A077415 -id:A077415 - cp's OEIS frontend

A006331 a(n) = n(n+1)(2*n+1)/3.

0, 2, 10, 28, 60, 110, 182, 280, 408, 570, 770, 1012, 1300, 1638, 2030, 2480, 2992, 3570, 4218, 4940, 5740, 6622, 7590, 8648, 9800, 11050, 12402, 13860, 15428, 17110, 18910, 20832, 22880, 25058, 27370, 29820, 32412, 35150, 38038, 41080, 44280

Offset: 0

N. J. A. Sloane

Triangles in rhombic matchstick arrangement of side n.
Maximum accumulated number of electrons at energy level n. - Scott A. Brown, Feb 28 2000
Let M_n denote the n X n matrix M_n(i,j)=i^2+j^2; then the characteristic polynomial of M_n is x^n - a(n)x^(n-1) - .... - Michael Somos, Nov 14 2002
Convolution of odds (A005408) and evens (A005843). - Graeme McRae, Jun 06 2006
a(n) is the number of non-monotonic functions with domain {0,1,2} and codomain {0,1,...,n}. - Dennis P. Walsh, Apr 25 2011
For any odd number 2n+1, find Sum_{aJ. M. Bergot, Jul 16 2011
a(n) gives the number of (n+1) X (n+1) symmetric (0,1)-matrices containing three ones (see [Cameron]). - L. Edson Jeffery, Feb 18 2012
a(n) is the number of 4-tuples (w,x,y,z) with all terms in {0,...,n} and |w - x| < y. - Clark Kimberling, Jun 02 2012
Partial sums of A001105. - Omar E. Pol, Jan 12 2013
Total number of square diagonals (of any size) in an n X n square grid. - Wesley Ivan Hurt, Mar 24 2015
Number of diagonal attacks of two queens on (n+1) X (n+1) chessboard. - Antal Pinter, Sep 20 2015
a(n) is the minimum value obtainable by partitioning either the set {x in the natural numbers | 1 <= x <= 2n} or the set {x in the natural numbers | 0 <= x <= 2n+1} into pairs, taking the product of all such pairs, and taking the sum of all such products. - Thomas Anton, Oct 21 2020
a(n) is the irregularity of the n-th power of a path of length at least 3*n.  (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 16 2023
a(n) is the maximum possible total number of inversions in all rows and all columns of a Latin square of order n+1. - Ivaylo Kortezov, Jun 28 2025

			For n=2, a(2)=10 since there are 10 non-monotonic functions f from {0,1,2} to {0,1,2}, namely, functions f = <f(1),f(2),f(3)> given by <0,1,0>, <0,2,0>, <0,2,1>, <1,0,1>, <1,0,2>, <1,2,0>, <1,2,1>, <2,0,1>, <2,0,2>, and <2,1,2>. - _Dennis P. Walsh_, Apr 25 2011
Let n=4, 2*n+1 = 9. Since 9 = 1+8 = 3+6 = 5+4 = 7+2, a(4) = 1*8 + 3*6 + 5*4 + 7*2 = 60. - _Vladimir Shevelev_, May 11 2012

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., Vol. 2 (1931), pp. 355-359.
J. L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., Vol. 2 (1931), pp. 355-359. [Annotated scanned copy]
Rowan Beckworth, Basic atomic information.
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
P. Cameron, T. Prellberg and D. Stark, Asymptotics for incidence matrix classes, Electron. J. Combin. 13 (2006), #R85, p. 11.
Jose Manuel Garcia Calcines, Luis Javier Hernandez Paricio, and Maria Teresa Rivas Rodriguez, Semi-simplicial combinatorics of cyclinders and subdivisions, arXiv:2307.13749 [math.CO], 2023. See p. 25.
N. S. S. Gu, H. Prodinger and S. Wagner, Bijections for a class of labeled plane trees, Eur. J. Combinat., Vol. 31 (2010), pp. 720-732, doi|10.1016/j.ejc.2009.10.007, Theorem 2 at n=3.
JBMO 2025, 29th Junior Balkan Mathematical Olympiad, Problem 4, author: Boris Mihov
Germain Kreweras, Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, Vol. 6 (1965), circa p. 82.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Dennis Walsh, Notes on finite monotonic and non-monotonic functions.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

