A086020 -id:A086020 - cp's OEIS frontend

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422

Offset: 0

Clark Kimberling

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 0..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 13, 15.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7.
Alexander R. Povolotsky, Problem 1147, Pi Mu Epsilon Fall 2006 Problems.
Alexander R. Povolotsky, Problem, Pi Mu Epsilon Spring 2007 Problems.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Cf. A000292, A000332, A000389, A000579, A000580, A024166, A027555, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127.

Cf. A000330, A000537 (first diffs), A001286, A003215, A100536, A101094 (partial sums), A101097, A101102, A222716.

a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
-- Reinhard Zumkeller, Oct 14 2001

[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017

A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017

Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)

a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008

for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A087127 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of triangular numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 2p-1, where a(i,p) satisfies Sum_{i=1..n} C(i+1,2)^p = 3 C(n+2,3) * Sum_{i=1..2p-1} a(i,p) C(n-1,i-1)/(i+2).

Original entry on oeis.org

1, 1, 2, 1, 1, 8, 19, 18, 6, 1, 26, 163, 432, 564, 360, 90, 1, 80, 1135, 6354, 18078, 28800, 26100, 12600, 2520, 1, 242, 7291, 77400, 405060, 1210680, 2211570, 2520000, 1751400, 680400, 113400, 1, 728, 45199, 862218, 7667646, 38350080, 118848420

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 08 2018: (Start)
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+2,2)^p of degree 2*p in terms of falling factorials: C(x+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the triangular numbers is also given by Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 2..2*p} A122193(p,k)*C(n+2,k+1) for p >= 1. (End)

			Row 3 contains 1,8,19,18,6, so Sum_{i=1..n} C(i+1,2)^3 = (n+2) * C(n+1,2) * [ a(1,3)/3 + a(2,3)*C(n-1,1)/4 + a(3,3)*C(n-1,2)/5 + a(4,3)*C(n-1,3)/6 + a(5,3)*C(n-1,4)/7 ] = [ (n+2)*(n+1)*n/2 ] * [ 1/3 + (8/4)*C(n-1,1) + (19/5)*C(n-1,2) + (18/6)*C(n-1,3) + (6/7)*C(n-1,4). Cf. A085438 for more details.
From _Peter Bala_, Mar 08 2018: (Start)
Table begins
n=0 |1
n=1 |1   2     1
n=2 |1   8    19    18      6
n=3 |1  26   163   432    564    360     90
n=4 |1  80  1135  6354  18078  28800  26100  12600  2520
...
Row 2: C(i+2,2)^2 = C(i,0) + 8*C(i,1) + 19*C(i,2) + 18*C(i,3) + 6*C(i,4). Hence, Sum_{i = 0..n-1} C(i+2,2)^2 =  C(n,1) + 8*C(n,2) + 19*C(n,3) + 18*C(n,4) + 6*C(n,5). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
M. Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A000217, A122193.

Flat(List([0..6],n->List([0..2*n],k->Sum([0..k],i->(-1)^(k-i)*Binomial(k,i)*Binomial(i+2,2)^n)))); # Muniru A Asiru, Mar 22 2018

