A000217 -id:A000217 - cp's OEIS frontend

A274830 Numbers k such that 7*k+1 is a triangular number (A000217).

0, 2, 5, 11, 17, 27, 36, 50, 62, 80, 95, 117, 135, 161, 182, 212, 236, 270, 297, 335, 365, 407, 440, 486, 522, 572, 611, 665, 707, 765, 810, 872, 920, 986, 1037, 1107, 1161, 1235, 1292, 1370, 1430, 1512, 1575, 1661, 1727, 1817, 1886, 1980, 2052, 2150, 2225

Offset: 1

Colin Barker, Jul 08 2016

From Peter Bala, Nov 21 2024: (Start)
Numbers of the form n*(7*n + 3)/2 for n in Z. Cf. A057570.
The sequence terms occur as the exponents in the expansion of Product_{n >= 1} (1 - x^(7*n)) * (1 + x^(7*n-2)) * (1 + x^(7*n-5)) = 1 + x^2 + x^5 + x^11 + x^17 + x^27 + x^36 + .... Cf. A363800. (End)

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A057570, A363800.

Cf. similar sequences where k*n+1 is a triangular number: A000096 (k=1), A074377 (k=2), A045943 (k=3), A274681 (k=4), A085787 (k=5), A274757 (k=6).

Table[(14 (n - 1) n + (2 n - 1) (-1)^n + 1)/16, {n, 1, 60}] (* Bruno Berselli, Jul 08 2016 *)

select(n->ispolygonal(7*n+1, 3), vector(3000, n, n-1))

concat(0, Vec(x^2*(2+3*x+2*x^2)/((1-x)^3*(1+x)^2) + O(x^100)))

G.f.: x^2*(2 + 3*x + 2*x^2) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.
a(n) = (14*(n - 1)*n + (2*n - 1)*(-1)^n + 1)/16. Therefore:
a(n) = n*(7*n - 6)/8 for n even,
a(n) = (n - 1)*(7*n - 1)/8 for n odd.
E.g.f.: (x*(7*x -1)*cosh(x) + (7*x^2 + x + 1)*sinh(x))/8. - Stefano Spezia, Nov 26 2024

Edited by Bruno Berselli, Jul 08 2016

A276599 Values of n such that n^2 + 5 is a triangular number (A000217).

Original entry on oeis.org

1, 4, 10, 25, 59, 146, 344, 851, 2005, 4960, 11686, 28909, 68111, 168494, 396980, 982055, 2313769, 5723836, 13485634, 33360961, 78600035, 194441930, 458114576, 1133290619, 2670087421, 6605301784, 15562409950, 38498520085, 90704372279, 224385818726

Offset: 1

Colin Barker, Sep 07 2016

			4 is in the sequence because 4^2 + 5 = 21, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276600 (k=6), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.

I:=[1,4,10,25]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[(1+x)*(1+3*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
LinearRecurrence[{0,6,0,-1},{1,4,10,25},30] (* Harvey P. Dale, Feb 13 2017 *)

Vec(x*(1+x)*(1+3*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))

a(n)=([0,1;-1,6]^(n\2)*if(n%2,[1;10],[-1;4]))[1,1] \\ Charles R Greathouse IV, Sep 07 2016

def P(n): return lucas_number1(n, 2, -1)
[(1/4)*(P(n+2) + P(n+1) + (-1)^n*(3*P(n) - 7*P(n-1))) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = 6*a(n-2) - a(n-4) for n>4.
G.f.: x*(1+x)*(1+3*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).
a(n) = (1/4)*(P(n+2) + P(n+1) + (-1)^n*(3*P(n) - 7*P(n-1))), where P(n) = A000129(n). - G. C. Greubel, Sep 15 2021

A276601 Values of k such that k^2 + 9 is a triangular number (A000217).

Original entry on oeis.org

1, 6, 12, 37, 71, 216, 414, 1259, 2413, 7338, 14064, 42769, 81971, 249276, 477762, 1452887, 2784601, 8468046, 16229844, 49355389, 94594463, 287664288, 551336934, 1676630339, 3213427141, 9772117746, 18729225912, 56956076137, 109161928331, 331964339076

Offset: 1

Colin Barker, Sep 07 2016

			6 is in the sequence because 6^2+9 = 45, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276600 (k=6), A276602 (k=10), where k is the value added to n^2.

