A000217 -id:A000217 - cp's OEIS frontend

A126883 a(n) = (2^0)(2^1)(2^2)*(2^3)...(2^n)-1 = 2^T(n) - 1 where T(n) = A000217(n) is the n-th triangular number.

0, 1, 7, 63, 1023, 32767, 2097151, 268435455, 68719476735, 35184372088831, 36028797018963967, 73786976294838206463, 302231454903657293676543, 2475880078570760549798248447, 40564819207303340847894502572031, 1329227995784915872903807060280344575

Offset: 0

Marco Matosic, Dec 29 2006

nonn

For n > 2, a(n) and a(n-1) share at least one prime factor.

Shows how many patterns can be created with 1-color thread while sewing on a button with buttonholes located on the vertices of a convex n-gon. - Ivan N. Ianakiev, Feb 09 2012

Masha Gessen, Perfect Rigor, A Genius and the Mathematical Breakthrough of the Century, Houghton Mifflin Harcourt, 2009, page 38.

Muniru A Asiru, Table of n, a(n) for n = 0..80

Cf. A000217, A006125.

List([-1..15],n->2^(Binomial(2+n,n))-1); # Muniru A Asiru, Feb 21 2019

seq(2^(binomial(n+1, 2))-1, n=0..12); # Zerinvary Lajos, Jun 12 2007

FoldList[Times,2^Range[0,20]]-1 (* Harvey P. Dale, Sep 09 2015 *)
2^Accumulate[Range[0,20]]-1 (* Harvey P. Dale, Jun 03 2019 *)

a(n) = A006125(n+1) - 1. - Zerinvary Lajos, Jun 12 2007

Corrected and extended by Harvey P. Dale, Sep 09 2015

A153007 Triangular number A000217(n) minus toothpick number A153006(n).

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 1, 0, 3, 8, 11, 13, 15, 13, 5, 0, 7, 20, 31, 41, 51, 57, 57, 59, 69, 79, 82, 81, 74, 51, 17, 0, 15, 44, 71, 97, 123, 145, 161, 179, 205, 231, 250, 265, 274, 267, 249, 247, 273, 307, 334, 357, 374, 375, 364, 363, 376, 380, 364, 332, 270, 163, 49, 0, 31, 92, 151

Offset: 0

Omar E. Pol, Dec 19 2008, May 27 2009

nonn

David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
OEIS (Plot 2), Triangular numbers (A000217) and toothpick numbers (A153006) vs n [From _Omar E. Pol_, Apr 29 2009]

Cf. A000217, A000225, A000396, A000668, A006516, A139250, A139251, A152980, A153006.

a(n) = A000217(n)-A153006(n).

More terms from R. J. Mathar, Jul 13 2009

A185096 Let T(n) = n(n+1)/2 be the n-th triangular number (A000217); a(n) = T(8T(n)).

Original entry on oeis.org

0, 36, 300, 1176, 3240, 7260, 14196, 25200, 41616, 64980, 97020, 139656, 195000, 265356, 353220, 461280, 592416, 749700, 936396, 1155960, 1412040, 1708476, 2049300, 2438736, 2881200, 3381300, 3943836, 4573800, 5276376, 6056940, 6921060, 7874496, 8923200, 10073316, 11331180, 12703320, 14196456, 15817500, 17573556, 19471920, 21520080

Offset: 0

N. J. A. Sloane, Feb 18 2011

Claudi Alsina and Roger B. Nelsen, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 4.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A180324, A185097, A322677, A339483.

Table[2*n*(n + 1)*(2*n + 1)^2, {n, 0, 50}] (* G. C. Greubel, Jun 22 2017 *)

for(n=0,50, print1(2*n*(n+1)*(2*n+1)^2, ", ")) \\ G. C. Greubel, Jun 22 2017

From G. C. Greubel, Jun 22 2017: (Start)
a(n) = 2*n*(n + 1)*(2*n + 1)^2.
G.f.: 12*x*(3 + 10*x + 3*x^2)/(1 - x)^5.
E.g.f.: 2*x*(18 + 57*x + 32*x^2 + 4*x^3)*exp(x). (End)
From Elmo R. Oliveira, Sep 08 2025: (Start)
a(n) = 12*A180324(n) = 4*A339483(n) = A322677(n)/8.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)

A196421 a(n) = prime(n)*T(n), where T = A000217.

Original entry on oeis.org

2, 9, 30, 70, 165, 273, 476, 684, 1035, 1595, 2046, 2886, 3731, 4515, 5640, 7208, 9027, 10431, 12730, 14910, 16863, 19987, 22908, 26700, 31525, 35451, 38934, 43442, 47415, 52545, 62992, 69168, 76857, 82705, 93870, 100566, 110371, 120783, 130260, 141860

Offset: 1

Harvey P. Dale, Oct 15 2011

nonn

This sequence is mentioned in A077320. - Omar E. Pol, Mar 12 2012

			The 4th prime is 7, the 4th triangular number is 10, therefore a(4) = 7*10 = 70.