A row of A132339.
Cf. A002378, A048147, A098077, A163102.
Cf. A005900, A084980, A077415, A112524, A188475, A100178, A035597, A096000.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

a006331 n = sum $ zipWith (*) [2*n-1, 2*n-3 .. 1] [2, 4 ..]
-- Reinhard Zumkeller, Feb 11 2012

[n*(n+1)*(2*n+1)/3: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011

A006331 := proc(n)
    n*(n+1)*(2*n+1)/3 ;
end proc:
seq(A006331(n),n=0..80) ; # R. J. Mathar, Sep 27 2013

Table[n(n+1)(2n+1)/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,2,10,28},50] (* Harvey P. Dale, Apr 12 2013 *)

PARI
```
a(n)=if(n<0,0,n*(n+1)*(2*n+1)/3)
```

G.f.: 2*x*(1 + x)/(1 - x)^4. - Simon Plouffe (in his 1992 dissertation)
a(n) = 2*binomial(n+1,3) + 2*binomial(n+2,3).
a(n) = 2*A000330(n) = A002492(n)/2.
a(n) = Sum_{i=0..n} T(i,n-i), array T as in A048147. - N. J. A. Sloane, Dec 11 1999
From the formula for the sum of squares of positive integers 1^2 + 2^2 + 3^2 + ... + n^2 = n*(n+1)(2*n+1)/6, if we multiply both sides by 2 we get Sum_{k=0..n} 2*k^2 = n*(n+1)*(2*n+1)/3, which is an alternative formula for this sequence. - Mike Warburton, Sep 08 2007
10*a(n) = A016755(n) - A001845(n); since A016755 are odd cubes and A001845 centered octahedral numbers, 10*a(n) are the "odd cubes without their octahedral contents." - Damien Pras, Mar 19 2011
a(n) = sum(a*b), where the summing is over all unordered partitions 2*n+1=a+b. - Vladimir Shevelev, May 11 2012
a(n) = binomial(2*n+2, 3)/2. - Ronan Flatley, Dec 13 2012
a(n) = A000292(n) + A002411(n). - Omar E. Pol, Jan 11 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3, with a(0)=0, a(1)=2, a(2)=10, a(3)=28. - Harvey P. Dale, Apr 12 2013
a(n) = A208532(n+1,2). - Philippe Deléham, Dec 05 2013
Sum_{n>0} 1/a(n) = 9 - 12*log(2). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A000292(n-1) + (n+1)*A000217(n). - J. M. Bergot, Sep 02 2015
a(n) = 2*(A000332(n+3) - A000332(n+1)). - Antal Pinter, Sep 20 2015
From Bruno Berselli, May 17 2018: (Start)
a(n) = n*A002378(n) - Sum_{k=0..n-1} A002378(k) for n>0, a(0)=0. Also:
A163102(n) = n*a(n) - Sum_{k=0..n-1} a(k) for n>0, A163102(0)=0. (End)
a(n) = A005900(n) - A000290(n) = A096000(n) - A000578(n+1) = A000578(n+1) - A084980(n+1) = A000578(n+1) - A077415(n)-1 = A112524(n) + 1 = A188475(n) - 1 = A061317(n) - A100178(n) = A035597(n+1) - A006331(n+1). - Bruce J. Nicholson, Jun 24 2018
E.g.f.: (1/3)*exp(x)*x*(6 + 9*x + 2*x^2). - Stefano Spezia, Jan 05 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi - 9. - Amiram Eldar, Jan 04 2022

A084990 a(n) = n(n^2+3n-1)/3.