I:=[1,6,12,37]; [n le 2 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[(1+x)*(1+5*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
LinearRecurrence[{0,6,0,-1}, {1,6,12,37}, 31] (* G. C. Greubel, Sep 15 2021 *)

Vec(x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))

def P(n): return lucas_number1(n, 2, -1)
[(1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = 6*a(n-2) - a(n-4) for n>4.
G.f.: x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).
a(n) = (1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))), where P(n) = A000129(n). - G. C. Greubel, Sep 15 2021

A276602 Values of k such that k^2 + 10 is a triangular number (A000217).

Original entry on oeis.org

0, 9, 54, 315, 1836, 10701, 62370, 363519, 2118744, 12348945, 71974926, 419500611, 2445028740, 14250671829, 83059002234, 484103341575, 2821561047216, 16445262941721, 95850016603110, 558654836676939, 3256079003458524, 18977819184074205, 110610836100986706

Offset: 1

Colin Barker, Sep 07 2016

			9 is in the sequence because 9^2+10 = 91, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Soumeya M. Tebtoub, Hacène Belbachir and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276600 (k=6), A276601 (k=9), where k is the value added to n^2.

[n le 2 select 9*(n-1) else 6*Self(n-1) - Self(n-2): n in [1..31]]; // G. C. Greubel, Sep 15 2021

CoefficientList[Series[9*x/(1 - 6*x + x^2), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
(9/2)*Fibonacci[2*(Range[30] -1), 2] (* G. C. Greubel, Sep 15 2021 *)

concat(0, Vec(9*x^2/(1-6*x+x^2) + O(x^30)))

[(9/2)*lucas_number1(2*n-2, 2, -1) for n in (1..30)] # G. C. Greubel, Sep 15 2021

a(n) = (9/(4*sqrt(2))*( (3 - 2*sqrt(2))*(3 + 2*sqrt(2))^n - (3 + 2*sqrt(2))*(3 - 2*sqrt(2))^n) ).
a(n) = 9*A001109(n-1).
a(n) = 6*a(n-1) - a(n-2) for n>2.
G.f.: 9*x^2 / (1-6*x+x^2).
a(n) = (9/2)*A000129(2*n-2). - G. C. Greubel, Sep 15 2021

A341885 a(n) is the sum of A000217(p) over the prime factors p of n, counted with multiplicity.

Original entry on oeis.org

0, 3, 6, 6, 15, 9, 28, 9, 12, 18, 66, 12, 91, 31, 21, 12, 153, 15, 190, 21, 34, 69, 276, 15, 30, 94, 18, 34, 435, 24, 496, 15, 72, 156, 43, 18, 703, 193, 97, 24, 861, 37, 946, 72, 27, 279, 1128, 18, 56, 33, 159, 97, 1431, 21, 81, 37, 196, 438, 1770, 27, 1891, 499, 40, 18, 106, 75, 2278, 159, 282

Offset: 1

J. M. Bergot and Robert Israel, Feb 22 2021

By definition, this sequence is completely additive. - Peter Munn, Aug 14 2022

			18 = 2*3*3 so a(18) = 2*3/2 + 3*4/2 + 3*4/2 = 15.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000217, A340834.

For other completely additive sequences with primes p mapped to a function of p, see A001414.

f:= proc(n) local t; add(t[1]*(t[1]+1)/2*t[2], t = ifactors(n)[2]) end proc:
map(f, [$1..100]);

Prepend[Array[Total@ PolygonalNumber@ Flatten[ConstantArray[#1, #2] & @@@ FactorInteger[#]] &, 68, 2], 0] (* Michael De Vlieger, Feb 22 2021 *)

a(n) = my(f=factor(n), p); sum(k=1, #f~, p=f[k,1]; f[k,2]*p*(p+1)/2); \\ Michel Marcus, Aug 14 2022

from sympy import factorint
def A341885(n): return sum(k*m*(m+1)//2 for m,k in factorint(n).items()) # Chai Wah Wu, Feb 25 2021

A008345 a(n+1) = a(n)-b(n+1) if a(n) >= b(n+1) else a(n)+b(n+1), where {b(n)} are the triangular numbers A000217.

