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* this sequence *)
LinearRecurrence[{4, -6, 5, -1}, {0, 0, 1, 4}, 41] (* G. C. Greubel, Apr 25 2023 *)
PROG
(PARI) concat(vector(2), Vec(x^2 / (1 - 4*x + 6*x^2 - 5*x^3 + x^4) + O(x^50))) \\ Colin Barker, Aug 22 2017 (Magma) I:=[0, 0, 1, 4]; [n le 4 select I[n] else 4*Self(n-1) -6*Self(n-2) +5*Self(n-3) -Self(n-4): n in [1..41]]; // G. C. Greubel, Apr 25 2023 (SageMath)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( x^2/(1-4*x+6*x^2-5*x^3+x^4) ).list()
p-INVERT of the Fibonacci numbers ( A000045, including 0), where p(S) = 1 - S - S^2.
+10
5
0, 1, 1, 4, 7, 18, 37, 85, 183, 407, 888, 1956, 4284, 9409, 20630, 45270, 99289, 217819, 477776, 1048053, 2298912, 5042783, 11061455, 24263687, 53223023, 116746272, 256086074, 561731936, 1232174181, 2702807740, 5928681960, 13004724921, 28526216361
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). See A290890 for a guide to related sequences.
FORMULA
G.f.: (x - x^2)/(1 - 2 x - 2 x^2 + 3 x^3 + x^4).
a(n) = 2*a(n-1) + 2*a(n-2) - 3*a(n-3) - a(n-4).
MATHEMATICA
z = 60; s = x^2/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289975 *)
p-INVERT of (0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by one zero, where p(S) = 1 - S - S^2.
+10
5
0, 0, 1, 2, 3, 6, 13, 26, 50, 96, 184, 351, 669, 1278, 2447, 4692, 9004, 17285, 33182, 63687, 122208, 234461, 449774, 862776, 1655010, 3174766, 6090231, 11683285, 22413104, 42997349, 82486280, 158241688, 303570021, 582365698, 1117202719, 2143225358
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453). See A290890 for a guide to related sequences.
FORMULA
a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 3*a(n-4) + a(n-5) + a(n-6).
G.f.: x^2*(1 - 2*x + x^2 + x^3) / (1 - 4*x + 6*x^2 - 5*x^3 + 3*x^4 - x^5 - x^6). - Colin Barker, Aug 24 2017
MATHEMATICA
z = 60; s = x^3/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0, 0, 1, 2, 3, 4, 5, ... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290991 *)
PROG
(PARI) concat(vector(2), Vec(x^2*(1 - 2*x + x^2 + x^3) / (1 - 4*x + 6*x^2 - 5*x^3 + 3*x^4 - x^5 - x^6) + O(x^40))) \\ Colin Barker, Aug 24 2017
p-INVERT of the even positive integers ( A005843), where p(S) = 1 - S - S^2.
+10
4
2, 12, 62, 312, 1570, 7908, 39838, 200688, 1010978, 5092860, 25655582, 129241512, 651061762, 3279762132, 16521995710, 83230530528, 419278719938, 2112141348588, 10640036959358, 53599815453720, 270012240337762, 1360202629711812, 6852101192007262
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453). See A289780 for a guide to related sequences.
FORMULA
G.f.: (2 (1 + x^2))/(1 - 6 x + 6 x^2 - 6 x^3 + x^4).
a(n) = 6*a(n-1) - 6*a(n-2) + 6*a(n-3) - a(n-4).
MATHEMATICA
z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *) u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)
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