Harvey P. Dale, Table of n, a(n) for n = 1..10000

Cf. A000040, A000217.
Row sums of triangle A077320. - Omar E. Pol, Mar 12 2012
Subsequence of A085783. - Michel Marcus, May 15 2018

With[{nn=60},Prime[Range[nn]]Accumulate[Range[nn]]]

a(n)=prime(n)*binomial(n+1,2) \\ Charles R Greathouse IV, Nov 22 2011

from sympy import prime
def a(n): return prime(n) * n*(n+1)//2
print([a(n) for n in range(1, 41)]) # Michael S. Branicky, Sep 01 2022

a(n) ~ 0.5 n^3 log n. - Charles R Greathouse IV, Nov 22 2011

a(n) = A000040(n)*A000217(n). - Omar E. Pol, Mar 12 2012

A207339 Triangular numbers T from A000217 such that (4*T+1)/5 is prime.

Original entry on oeis.org

6, 21, 36, 66, 91, 136, 171, 351, 496, 561, 741, 946, 1176, 1326, 1596, 2016, 2346, 2701, 2926, 3321, 3486, 4851, 6216, 6441, 7626, 8646, 8911, 9591, 10011, 10296, 11026, 11476, 13041, 15051, 16471, 16836, 17391, 18336, 19701, 21736

Offset: 1

Wolfdieter Lang, Feb 27 2012

nonn

The corresponding primes are gven in A207337, where also equivalent formulations are found.

The indices of these triangular numbers are given by (A002733(n)-1)/2.

			a(3) = 36 = T((17-1)/2) = T(8)=A000217(8). (4*36+1)/5 = 29 = A207337(3).

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A207337, A129307, A027862, A002731.

Select[Accumulate[Range[300]],PrimeQ[(4#+1)/5]&] (* Harvey P. Dale, Sep 18 2019 *)

a(n) = T(K(n)):= A000217(K(n)) with K(n)=(m(n)-1)/2, and m(n) given in A002733(n).

A230405 a(n) = A000217(A230404(n+1)); the first differences of A219650.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 10, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3

Offset: 0

Antti Karttunen, Oct 31 2013

nonn

Construction: Count the trailing zeros in the factorial base representation (A007623) of 2n+2 (2, 4, 6, 8, ...) and then take the corresponding triangular number from A000217.

Antti Karttunen, Table of n, a(n) for n = 0..10079

First differences of A219650. Can be used to compute A219650 and A230412.
Cf. A000217, A007623, A230404.
Cf. also A230413.

a(n) = A000217(A230404(n+1)).
a(n) = A219650(n+1) - A219650(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=0..m} a(k) = 2. - Amiram Eldar, Jan 05 2024

A253880 Triangular numbers (A000217) that are also centered heptagonal numbers (A069099).

Original entry on oeis.org

1, 253, 64261, 16322041, 4145734153, 1053000152821, 267457893082381, 67933251842771953, 17254778510170993681, 4382645808331589623021, 1113174780537713593253653, 282742011610770921096804841, 71815357774355276244995175961, 18240818132674629395307677889253

Offset: 1

Colin Barker, Jan 17 2015

			253 is in the sequence because it is the 22nd triangular number and the 9th centered heptagonal number.

Colin Barker, Table of n, a(n) for n = 1..416
Index entries for linear recurrences with constant coefficients, signature (254,-1).

Cf. A000217, A069099, A253878, A253879.

Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=8.

LinearRecurrence[{254,-1},{1,253},20] (* Harvey P. Dale, May 17 2017 *)

PARI
```
Vec(-x*(x-1)/(x^2-254*x+1) + O(x^100))
```

a(n) = 254*a(n-1) - a(n-2).
G.f.: -x*(x-1) / (x^2 - 254*x + 1).
a(n) = (1/8)*T(2*n-1, 8), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022

A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

Original entry on oeis.org

1, 2, 11, 28, 189, 494, 3383, 8856, 60697, 158906, 1089155, 2851444, 19544085, 51167078, 350704367, 918155952, 6293134513, 16475640050, 112925716859, 295643364940, 2026369768941, 5305104928862, 36361730124071, 95196245354568, 652484772464329

Offset: 1

Colin Barker, Feb 03 2015

Also positive integers y in the solutions to x^2 - 5*y^2 + x + 5*y - 2 = 0, the corresponding values of x being A254626.

Also indices of centered pentagonal numbers (A005891) that are also hexagonal numbers (A000384). - Colin Barker, Feb 11 2015

			2 is in the sequence because the 2nd centered pentagonal number is 6, which is also the 3rd triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Cf. A000217, A005891, A254626, A254628.

Cf. A000384, A254962.

