Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
Search: a218722 -id:a218722
     Sort: relevance | references | number | modified | created      Format: long | short | data
Partial sums of powers of 16.
+10
52
1, 17, 273, 4369, 69905, 1118481, 17895697, 286331153, 4581298449, 73300775185, 1172812402961, 18764998447377, 300239975158033, 4803839602528529, 76861433640456465, 1229782938247303441, 19676527011956855057, 314824432191309680913, 5037190915060954894609
OFFSET
0,2
COMMENTS
16 = 2^4 is the growth measure for the Jacobsthal spiral (compare with phi^4 for the Fibonacci spiral). - Paul Barry, Mar 07 2008
Second quadrisection of A115451. - Paul Curtz, May 21 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=16, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, Feb 21 2010
Partial sums are in A014899. Also, the sequence is related to A014931 by A014931(n+1) = (n+1)*a(n) - Sum_{i=0..n-1} a(i) for n>0. - Bruno Berselli, Nov 07 2012
a(n) is the total number of holes in a certain box fractal (start with 16 boxes, 1 hole) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 28 2015
Except for 1 and 17, all terms are Brazilian repunits numbers in base 16, and so belong to A125134. All terms >= 273 are composite because a(n) = ((4^(n+1) + 1) * (4^(n+1) - 1))/15. - Bernard Schott, Jun 06 2017
The sequence in binary is 1, 10001, 100010001, 1000100010001, 10001000100010001, ... cf. Plouffe link, A330135. - Frank Ellermann, Mar 05 2020
LINKS
FORMULA
a(n) = if n=0 then 1 else a(n-1) + A001025(n).
for n > 0: A131851(a(n)) = n and abs(A131851(m)) < n for m < a(n).
a(n) = A098704(n+2)/2.
a(n) = (16^(n+1) - 1)/15. - Bernard Schott, Jun 06 2017
a(n) = (A001025(n+1) - 1)/15.
a(n) = 16*a(n-1) + 1. - Paul Curtz, May 20 2008
G.f.: 1 / ( (16*x-1)*(x-1) ). - R. J. Mathar, Feb 06 2011
E.g.f.: exp(x)*(16*exp(15*x) - 1)/15. - Stefano Spezia, Mar 06 2020
EXAMPLE
a(3) = 1 + 16 + 256 + 4096 = 4369 = in binary: 1000100010001.
a(4) = (16^5 - 1)/15 = (4^5 + 1) * (4^5 - 1)/15 = 1025 * 1023/15 = 205 * 341 = 69905 = 11111_16. - Bernard Schott, Jun 06 2017
MAPLE
A131865:=n->(16^(n+1)-1)/15: seq(A131865(n), n=0..30); # Wesley Ivan Hurt, Apr 29 2017
MATHEMATICA
Table[(2^(4 n) - 1)/15, {n, 16}] (* Robert G. Wilson v, Aug 22 2007 *)
Accumulate[16^Range[0, 20]] (* or *) LinearRecurrence[{17, -16}, {1, 17}, 20] (* Harvey P. Dale, Jul 19 2019 *)
PROG
(Sage) [gaussian_binomial(n, 1, 16) for n in range(1, 18)] # Zerinvary Lajos, May 28 2009
(Magma) [(16^(n+1)-1)/15: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011
(Maxima)
a[0]:0$
a[n]:=16*a[n-1]+1$
A131865(n):=a[n]$
makelist(A131865(n), n, 1, 30); /* Martin Ettl, Nov 05 2012 */
(PARI) A131865(n)=16^n\15 \\ M. F. Hasler, Nov 05 2012
(Python)
def A131865(n): return (1<<(n+1<<2))//15 # Chai Wah Wu, Nov 10 2022
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Jul 22 2007
STATUS
approved
Partial sums of powers of 17 (A001026).
+10
41
1, 18, 307, 5220, 88741, 1508598, 25646167, 435984840, 7411742281, 125999618778, 2141993519227, 36413889826860, 619036127056621, 10523614159962558, 178901440719363487, 3041324492229179280, 51702516367896047761
OFFSET
1,2
COMMENTS
17^a(n) is largest power of 17 dividing (17^n)!.