seq(seq(add( (-1)^(k-i)*binomial(k,i)*binomial(i+2,2)^n, i = 0..k), k = 0..2*n), n = 0..8); # Peter Bala, Mar 08 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 3, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 2, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 2*p - 1}]//Flatten (* G. C. Greubel, Nov 23 2017 *)
a[i_,p_]:=(-1)^i HypergeometricPFQ[ConstantArray[3,p]~Join~{2-i},ConstantArray[1,p],1];Table[a[i,p],{p,0,10},{i,2,2 p+2}]//Flatten (* Jonathan Burns, Mar 20 2018 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 3, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 2, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 2*p-1, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(1, p) = 1, a(2, p) = 3^(p-1)-1, a(3, p) = 3^(p-1)*[2^(p-1)-2]+1, ..., a(2*p-3, p) = [ (6*p^4-20*p^3+21*p^2-7*p)*(2*p-4)! ]/[3*2^(p-1)], a(2*p-2, p) = [ (p^2-p)*(2*p-3)! ]/2^(p-2), a(2*p-1, p) = [ (p-1)*(2*p-3)! ]/2^(p-2).
a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+3, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+2, i-2*k)^(p-1) ]
From Peter Bala, Mar 08 2018: (Start)
The following remarks assume row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*C(i+2,2)^n. Equivalently, let v_n denote the sequence (1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318. Cf. A122193.
T(n+1,k) = C(k+2,2)*T(n,k) + 2*C(k+1,2)*T(n,k-1) + C(k,2)*T(n,k-2), with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 2*n.
Let R(n,x) denote the n-th row polynomial.
R(n+1,x) = 1/2!*(1 + x)^2*(d/dx)^2 (x^2*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+2,2)^n*x^i/(1 + x)^(i+1).
R(n,x) = (1 + x)^2 o (1 + x)^2 o ... o (1 + x)^2 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^2 o ... o x^2 (n factors) is the n-th row polynomial of A122193.
x^2*R(n,x) = (1 + x)^2 * the n-th row polynomial of A122193 (End)

Edited by Dean Hickerson, Aug 16 2003

A000540 Sum of 6th powers: 0^6 + 1^6 + 2^6 + ... + n^6.

Original entry on oeis.org

0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, 1978405, 3749966, 6735950, 11562759, 19092295, 30482920, 47260136, 71397705, 105409929, 152455810, 216455810, 302221931, 415601835, 563637724, 754740700, 998881325, 1307797101, 1695217590

Offset: 0

N. J. A. Sloane

This sequence is related to A000539 by a(n) = n*A000539(n)-sum(A000539(i), i=0..n-1). - Bruno Berselli, Apr 26 2010

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, (2008), p. 289.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
J. L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359. [Annotated scanned copy]
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
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
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1)

Cf. A101093, A000539.
Row 6 of array A103438.
Partial sums of A001014.

a000540 n = a000540_list !! n
a000540_list = scanl1 (+) a001014_list -- Reinhard Zumkeller, Dec 04 2011

[n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015

a:=n->sum (j^6,j=0..n): seq(a(n),n=0..27); # Zerinvary Lajos, Jun 27 2007
A000540:=(z+1)*(z**4+56*z**3+246*z**2+56*z+1)/(z-1)**8; # g.f. by Simon Plouffe in his 1992 dissertation, without the leading 0.
A000540 := proc(n) n^7/7+n^6/2+n^5/2-n^3/6+n/42 ; end proc: # R. J. Mathar

Accumulate[Range[0,30]^6] (* Harvey P. Dale, Jul 30 2009 *)
LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 1, 65, 794, 4890, 20515, 67171, 184820}, 31] (* Jean-François Alcover, Feb 09 2016 *)

a(n)=n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42 \\ Edward Jiang, Sep 10 2014

a(n)=sum(i=1, n, i^6); \\ Michel Marcus, Sep 11 2014

A000540_list, m = [0], [720, -1800, 1560, -540, 62, -1, 0, 0]
for _ in range(10**2):
    for i in range(7):
        m[i+1] += m[i]
    A000540_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

[bernoulli_polynomial(n,7)/7 for n in range(1, 29)]# Zerinvary Lajos, May 17 2009