[(2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4 : n in [1..30]]; // G. C. Greubel, Apr 19 2019

CoefficientList[Series[x (x^3 + 9 x^2 - x - 1)/((x - 1) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, 25}], x] (* Michael De Vlieger, Jun 06 2016 *)
LinearRecurrence[{1,18,-18,-1,1},{1,2,11,28,189},30] (* Harvey P. Dale, Apr 23 2017 *)

Vec(x*(x^3+9*x^2-x-1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^30))

{a(n) = (2 +(1+3*(-1)^n)*fibonacci(3*n) - 2*(-1)^n*fibonacci(3*n+1))/4}; \\ G. C. Greubel, Apr 19 2019

[(2 +(1+3*(-1)^n)*fibonacci(3*n) -2*(-1)^n*fibonacci(3*n+1))/4 for n in (1..30)] # G. C. Greubel, Apr 19 2019

a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+x-9*x^2-x^3)/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)).
a(n) = (10 - sqrt(5)*(2-sqrt(5))^n - 5*(-2+sqrt(5))^n - 2*sqrt(5)*(-2+sqrt(5))^n + sqrt(5)*(2+sqrt(5))^n + (-2-sqrt(5))^n*(-5+2*sqrt(5)))/20. - Colin Barker, Jun 06 2016
a(2*n+2) = A232970(2*n+1); a(2*n+1) = A110679(2*n). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
a(n) = (2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4. - G. C. Greubel, Apr 19 2019

A257712 Triangular numbers (A000217) that are the sum of eight consecutive triangular numbers.

Original entry on oeis.org

120, 276, 1176, 28920, 126756, 306936, 1345620, 33362196, 146264856, 354192420, 1552832856, 38499933816, 168789505620, 408737734296, 1791967758756, 44428890250020, 194782943209176, 471682991173716, 2067929240760120, 51270900848577816, 224779347673872036

Offset: 1

Colin Barker, May 05 2015

			120 is in the sequence because T(15) = 120 = 1+3+6+10+15+21+28+36 = T(1)+ ... +T(8).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1154,-1154,0,0,-1,1).

Cf. A000217, A001110, A129803, A131557, A257711, A257713, A259413, A259414, A259415.

LinearRecurrence[{1, 0, 0, 1154, -1154, 0, 0, -1, 1}, {120, 276, 1176, 28920, 126756, 306936, 1345620, 33362196, 146264856}, 30] (* Vincenzo Librandi, Jun 27 2015 *)
Select[Total/@Partition[Accumulate[Range[5*10^6]],8,1],OddQ[ Sqrt[ 1+8#]]&] (* The program generates the first 16 terms of the sequence *) (* Harvey P. Dale, Feb 27 2022 *)

Vec(-12*x*(3*x^8+7*x^6+13*x^5-3387*x^4+2312*x^3+75*x^2+13*x+10) / ((x-1)*(x^2-6*x+1)*(x^2+6*x+1)*(x^4+34*x^2+1)) + O(x^100))

G.f.: -12*x*(3*x^8+7*x^6+13*x^5-3387*x^4+2312*x^3+75*x^2+13*x+10) / ((x-1)*(x^2-6*x+1)*(x^2+6*x+1)*(x^4+34*x^2+1)).

A257713 Triangular numbers (A000217) that are the sum of ten consecutive triangular numbers.

Original entry on oeis.org

1485, 7260, 28920, 142845, 2112540, 10440165, 41673885, 205953660, 3046252485, 15054681960, 60093684540, 296985006165, 4392693942120, 21708840917445, 86655051404085, 428252172907560, 6334261618255845, 31304133548245020, 124956524030977320, 617539336347666645

Offset: 1

Colin Barker, May 05 2015

nonn

			1485 is in the sequence because T(54) = 1485 = 78+91+105+120+136+153+171+190+210+231 = T(12)+ ... +T(21).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1442,-1442,0,0,-1,1).

Cf. A000217, A001110, A129803, A131557, A257711, A257712, A259413, A259414, A259415.

 LinearRecurrence[{1, 0, 0, 1442, -1442, 0, 0, -1, 1}, {1485, 7260, 28920, 142845, 2112540, 10440165, 41673885, 205953660, 3046252485}, 30] (* Vincenzo Librandi, Jun 27 2015 *)

Vec(-15*x*(8*x^8-5*x^7+5*x^5-11445*x^4+7595*x^3+1444*x^2+385*x+99) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)*(x^4+38*x^2+1)) + O(x^100))

G.f.: -15*x*(8*x^8-5*x^7+5*x^5-11445*x^4+7595*x^3+1444*x^2+385*x+99) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)*(x^4+38*x^2+1)).

cp's OEIS Frontend

A126883 a(n) = (2^0)*(2^1)*(2^2)*(2^3)...(2^n)-1 = 2^T(n) - 1 where T(n) = A000217(n) is the n-th triangular number.

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A153007 Triangular number A000217(n) minus toothpick number A153006(n).

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A185096 Let T(n) = n(n+1)/2 be the n-th triangular number (A000217); a(n) = T(8T(n)).

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A196421 a(n) = prime(n)*T(n), where T = A000217.

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A207339 Triangular numbers T from A000217 such that (4*T+1)/5 is prime.

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A230405 a(n) = A000217(A230404(n+1)); the first differences of A219650.

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A253880 Triangular numbers (A000217) that are also centered heptagonal numbers (A069099).

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Mathematica

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A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

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A126883 a(n) = (2^0)(2^1)(2^2)*(2^3)...(2^n)-1 = 2^T(n) - 1 where T(n) = A000217(n) is the n-th triangular number.