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=17, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
FORMULA
a(n) = Sum_{k=0..n-1} 17^k = (17^n - 1)/16.
G.f.: x/((1 - 17*x)*(1 - x))= (1/(1 - 17*x) - 1/(1 - x))/16.
a(n) = 17*a(n-1)+1 (with a(1)=1). - Vincenzo Librandi, Nov 16 2010
E.g.f.: exp(9*x)*sinh(8*x)/8. - Stefano Spezia, Mar 11 2023
MAPLE
ListTools:-PartialSums([seq(17^k, k=0..30)]); # Robert Israel, Feb 18 2018
MATHEMATICA
Table[17^n, {n, 0, 16}] // Accumulate (* Jean-François Alcover, Jul 05 2013 *)
PROG
(Sage) [gaussian_binomial(n, 1, 17) for n in range(1, 18)] # Zerinvary Lajos, May 28 2009
(Maxima) makelist(sum(17^k, k, 0, n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Magma) [&+[17^i: i in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Feb 19 2018
CROSSREFS
Cf. similar sequences of the form (k^n-1)/(k-1) with k prime: A000225 (k=2), A003462 (k=3), A003463 (k=5), A023000 (k=7), A016123 (k=11), A091030 (k=13), this sequence (k=17), A218722 (k=19), A218726 (k=23), A218732 (k=29), A218734 (k=31), A218740 (k=37), A218744 (k=41), A218746 (k=43), A218750 (k=47).
Cf. A001026.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 23 2004
STATUS
approved
a(n) = (20^n - 1)/19.
+10
39
0, 1, 21, 421, 8421, 168421, 3368421, 67368421, 1347368421, 26947368421, 538947368421, 10778947368421, 215578947368421, 4311578947368421, 86231578947368421, 1724631578947368421, 34492631578947368421, 689852631578947368421, 13797052631578947368421, 275941052631578947368421
OFFSET
0,3
COMMENTS
Partial sums of powers of 20 (A009964), q-integers for q=20: diagonal k=1 in triangle A022184.
Partial sums are in A014904. Also, the sequence is related to A014937 by A014937(n) = n*a(n)-Sum_{i=0..n-1} a(i), for n>0. - Bruno Berselli, Nov 06 2012
For n >= 1, a(n) is the total number of holes in a certain box fractal (start with 20 boxes, 1 hole) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 28 2015
FORMULA
a(n) = 20*a(n-1) + 1, with a(0)=0. - Vincenzo Librandi, Aug 07 2010
a(0)=0, a(1)=1, a(n) = 21*a(n-1) - 20*a(n-2). - Harvey P. Dale, Oct 04 2012
a(n) = floor(20^n/19). - M. F. Hasler, Nov 04 2012
G.f.: x/((1 - x)*(1 - 20*x)). - Bruno Berselli, Nov 06 2012
E.g.f.: exp(x)*(exp(19*x) - 1)/19. - Stefano Spezia, Mar 23 2023
EXAMPLE
From N. J. A. Sloane, Nov 04 2014: Can also be obtained by writing powers of 2 in a staggered array and adding them (cf. A249604). For example, a(9) is:
..........1
.........2
........4
.......8
.....16
....32
...64
.128
256
-----------
26947368421
MAPLE
a:=n->sum(20^(n-j), j=0..n): seq(a(n), n=0..15); # Zerinvary Lajos, Feb 11 2007
MATHEMATICA
(20^Range[20]-1)/19 (* or *) NestList[20#+1&, 1, 20] (* Harvey P. Dale, Oct 04 2012 *)
PROG
(Sage) [gaussian_binomial(n, 1, 20) for n in range(1, 17)] # Zerinvary Lajos, May 29 2009
(PARI) for (n=0, 100, write("b064108.txt", n, " ", (20^n - 1)/19)) \\ Harry J. Smith, Sep 07 2009
(PARI) A064108(n)=20^n\19 \\ M. F. Hasler, Nov 04 2012
(Maxima) A064108(n):=(20^n-1)/19$ makelist(A064108(n), n, 1, 30); /* Martin Ettl, Nov 05 2012 */
KEYWORD
nonn,easy
AUTHOR
Jason Earls, Sep 17 2001
EXTENSIONS
Edited and extended to offset 0 by M. F. Hasler, Nov 04 2012
STATUS
approved
a(n) = (21^n - 1)/20.