a(n) = n*(n+1)*(2*n+1)*(3*n^4+6*n^3-3*n+1)/42.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^6). - Alexander Adamchuk, Oct 26 2004
G.f.: A(x) = 3*x/7*G(0); with G(k) = 1 + 2/(k+1+(k+1)/(2*k^2 + 4*k + 1 + 2*(k+1)^2/(3*k + 2 - 9*x*(k+1)*(k+2)^4*(k+3)*(2*k+5)/(3*x*(k+2)^4*(k+3)*(2*k+5)+(k+1)*(2*k+3)/G(k+1))))); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
G.f.: x*(1+x)*(x^4 + 56*x^3 + 246*x^2 + 56*x + 1) / (x-1)^8 . - R. J. Mathar, Aug 07 2012
a(n) = Sum_{i=1..n} J_6(i)*floor(n/i), where J_6 is A069091. - Enrique Pérez Herrero, Mar 09 2013
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) + 720. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..6} j*Stirling1(n+1,n+1-j)*Stirling2(n+6-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = 84*Pi*(8*cos(sqrt((sqrt(93) + 9)/6)*Pi) + 15*cos(sqrt((sqrt(93) + 9)/6)*Pi/2) * cosh(sqrt((sqrt(93) - 9)/6)*Pi/2) + 8*cosh(sqrt((sqrt(93) - 9)/6)*Pi) - 7*sqrt(3)*sin(sqrt((sqrt(93) + 9)/6)*Pi/2) * sinh(sqrt((sqrt(93) - 9)/6)*Pi/2)) / (31*(cos(sqrt((sqrt(93) + 9)/6)*Pi) + cosh(sqrt((sqrt(93) - 9)/6)*Pi))) = 0.985708051237101247832970793342271511... . - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + z)*(n + 1/2 - z)*(n + 1/2 + zbar)*(n + 1/2 - zbar)/7, with I^2 = -1 and z = 2^(-3/2)*3^(-1/4)*(sqrt(sqrt(31) + 3*sqrt(3)) + I*sqrt(sqrt(31) - 3*sqrt(3))), and zbar is the complex conjugate of z. See the Graham et al. reference, eq. (6.98), pp. 288-289 (with n -> n+1). (There was a typo in the first edition, which was corrected in the second edition.) - Wolfdieter Lang, Apr 03 2015
a(n+2) = 36*A086020(n+1) + 24*A005585(n+1) + A000330(n+2). - Yasser Arath Chavez Reyes, Apr 16 2024

A085438 a(n) = Sum_{i=1..n} binomial(i+1,2)^3.

Original entry on oeis.org

1, 28, 244, 1244, 4619, 13880, 35832, 82488, 173613, 339988, 627484, 1102036, 1855607, 3013232, 4741232, 7256688, 10838265, 15838476, 22697476, 31958476, 44284867, 60479144, 81503720, 108503720, 142831845, 186075396, 240085548, 307008964, 389321839

Offset: 1

André F. Labossière, Jun 30 2003

			a(10) = (90*(10^7)+630*(10^6)+1638*(10^5)+1890*(10^4)+840*(10^3)-48*(10))/5040 = 339988.

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Column k=3 of A334781.

Cf. A000292, A087127, A024166, A024166, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/ Factorial(7): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n)/7!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)

Vec(x*(x^4+20*x^3+48*x^2+20*x+1)/(x-1)^8 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = sum(i=1, n, binomial(i+1, 2)^3); \\ Michel Marcus, Nov 22 2017

a(n) = (90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/7!.
a(n) = (C(n+2, 3)/35)*(35 +210*C(n-1, 1) +399*C(n-1, 2) +315*C(n-1, 3) +90*C(n-1, 4)).
G.f.: x*(x^4+20*x^3+48*x^2+20*x+1) / (x-1)^8. - Colin Barker, May 02 2014

More terms from Colin Barker, May 02 2014

Formula and example edited by Colin Barker, May 02 2014

A085442 a(n) = Sum_{i=1..n} binomial(i+1,2)^7.