Original entry on oeis.org

0, 1, 4, 10, 0, 15, 36, 8, 44, 89, 34, 100, 22, 113, 8, 128, 264, 111, 282, 92, 302, 71, 324, 48, 348, 23, 374, 752, 346, 781, 316, 812, 284, 845, 250, 880, 214, 917, 176, 956, 136, 997, 94, 1040, 50, 1085, 4, 1132, 2308, 1083, 2358, 1032, 2410, 979, 2464, 924

Offset: 0

N. J. A. Sloane and J. H. Conway

a(0) and a(4) are both zero. Are there any other zero values? - N. J. A. Sloane, Sep 12 2019

Rémy Sigrist, Table of n, a(n) for n = 0..25000 (first 1001 terms from Franklin T. Adams-Watters)

Cf. A000217, A008344, A076042.

A008345 := proc(n) option remember; if n = 1 then n-1 elif A008345(n-1) >= n*(n+1)/2 then A008345(n-1)-n*(n+1)/2 else A008345(n-1)+n*(n+1)/2; fi; end;

nxt[{n_,a_}]:=Module[{tr=((n+1)(n+2))/2},{n+1,If[a>=tr,a-tr,a+tr]}]; Transpose[NestList[nxt,{0,0},50]][[2]] (* Harvey P. Dale, Jun 19 2013 *)

A051543 Quotients of consecutive values of lcm of first n triangular numbers (A000217).

Original entry on oeis.org

3, 2, 5, 1, 7, 2, 3, 1, 11, 1, 13, 1, 1, 2, 17, 1, 19, 1, 1, 1, 23, 1, 5, 1, 3, 1, 29, 1, 31, 2, 1, 1, 1, 1, 37, 1, 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 7, 1, 1, 1, 53, 1, 1, 1, 1, 1, 59, 1, 61, 1, 1, 2, 1, 1, 67, 1, 1, 1, 71, 1, 73, 1, 1, 1, 1, 1, 79, 1, 3, 1, 83, 1, 1, 1, 1, 1, 89, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Asher Auel

			a(5) = A025555(6)/A025555(5) = 210/30 = 7

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A025555.

Cf. A051542.

a051543 n = a051542_list !! (n-1)
a051543_list = zipWith div (tail a025555_list) a025555_list
-- Reinhard Zumkeller, Mar 12 2014

a(n) = A025555(n+1)/A025555(n)

Corrected and extended by James Sellers

A058395 Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 3, 1, 1, 0, 3, 2, 1, 6, 3, 4, 3, 1, 0, 6, 6, 6, 4, 1, 10, 6, 9, 10, 9, 5, 1, 0, 10, 12, 15, 16, 13, 6, 1, 15, 10, 16, 21, 25, 25, 18, 7, 1, 0, 15, 20, 28, 36, 41, 38, 24, 8, 1, 21, 15, 25, 36, 49, 61, 66, 56, 31, 9, 1, 0, 21, 30, 45, 64, 85, 102, 104, 80, 39, 10, 1, 28, 21, 36, 55, 81, 113, 146, 168, 160, 111, 48, 11, 1

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(2n, 0) = T(n, 3) with T(2n, 0) = T(n, m) for some other value of m would change the generating function to the coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^m. This would produce A058393, A058394, A057884 (and effectively A007318).

			The array T(n, k) starts:
[0] 1, 0,  3,   0,   6,   0,  10,    0,   15,    0, ...
[1] 1, 1,  3,   3,   6,   6,  10,   10,   15,   15, ...
[2] 1, 2,  4,   6,   9,  12,  16,   20,   25,   30, ...
[3] 1, 3,  6,  10,  15,  21,  28,   36,   45,   55, ...
[4] 1, 4,  9,  16,  25,  36,  49,   64,   81,  100, ...
[5] 1, 5, 13,  25,  41,  61,  85,  113,  145,  181, ...
[6] 1, 6, 18,  38,  66, 102, 146,  198,  258,  326, ...
[7] 1, 7, 24,  56, 104, 168, 248,  344,  456,  584, ...
[8] 1, 8, 31,  80, 160, 272, 416,  592,  800, 1040, ...
[9] 1, 9, 39, 111, 240, 432, 688, 1008, 1392, 1840, ...