+10
37
0, 1, 22, 463, 9724, 204205, 4288306, 90054427, 1891142968, 39714002329, 833994048910, 17513875027111, 367791375569332, 7723618886955973, 162195996626075434, 3406115929147584115, 71528434512099266416, 1502097124754084594737, 31544039619835776489478
OFFSET
0,3
COMMENTS
Partial sums of powers of 21 (A009965); q-integers for q=21: diagonal k=1 in triangle A022185.
Partial sums are in A014905. Also, the sequence is related to A014938 by A014938(n) = n*a(n) - Sum_{i=0..n-1} a(i) for n > 0. - Bruno Berselli, Nov 06 2012
For n >= 1, 4*a(n) is the total number of holes in a certain box fractal (start with 21 boxes, 4 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 27 2015
FORMULA
a(n) = floor(21^n/20).
G.f.: x/((1-x)*(1-21*x)). - Bruno Berselli, Nov 06 2012
a(n) = 22*a(n-1) - 21*a(n-2). - Vincenzo Librandi, Nov 07 2012
a(n) = 21*a(n-1) + 1. - Kival Ngaokrajang, Jan 27 2015
a(n) = a(n-1) + 21^(n-1), n >= 1, a(0) = 0. - Wolfdieter Lang, Feb 02 2015
E.g.f.: exp(11*x)*sinh(10*x)/10. - Elmo R. Oliveira, Aug 29 2024
MATHEMATICA
LinearRecurrence[{22, -21}, {0, 1}, 40] (* Vincenzo Librandi, Nov 07 2012 *)
PROG
(PARI) A218724(n)=21^n\20
(Maxima) A218724(n):=(21^n-1)/20$ makelist(A218724(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Magma) [n le 2 select n-1 else 22*Self(n-1) - 21*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Nov 04 2012
STATUS
approved
a(n) = (31^n - 1)/30.
+10
37
0, 1, 32, 993, 30784, 954305, 29583456, 917087137, 28429701248, 881320738689, 27320942899360, 846949229880161, 26255426126284992, 813918209914834753, 25231464507359877344, 782175399728156197665, 24247437391572842127616, 751670559138758105956097
OFFSET
0,3
COMMENTS
Partial sums of powers of 31 (A009975).
LINKS
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 17.
FORMULA
From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 31*x)).
a(n) = 32*a(n-1) - 31*a(n-2) for n > 1.
a(n) = floor(31^n/30). (End)
E.g.f.: exp(16*x)*sinh(15*x)/15. - Stefano Spezia, Mar 11 2023
MATHEMATICA
LinearRecurrence[{32, -31}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)
PROG
(PARI) a(n)=31^n\30
(Magma) [n le 2 select n-1 else 32*Self(n-1)-31*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012
(Maxima) A218734(n):=(31^n-1)/30$
makelist(A218734(n), n, 0, 30); /* Martin Ettl, Nov 07 2012 */
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Nov 04 2012
STATUS
approved
a(n) = (2^(5*n) - 1)/31.
+10
35
0, 1, 33, 1057, 33825, 1082401, 34636833, 1108378657, 35468117025, 1134979744801, 36319351833633, 1162219258676257, 37191016277640225, 1190112520884487201, 38083600668303590433, 1218675221385714893857, 38997607084342876603425, 1247923426698972051309601
OFFSET
0,3
COMMENTS
Partial sums of powers of 32 (A009976), a.k.a. q-numbers for q=32. - M. F. Hasler, Nov 05 2012
REFERENCES
A. K. Devaraj, "Minimum Universal Exponent Generalisation of Fermat's Theorem", in ISSN #1550-3747, Proceedings of Hawaii Intl Conference on Statistics, Mathematics & Related Fields, 2004.