Original entry on oeis.org

0, 1, 6, 17, 36, 65, 106, 161, 232, 321, 430, 561, 716, 897, 1106, 1345, 1616, 1921, 2262, 2641, 3060, 3521, 4026, 4577, 5176, 5825, 6526, 7281, 8092, 8961, 9890, 10881, 11936, 13057, 14246, 15505, 16836, 18241, 19722, 21281, 22920, 24641, 26446, 28337, 30316

Offset: 0

Gary W. Adamson, Jul 16 2003

Row sums of triangle A131782 starting (1, 6, 17, 36, 65, 106, ...). - Gary W. Adamson, Jul 14 2007
a(n) is the number of triples (x,y,z) in {1,2,...,n}^3 with x <= y <= z or x >= y >= z. - Jack Kennedy, Mar 14 2009
a(2*n) is the difference between numbers of nonnegative multiples of 2*n+1 with even and odd digit sum in base 2*n in interval [0, 16*n^4). - Vladimir Shevelev, May 18 2012

			Let n=2. Consider nonnegative multiples of 5 up to 16*2^4 - 1 = 255. There are 52 such numbers and from them only 8 (namely, 35, 50, 55, 115, 140, 200, 205, 220) have an odd digit sum in base 4. Therefore, a(4) = (52 - 8) - 8 = 36. - _Vladimir Shevelev_, May 18 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem III. Partial queens, arXiv:1402.4886 [math.CO], 2014-2018. See Conjecture 4.4.
Vladimir Shevelev, On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m, arXiv:0710.3177 [math.NT], 2007.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000290, A000292, A028387, A064043, A077415, A131782.

I:=[0,1,6,17]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Mar 28 2014

A084990:=n->n*(n^2+3*n-1)/3; seq(A084990(k),k=0..100); # Wesley Ivan Hurt, Oct 19 2013

Table[n*(n^2+3*n-1)/3,{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Mar 08 2010 *)
CoefficientList[Series[(x + 2 x^2 - x^3)/((1 - x)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 28 2014 *)
LinearRecurrence[{4,-6,4,-1},{0,1,6,17},50] (* Harvey P. Dale, Aug 18 2015 *)

a(n) = n*(n^2+3*n-1)/3 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 2*A000292(n-1) - 1 (notice offset=-1 in A000292!).
a(n) = (n-1)*(n+1)*(n+3)/3 + 1. - Reinhard Zumkeller, Aug 20 2007
a(n) = A077415(n+1) + 1 for n > 0; a(n) = A000290(n) + A007290(n); a(n+1) = Sum_{k=0..n} A028387(k). - Reinhard Zumkeller, Aug 20 2007
a(2*n) = Sum_{i=0..16*n^4, i==0 (mod 2*n+1)} (-1)^s_(2*n)(i), where s_k(n) is the digit sum of n in base k. - Vladimir Shevelev, May 18 2012
a(2*n) = (2/(2*n+1))*Sum_{i=1..n} tan^4(Pi*i/(2*n+1)). - Vladimir Shevelev, May 23 2012
a(n) = Sum_{i=1..n} i*(i+1)-1. - Wesley Ivan Hurt, Oct 19 2013
G.f.: x*(1+2*x-x^2)/(1-x)^4. - Vincenzo Librandi, Mar 28 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Vincenzo Librandi, Mar 28 2014
a(n) = A064043(n)/3. - Alois P. Heinz, Jul 21 2017
E.g.f.: x*exp(x)*(x^2 + 6*x + 3)/3. - Stefano Spezia, Mar 06 2024

A176541 Numbers n such that there exist n consecutive triangular numbers which sum to a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 11, 13, 22, 23, 25, 27, 32, 37, 39, 46, 47, 48, 49, 50, 52, 59, 66, 71, 73, 83, 94, 98, 100, 104, 107, 109, 111, 118, 121, 128, 143, 146, 147, 148, 157, 167, 176, 179, 181, 183, 191, 192, 193, 194, 200, 214, 219, 227, 239, 241, 242, 243, 244, 253, 263