Original entry on oeis.org

1, 2188, 282124, 10282124, 181141499, 1982230040, 15475158552, 93839322648, 467508775773, 1989944010148, 7445104711204, 25010673566116, 76686775501847, 217396817767472, 575714897767472, 1436257466526768, 3398894618986905, 7674255436599996, 16612972826599996

Offset: 1

André F. Labossière, Jul 07 2003

T. D. Noe, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1).

Column k=7 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085441, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[Sum[Binomial[k+1,2]^7, {k,1,n}], {n,1,30}] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1},{1,2188,282124,10282124,181141499,1982230040,15475158552,93839322648,467508775773,1989944010148,7445104711204,25010673566116,76686775501847,217396817767472,575714897767472,1436257466526768},20] (* Harvey P. Dale, May 11 2022 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^7), ", ")) \\ G. C. Greubel, Nov 22 2017

a(n) = (1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568). - Vladeta Jovovic, Jul 07 2003

G.f.: x*(x^12 +2172*x^11 +247236*x^10 +6030140*x^9 +49258935*x^8 +163809288*x^7 +242384856*x^6 +163809288*x^5 +49258935*x^4 +6030140*x^3 +247236*x^2 +2172*x+ 1) / (x -1)^16. - Colin Barker, May 02 2014

A086030 a(n) = Sum_{i=1..n} C(i+6,7)^3.

Original entry on oeis.org

1, 513, 47169, 1775169, 37712169, 534505257, 5587534953, 46011772521, 312480135396, 1809674119396, 9165388162788, 41395684407012, 169328324418084, 635173167426084, 2207399512578084, 7167715400927268, 21902130296812161, 63361228916945025, 174437774859945025

Offset: 1

André F. Labossière, Jul 11 2003

			a(3) = Sum_{i=1..3} C(6+i,7)^3 = C(10,8)*(5*C(16,14) + 210*C(15,14) + 1491*C(14,14))/5 = 47169.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (23, -253, 1771, -8855, 33649, -100947, 245157, -490314, 817190, -1144066, 1352078, -1352078, 1144066, -817190, 490314, -245157, 100947, -33649, 8855, -1771, 253, -23, 1).

Cf. A087127, A024166, A085438 - A085442, A086020 - A086029.

[(1/28165294080000)*n^2*(1 + n)^2*(2 + n)^2*(3 + n)^2*(4 + n)^2*(5 + n)^2*(6 + n)^2*(7 + n)^2*(-3 + 3234*n + 6979*n^2 + 5292*n^3 + 1603*n^4 + 210*n^5 + 10*n^6): n in [1..30]]; // G. C. Greubel, Nov 22 2017

A086030:=n->add(binomial(i+6,7)^3, i=1..n); seq(A086030(n), n=1..30); # Wesley Ivan Hurt, Dec 22 2013

Table[Sum[Binomial[i + 6, 7]^3, {i, n}], {n, 30}] (* Wesley Ivan Hurt, Dec 22 2013 *)

a(n) = sum(i=1, n, binomial(i+6, 7)^3); \\ Michel Marcus, Dec 22 2013

a(n) = C(n+7,8)*(5*C(n+13,14) + 210*C(n+12,14) + 1491*C(n+11,14) + 2828*C(n+10,14) + 1491*C(n+9,14) + 210*C(n+8,14) + 5*C(n+7,14))/5. - Yahia Kahloune, Dec 22 2013
-(n-1)^3*a(n) +(2*n+5)*(n^2+5*n+43)*a(n-1) -(n+6)^3*a(n-2)=0. - R. J. Mathar, Dec 22 2013
G.f.: -x*(x^14 + 490*x^13 + 35623*x^12 + 818300*x^11 + 7917371*x^10 + 37215794*x^9 + 91789005*x^8 + 123519792*x^7 + 91789005*x^6 + 37215794*x^5 + 7917371*x^4 + 818300*x^3 + 35623*x^2 + 490*x + 1)/(x-1)^23. - Vaclav Kotesovec, Dec 23 2013
a(n) = (1/28165294080000)*n^2*(1 + n)^2*(2 + n)^2*(3 + n)^2*(4 + n)^2*(5 + n)^2*(6 + n)^2*(7 + n)^2*(-3 + 3234*n + 6979*n^2 + 5292*n^3 + 1603*n^4 + 210*n^5 + 10*n^6). - G. C. Greubel, Nov 22 2017

A086023 a(n) = Sum_{i=1..n} C(i+3,4)^2.