FORMULA
a(n) = (32^n - 1)/31 = floor(32^n/31) = Sum_{k=0..n} 32^k. - M. F. Hasler, Nov 05 2012
G.f.: x/((1 - x)*(1 - 32*x)). - Bruno Berselli, Nov 06 2012
E.g.f.: exp(x)*(exp(31*x) - 1)/31. - Stefano Spezia, Mar 23 2023
MATHEMATICA
Table[(2^(5 n) - 1)/31, {n, 16}] (* Robert G. Wilson v *)
LinearRecurrence[{33, -32}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)
PROG
(Sage) [gaussian_binomial(5*n, 1, 2)/31 for n in range(1, 17)] # Zerinvary Lajos, May 28 2009
(Magma) [n le 2 select n-1 else 33*Self(n-1) - 32*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012
(PARI) A132469(n)=32^n\31 \\ M. F. Hasler, Nov 07 2012
(Maxima) A132469(n):=(32^n-1)/31$
makelist(A132469(n), n, 0, 30); /* Martin Ettl, Nov 07 2012 */
KEYWORD
nonn,easy
AUTHOR
A.K. Devaraj, Aug 22 2007
EXTENSIONS
Edited and extended by Robert G. Wilson v, Aug 22 2007
Edited and extended to offset 0 by M. F. Hasler, Nov 05 2012
STATUS
approved
a(n) = (18^n-1)/17.
+10
35
0, 1, 19, 343, 6175, 111151, 2000719, 36012943, 648232975, 11668193551, 210027483919, 3780494710543, 68048904789775, 1224880286215951, 22047845151887119, 396861212733968143, 7143501829211426575, 128583032925805678351
OFFSET
0,3
COMMENTS
Partial sums of powers of 18 (A001027), q-integers for q=18: diagonal k=1 in triangle A022182.
Partial sums are in A014901. Also, the sequence is related to A014935 by A014935(n) = n*a(n) - Sum_{i=0..n-1} a(i), for n>0. - Bruno Berselli, Nov 06 2012
From Bernard Schott, May 06 2017: (Start)
Except for 0, 1 and 19, all terms are Brazilian repunits numbers in base 18, and so belong to A125134. From n = 3 to n = 8286, all terms are composite. See link "Generalized repunit primes".
As explained in the extensions of A128164, a(25667) = (18^25667 - 1)/17 would be (is) the smallest prime in base 18. (End)
FORMULA
a(n) = floor(18^n/17).
G.f.: x/((1-x)*(1-18*x)). - Bruno Berselli, Nov 06 2012
a(n) = 19*a(n-1) - 18*a(n-2). - Vincenzo Librandi, Nov 07 2012
E.g.f.: exp(x)*(exp(17*x) - 1)/17. - Stefano Spezia, Mar 11 2023
EXAMPLE
a(3) = (18^3 - 1)/17 = 343 = 7 * 49; a(6) = (18^6 - 1)/17 = 2000719 = 931 * 2149. - Bernard Schott, May 01 2017
MATHEMATICA
LinearRecurrence[{19, -18}, {0, 1}, 40] (* Vincenzo Librandi, Nov 07 2012 *)
Join[{0}, Accumulate[18^Range[0, 20]]] (* Harvey P. Dale, Nov 08 2012 *)
PROG
(PARI) A218721(n)=18^n\17
(Maxima) A218721(n):=(18^n-1)/17$ makelist(A218721(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Magma) [n le 2 select n-1 else 19*Self(n-1)-18*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Nov 04 2012
STATUS
approved
a(n) = (49^n - 1)/48.
+10
35
0, 1, 50, 2451, 120100, 5884901, 288360150, 14129647351, 692352720200, 33925283289801, 1662338881200250, 81454605178812251, 3991275653761800300, 195572507034328214701, 9583052844682082520350, 469569589389422043497151, 23008909880081680131360400
OFFSET
0,3
COMMENTS
Partial sums of powers of 49 (A087752).