Offset: 1

Andrew Weimholt, Apr 20 2010

nonn

Numbers n such that there exists some x >= 0 such that A000292(x+n) - A000292(x) is a square. Terms of this sequence, for which only a finite number of solutions x exist, are given in A176542.
Integer n is in the sequence if there exist non-degenerate solutions to the Diophantine equation: 8x^2 - n*y^2 - A077415(n) = 0. A degenerate solution is one involving triangular numbers with negative indexes.
The sum of n consecutive triangular numbers starting at the j-th is Sum_{k=j..j+n-1} A000217(k) = n*(n^2 + 3*j*n + 3*j^2 - 1)/6, see A143037. - R. J. Mathar, May 06 2015

			0 is in the sequence because the sum of 0 consecutive triangular numbers is 0 (a square).
1 is in the sequence because there exist triangular numbers which are squares (cf. A001110).
2 is in the sequence because ANY 2 consecutive triangular numbers sum to a square.
3 is in the sequence because there are infinitely many solutions (cf. A165517).
4 is in the sequence because there infinitely many solutions (cf. A202391).
5 is NOT in the sequence because no 5 consecutive triangular numbers sum to a square.
For n=8, solutions to the Diophantine equation exist, but start at A000217(-2) and A000217(-6): 1 + 0 + 0 + 1 + 3 + 6 + 10 + 15 = 36 and 15 + 10 + 6 + 3 + 1 + 0 + 0 + 1 = 36. There are no non-degenerate solutions for n=8. Hence, 8 is not included in the sequence.
For n=11, there exist infinitely many solutions (cf. A116476), so 11 is in the sequence.

Cf. A176542, A000217, A000292, A001110, A077415.

More terms from Max Alekseyev, May 10 2010

A176542 Numbers n such that there are only a finite nonzero number of sets of n consecutive triangular numbers that sum to a square.

Original entry on oeis.org

32, 50, 98, 128, 200, 242, 338, 392, 512, 578, 722, 800, 968, 1058, 1250, 1352, 1568, 1682, 1922, 2048, 2312, 2450, 2738, 2888, 3200, 3362, 3698, 3872, 4232, 4418, 4802, 5000, 5408, 5618, 6050, 6272, 6728, 6962, 7442, 7688, 8192, 8450, 8978, 9248, 9800

Offset: 1

Andrew Weimholt, Apr 20 2010

nonn

Members of A176541, for which there are only a finite number of solutions.
Integer n is in this sequence if n=2*m^2 and the equation (2*x-m*y)*(2*x+m*y)=A077415(n)/2 has integer solutions with y>=n. - Max Alekseyev, May 10 2010
It seems that a(n) = 2*A001651(n+2)^2. - Colin Barker, Sep 25 2015

			32 is in this sequence because there is only one set of 32 consecutive triangular numbers that sum to a square (namely, A000217(26) thru A000217(57), which sum to 29584 = 172^2).
3 is NOT in this sequence, because there are infinitely many sets of 3 consecutive triangular numbers that sum to a square (cf. A165517).
4 is NOT in this sequence, because there are infinitely many sets of 4 consecutive triangular numbers that sum to a square (cf. A202391).
5 is NOT in this sequence, because there are NO sets of 5 consecutive triangular numbers that sum to a square.
11 is NOT in this sequence, since there are infinitely many sets of 11 consecutive triangular numbers that sum to a square (cf. A116476).

Cf. A176541, A000217, A000292, A077415.

Conjectures from Colin Barker, Sep 24 2015: (Start)
a(n) = (9*n^2+24*n+16)/2 for n even.
a(n) = (9*n^2+30*n+25)/2 for n odd.
a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>5.
G.f.: -2*x*(4*x^4-3*x^3-8*x^2+9*x+16) / ((x-1)^3*(x+1)^2).
(End)

Terms a(6) onward from Max Alekseyev, May 10 2010

A257293 Numbers n such that T(n) + T(n+1) + ... + T(n+12) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