Rows are A000217 with zeros, A008805, A002620, A000217, A000290, A001844, A005899.
Columns are A000012, A001477, A016028.
Diagonals include A058396, A049611, A001793, A001788, A055580, A055581, A055582.
The triangle A055252 also appears in half of the array.

gf := n -> (1 + x)^n / (1 - x^2)^3: ser := n -> series(gf(n), x, 20):
seq(lprint([n], seq(coeff(ser(n), x, k), k = 0..9)), n = 0..9); # Peter Luschny, Apr 12 2023

T[0, k_] := If[OddQ[k], 0, (k+2)(k+4)/8];
T[n_, k_] := T[n, k] = If[k == 0, 1, T[n-1, k-1] + T[n-1, k]];
Table[T[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Apr 13 2023 *)

T(n, k) = T(n-1, k-1) + T(n, k-1) with T(0, k) = 1, T(2*n, 0) = T(n, 3) and T(2*n + 1, 0) = 0. Coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^3.

A085741 a(n) = T(n)^n, where T() are the triangular numbers (A000217).

Original entry on oeis.org

1, 1, 9, 216, 10000, 759375, 85766121, 13492928512, 2821109907456, 756680642578125, 253295162119140625, 103510234140112521216, 50714860157241037295616, 29345269354638035222576971

Offset: 0

Jon Perry, Jul 21 2003

			a(3) = T(3)^3 = 6^3 = 216.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000217.
Essentially the same as A061718.
Cf. A000312, A062206, A212333.

[((n*(n+1))/2)^n: n in [0..20]]; // Vincenzo Librandi, Sep 14 2011

a:=n->mul(sum(j, j=1..n),k=1..n): seq(a(n), n=0..13); # Zerinvary Lajos, Jun 02 2007

With[{rnn=Range[20]},Join[{1},First[#]^Last[#]&/@Thread[ {Accumulate[ rnn],  rnn}]]] (* Harvey P. Dale, Dec 08 2013 *)

a(n) = (n*(n+1)/2)^n; \\ Michel Marcus, Feb 19 2019

a(n) = ((n*(n+1))/2)^n. - Vincenzo Librandi, Sep 14 2011

More terms from Ray Chandler, Nov 09 2003

A085744 a(n) = A000217(n^3) - n^3.

Original entry on oeis.org

0, 0, 28, 351, 2016, 7750, 23220, 58653, 130816, 265356, 499500, 885115, 1492128, 2412306, 3763396, 5693625, 8386560, 12066328, 17003196, 23519511, 31996000, 42878430, 56684628, 74011861, 95544576, 122062500, 154449100, 193700403, 240934176, 297399466, 364486500

Offset: 0

Jon Perry, Jul 21 2003

			a(3) = T(3^3) - 3^3 = T(27) - 27 = 378 - 27 = 351.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000217.

Cf. A000578, A071232.

[n^3*(n^3-1)/2: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

Table[(n^6-n^3)/2,{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Feb 28 2011 *)
(#(#-1))/2&/@(Range[0,30]^3) (* Harvey P. Dale, Dec 26 2021 *)

t(n)=n*(n+1)/2; for(n=0,30,print1(t(n^3)-n^3","))

a(n) = n^3*(n^3 - 1)/2. - Vincenzo Librandi, Sep 14 2011
From Chai Wah Wu, Aug 08 2022: (Start)
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n > 6.
G.f.: -x^2*(x^4 + 29*x^3 + 147*x^2 + 155*x + 28)/(x - 1)^7. (End)
a(n) = A071232(n) - A000578(n). - J.S. Seneschal, Jul 08 2025

cp's OEIS Frontend

A274830 Numbers k such that 7*k+1 is a triangular number (A000217).

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A276599 Values of n such that n^2 + 5 is a triangular number (A000217).

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A276601 Values of k such that k^2 + 9 is a triangular number (A000217).

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A276602 Values of k such that k^2 + 10 is a triangular number (A000217).

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A341885 a(n) is the sum of A000217(p) over the prime factors p of n, counted with multiplicity.

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A008345 a(n+1) = a(n)-b(n+1) if a(n) >= b(n+1) else a(n)+b(n+1), where {b(n)} are the triangular numbers A000217.

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A051543 Quotients of consecutive values of lcm of first n triangular numbers (A000217).

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