FORMULA
G.f.: x/((1-x)*(1-49*x)). - Vincenzo Librandi, Nov 08 2012
a(n) = 50*a(n-1) - 49*a(n-2) with a(0)=0, a(1)=1. - Vincenzo Librandi, Nov 08 2012
a(n) = 49*a(n-1) + 1 with a(0)=0. - Vincenzo Librandi, Nov 08 2012
a(n) = floor(49^n/48). - Vincenzo Librandi, Nov 08 2012
E.g.f.: exp(25*x)*sinh(24*x)/24. - Elmo R. Oliveira, Aug 27 2024
MATHEMATICA
LinearRecurrence[{50, -49}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
Join[{0}, Accumulate[49^Range[0, 20]]] (* Harvey P. Dale, Apr 14 2023 *)
PROG
(PARI) A218753(n)=49^n\48
(Maxima) A218753(n):=floor(49^n/48)$ makelist(A218753(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Magma) [n le 2 select n-1 else 50*Self(n-1) - 49*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 08 2012
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Nov 04 2012
STATUS
approved
a(n) = (33^n - 1)/32.
+10
34
0, 1, 34, 1123, 37060, 1222981, 40358374, 1331826343, 43950269320, 1450358887561, 47861843289514, 1579440828553963, 52121547342280780, 1720011062295265741, 56760365055743769454, 1873092046839544391983, 61812037545704964935440, 2039797239008263842869521
OFFSET
0,3
COMMENTS
Partial sums of powers of 33 (A009977).
FORMULA
From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 33*x)).
a(n) = 34*a(n-1) - 33*a(n-2).
a(n) = floor(33^n/32). (End)
E.g.f.: exp(x)*(exp(32*x) - 1)/32. - Stefano Spezia, Mar 24 2023
MATHEMATICA
LinearRecurrence[{34, -33}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)
PROG
(PARI) A218736(n)=33^n>>5
(Magma) [n le 2 select n-1 else 34*Self(n-1)-33*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012
(Maxima) A218736(n):=(33^n-1)/32$
makelist(A218736(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Nov 04 2012
STATUS
approved
Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.
+10
15
1, 3, 4, 7, 13, 21, 15, 40, 85, 156, 31, 121, 341, 781, 1555, 63, 364, 1365, 3906, 9331, 19608, 127, 1093, 5461, 19531, 55987, 137257, 299593, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111
OFFSET
1,2
LINKS
Eric Weisstein's World of Mathematics, Repunit
FORMULA
T(n, k) = Sum_{i=0..n-1} (k+1)^i.
T(n+1, k) = (k+1)*T(n, k) + 1.
Sum_{k=1..n} T(n, k) = A125120(n).
T(2*n-1, n) = A125119(n).
T(n, 1) = A000225(n).
T(n, 2) = A003462(n) for n>1.
T(n, 3) = A002450(n) for n>2.
T(n, 4) = A003463(n) for n>3.
T(n, 5) = A003464(n) for n>4.
T(n, 9) = A002275(n) for n>8.
T(n, n) = A060072(n+1).
T(n, n-1) = A023037(n) for n>1.
T(n, n-2) = A031973(n) for n>2.
T(n, k) = A055129(n, k+1) = A104878(n+k, k+1), 1<=k<=n. - Mathew Englander, Dec 19 2020
EXAMPLE
First 4 rows:
1: [1]_2
2: [11]_2 ........ [11]_3
3: [111]_2 ....... [111]_3 ....... [111]_4
4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5
_
1: 1
2: 2+1 ........... 3+1
3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1
4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1 ((5+1)*5+1)*5+1.
MATHEMATICA
Table[((k+1)^n -1)/k, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Aug 15 2022 *)
PROG
(Magma) [((k+1)^n -1)/k : k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 15 2022
(SageMath)
def A125118(n, k): return ((k+1)^n -1)/k
flatten([[A125118(n, k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Aug 15 2022
CROSSREFS
This triangle shares some features with triangle A104878.
This triangle is a portion of rectangle A055129.
Each term of A110737 comes from the corresponding row of this triangle.
Diagonals (adjusting offset as necessary): A060072, A023037, A031973, A173468.
Cf. A023037, A031973, A125119, A125120 (row sums).
KEYWORD
nonn,tabl,base
AUTHOR
Reinhard Zumkeller, Nov 21 2006
STATUS
approved

Search completed in 0.021 seconds