3, 29, 75, 432, 998, 3624, 8310, 44717, 102443, 370269, 848195, 4561352, 10448838, 37764464, 86508230, 465213837, 1065679683, 3851605709, 8822991915, 47447250672, 108688879478, 392826018504, 899858667750, 4839154355357, 11085200027723, 40064402282349

Offset: 1

M. F. Hasler, May 04 2015

It is well known that T(n)+T(n+1) is always a square. T(n)+T(n+1)+T(n+2) is a square for n in A165517. T(n)+T(n+1)+T(n+2)+T(n+3) is a square for n in A202391. There is no sequence of 5, 6, 7, 8, 9 or 10 consecutive T(i)'s which sum to a square, cf. A176541. The next possible length is 11, see A116476. Then comes this sequence, corresponding to length 13.

Positive integers y in the solutions to 2*x^2-13*y^2-169*y-728 = 0. - Colin Barker, May 04 2015

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,102,-102,0,0,-1,1).

Cf. A176541, A176542, A000217, A000292, A001110, A077415.

Cf. A116476 (length 11).

I:=[3,29,75,432,998,3624,8310,44717,102443]; [n le 9 select I[n] else Self(n-1)+102*Self(n-4)-102*Self(n-5)-Self(n-8)+Self(n-9): n in [1..40]]; // Vincenzo Librandi, May 05 2015

Select[Range[10^5],IntegerQ[Sqrt[(#^2+13*#+56)*13/2]]&] (* Ivan N. Ianakiev, May 04 2015 *)
LinearRecurrence[{1, 0, 0, 102, -102, 0, 0, -1, 1}, {3, 29, 75, 432, 998, 3624, 8310, 44717, 102443}, 50] (* Vincenzo Librandi, May 05 2015 *)

for(n=0,10^8,issquare(binomial(n+14,3)-binomial(n+1,3))&&print1(n","))

Vec(x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)) + O(x^100)) \\ Colin Barker, May 04 2015

G.f.: x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)). - Colin Barker, May 04 2015

A257707 Numbers n such that T(n) + T(n+1) + ... + T(n+22) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

56, 470, 1094, 7856, 128534, 201539, 3293081, 23435699, 53805155, 382911281, 6256309475, 9809462822, 160274811896, 1140616029542, 2618697452438, 18636292598096, 304494582579398, 477426555904883, 7800575092244921, 55513782134933123, 127452004956911987

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-23*y^2-529*y-4048 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,48670,-48670,0,0,0,0,-1,1).

Cf. A176541, A176542, A000217, A000292, A001110, A077415.

Cf. A116476 (length 11), A257293 (length 13), A257708 (length 25), A257709 (length 27), A257710 (length 37).

LinearRecurrence[{1, 0, 0, 0, 0, 48670, -48670, 0, 0, 0, 0, -1, 1}, {56, 470, 1094, 7856, 128534, 201539, 3293081, 23435699, 53805155, 382911281, 6256309475, 9809462822, 160274811896}, 50] (* Vincenzo Librandi, May 05 2015 *)

Vec(x*(10*x^12 +3*x^11 +66*x^10 +414*x^9 +624*x^8 +6762*x^7 -366022*x^6 -73005*x^5 -120678*x^4 -6762*x^3 -624*x^2 -414*x -56) / ((x -1)*(x^12 -48670*x^6 +1)) + O(x^100))

G.f.: x*(10*x^12 +3*x^11 +66*x^10 +414*x^9 +624*x^8 +6762*x^7 -366022*x^6 -73005*x^5 -120678*x^4 -6762*x^3 -624*x^2 -414*x -56) / ((x -1)*(x^12 -48670*x^6 +1)).

A257708 Numbers n such that T(n) + T(n+1) + ... + T(n+24) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

25, 55, 208, 382, 1273, 2287, 7480, 13390, 43657, 78103, 254512, 455278, 1483465, 2653615, 8646328, 15466462, 50394553, 90145207, 293721040, 525404830, 1711931737, 3062283823, 9977869432, 17848298158, 58155284905, 104027505175, 338953840048, 606316732942

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-25*y^2-625*y-5200 = 0.

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

Cf. A176541, A176542, A000217, A000292, A001110, A077415.

Cf. A116476 (length 11), A257293 (length 13), A257707 (length 23), A257709 (length 27), A257710 (length 37).

LinearRecurrence[{1, 6, -6, -1, 1}, {25, 55, 208, 382, 1273}, 50] (* Vincenzo Librandi, May 05 2015 *)

Vec(x*(x^2+4*x+5)*(2*x^2-2*x-5)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^100))

G.f.: x*(x^2+4*x+5)*(2*x^2-2*x-5) / ((x-1)*(x^2-2*x-1)*(x^2+2*x-1)).

A257709 Numbers n such that T(n) + T(n+1) + ... + T(n+26) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

8, 14, 39, 53, 103, 112, 206, 264, 509, 647, 1141, 1230, 2160, 2734, 5159, 6525, 11415, 12296, 21502, 27184, 51189, 64711, 113117, 121838, 212968, 269214, 506839, 640693, 1119863, 1206192, 2108286, 2665064, 5017309, 6342327, 11085621, 11940190, 20870000

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-27*y^2-729*y-6552 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,10,-10,0,0,0,0,-1,1).

Cf. A176541, A176542, A000217, A000292, A001110, A077415.

Cf. A116476 (length 11), A257293 (length 13), A257707 (length 23), A257708 (length 25), A257710 (length 37).

I:=[8,14,39,53,103,112,206,264,509,647,1141,1230, 2160]; [n le 13 select I[n] else Self(n-1)+10*Self(n-6)-10*Self(n-7)-Self(n-12)+Self(n-13): n in [1..40]]; // Vincenzo Librandi, May 05 2015

LinearRecurrence[{1, 0, 0, 0, 0, 10, -10, 0, 0, 0, 0, -1, 1}, {8, 14, 39, 53, 103, 112, 206, 264, 509, 647, 1141, 1230, 2160}, 50] (* Vincenzo Librandi, May 05 2015 *)
Position[Total/@Partition[Accumulate[Range[70000]],27,1],?(IntegerQ[ Sqrt[ #]]&)]//Flatten (* The program generates the first 22 terms of the sequence. To generate more, increase the Range constant but the program may take a long time to run. *) (* _Harvey P. Dale, Jul 27 2021 *)

Vec(x*(2*x^12+x^11+6*x^10+2*x^9+5*x^8+2*x^7-14*x^6-9*x^5-50*x^4-14*x^3-25*x^2-6*x-8) / ((x-1)*(x^12-10*x^6+1)) + O(x^100))

G.f.: x*(2*x^12+x^11+6*x^10+2*x^9+5*x^8+2*x^7-14*x^6-9*x^5-50*x^4-14*x^3-25*x^2-6*x-8) / ((x-1)*(x^12-10*x^6+1)).

A257710 Numbers n such that T(n) + T(n+1) + ... + T(n+36) is a square, where T = A000217 (triangular numbers).

Original entry on oeis.org

5, 32, 291, 661, 4102, 8515, 13685, 113558, 182368, 377701, 2290342, 5027232, 30483491, 63130838, 101378488, 840238915, 1349295285, 2794368792, 16944086651, 37191598501, 225516999142, 467042067835, 749998177365, 6216087516438, 9982086472888, 20672740082341

Offset: 1

Colin Barker, May 04 2015

Positive integers y in the solutions to 2*x^2-37*y^2-1369*y-16872 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,7398,-7398,0,0,0,0,0,0,-1,1).

Cf. A176541, A176542, A000217, A000292, A001110, A077415.

Cf. A116476 (length 11), A257293 (length 13), A257707 (length 23), A257708 (length 25), A257709 (length 27).

LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 7398, -7398, 0, 0, 0, 0, 0, 0, -1, 1}, {5, 32, 291, 661, 4102, 8515, 13685, 113558, 182368, 377701, 2290342, 5027232, 30483491, 63130838, 101378488, 840238915, 1349295285}, 50] (* Vincenzo Librandi, May 05 2015 *)

Vec(x*(5*x^16 +27*x^15 +10*x^14 +27*x^13 +259*x^12 +370*x^11 +3441*x^10 +4413*x^9 -31820*x^8 -99873*x^7 -5170*x^6 -4413*x^5 -3441*x^4 -370*x^3 -259*x^2 -27*x -5) / ((x -1)*(x^8 -86*x^4 -1)*(x^8 +86*x^4 -1)) + O(x^100))

G.f.: x*(5*x^16 +27*x^15 +10*x^14 +27*x^13 +259*x^12 +370*x^11 +3441*x^10 +4413*x^9 -31820*x^8 -99873*x^7 -5170*x^6 -4413*x^5 -3441*x^4 -370*x^3 -259*x^2 -27*x -5) / ((x -1)*(x^8 -86*x^4 -1)*(x^8 +86*x^4 -1)).

A162540 a(n) = (2n+1)(2n+3)(2*n+5)/3.

Original entry on oeis.org

5, 35, 105, 231, 429, 715, 1105, 1615, 2261, 3059, 4025, 5175, 6525, 8091, 9889, 11935, 14245, 16835, 19721, 22919, 26445, 30315, 34545, 39151, 44149, 49555, 55385, 61655, 68381, 75579, 83265, 91455, 100165, 109411, 119209, 129575, 140525, 152075, 164241

Offset: 0

Jacob Landon (jacoblandon(AT)aol.com), Jul 05 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
M. R. Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A061550, A077415, A093828.

[(2*n+1)*(2*n+3)*(2*n+5)/3: n in [0..40]]; // Vincenzo Librandi, Nov 16 2011

A162540:=n->(2*n+1)*(2*n+3)*(2*n+5)/3: seq(A162540(n), n=0..80); # Wesley Ivan Hurt, May 28 2016

Table[((2n+1)(2n+3)(2n+5))/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{5,35,105,231},40] (* Harvey P. Dale, Nov 06 2011 *)

Vec((5+15*x-5*x^2+x^3)/(x-1)^4 + O(x^100)) \\ Altug Alkan, Oct 26 2015

a(n) = A061550(n)/3 = A077415(2*n+3).
From R. J. Mathar, Jul 16 2009: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.
G.f.: (5 + 15*x - 5*x^2 + x^3)/(x-1)^4. (End)
a(n) = 5*Pochhammer(7/2,n)/Pochhammer(1/2,n). Hence e.g.f. is 5* 1F1(7/2;1/2;x), with 1F1 being the confluent hypergemetric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
E.g.f.: (8*x^3 + 60*x^2 + 90*x + 15)*exp(x)/3. - Robert Israel, May 27 2016
From Amiram Eldar, Jan 09 2021: (Start)
Sum_{n>=0} 1/a(n) = 1/4.
Sum_{n>=0} (-1)^n/a(n) = 3*Pi/8 - 1 = A093828 - 1. (End)

Offset corrected, definition clarified by R. J. Mathar, Jul 16 2009

cp's OEIS Frontend

A006331 a(n) = n*(n+1)*(2*n+1)/3.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Formula

A084990 a(n) = n*(n^2+3*n-1)/3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A176541 Numbers n such that there exist n consecutive triangular numbers which sum to a square.

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions

A176542 Numbers n such that there are only a finite nonzero number of sets of n consecutive triangular numbers that sum to a square.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A257293 Numbers n such that T(n) + T(n+1) + ... + T(n+12) is a square, where T = A000217 (triangular numbers).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

A257707 Numbers n such that T(n) + T(n+1) + ... + T(n+22) is a square, where T = A000217 (triangular numbers).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A257708 Numbers n such that T(n) + T(n+1) + ... + T(n+24) is a square, where T = A000217 (triangular numbers).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

A006331 a(n) = n(n+1)(2*n+1)/3.

A084990 a(n) = n(n^2+3n-1)/3.

A162540 a(n) = (2n+1)(2n+3)(2*n+5)/3.