Original entry on oeis.org

1, 26, 251, 1476, 6376, 22252, 66352, 175252, 420277, 931502, 1933503, 3796728, 7109128, 12773528, 22137128, 37160504, 60634529, 96454754, 149963979, 228375004, 341286880, 501309380, 724811880, 1032814380, 1452040005, 2016150006, 2767184031, 3757230256

Offset: 1

André F. Labossière, Jul 11 2003

G. C. Greubel, Table of n, a(n) for n = 1..5000
John Engbers and Christopher Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A087127, A024166, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086024, A086025, A086026, A086027, A086028, A086029, A086030.

[n*(n+1)*(n+2)*(n+3)*(n+4)*(35*n^4 +280*n^3 +685*n^2 +500*n +12 )/181440: n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(35*n^4 + 280*n^3 + 685*n^2 + 500*n + 12)/181440, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)
Accumulate[Binomial[Range[30]+3,4]^2] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,26,251,1476,6376,22252,66352,175252,420277,931502},30] (* Harvey P. Dale, May 06 2018 *)

a(n) = sum(i=1, n, binomial(i+3, 4)^2); \\ Michel Marcus, Sep 05 2013

Vec(x*(x^4+16*x^3+36*x^2+16*x+1)/(x-1)^10 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = ( C(n+4,5)/126 )*( 126 +420*C(n-1,1) +540*C(n-1,2) +315*C(n-1,3) +70*C(n-1,4) ).
a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(35*n^4 +280*n^3 +685*n^2 +500*n +12 )/181440. - Bruno Berselli, Sep 05 2013
G.f.: x*(x^4+16*x^3+36*x^2+16*x+1) / (x-1)^10. - Colin Barker, May 02 2014

More terms from Michel Marcus, Sep 05 2013

A086025 a(n) = Sum_{i=1..n} C(i+4,5)^2.

Original entry on oeis.org

1, 37, 478, 3614, 19490, 82994, 296438, 923702, 2580071, 6588075, 15606084, 34685508, 72976852, 146387476, 281597860, 521971876, 936053677, 1629533233, 2761788434, 4568378450, 7391175350, 11718183750, 18235516650, 27894475050, 41997225075, 62305185111

Offset: 1

André F. Labossière, Jul 11 2003

T. D. Noe, Table of n, a(n) for n = 1..1000
John Engbers and Christopher Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495, 220,-66,12,-1).

Cf. A087127, A024166, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086026, A086027, A086028, A086029, A086030.

[n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12)/19958400: n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12)/19958400, {n,1,30}] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{12,-66,220,-495,792,-924,792,-495,220,-66,12,-1},{1,37,478,3614,19490,82994,296438,923702,2580071,6588075,15606084,34685508},30] (* Harvey P. Dale, Dec 22 2024 *)

for(n=1,30, print1(sum(i=1,n, binomial(i+4,5)^2), ", ")) \\ G. C. Greubel, Nov 22 2017

From R. J. Mathar, Jun 16 2010: (Start)
G.f.: x*(1+x)*(x^4+24*x^3+76*x^2+24*x+1)/(x-1)^12.
a(n) = n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12) / 19958400. (End)

More terms from R. J. Mathar, Jun 16 2010

A086024 a(n) = Sum_{i=1..n} C(i+3,4)^3.

Original entry on oeis.org

1, 126, 3501, 46376, 389376, 2389752, 11650752, 47587752, 168875127, 534401002, 1537404003, 4080706128, 10109274128, 23590546128, 52243162128, 110473767504, 224205418629, 438589465254, 830009446129, 1524339072504, 2724140666880, 4748425291880, 8089787666880

Offset: 1

André F. Labossière, Jul 11 2003

T. D. Noe, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432, -3003,2002,-1001,364,-91,14,-1).

Cf. A087127, A024166, A085438 - A085442, A086020 - A086030.

[(n/69189120)*(13824 + 960960*n^2 + 5885880*n^3 + 14370356*n^4 + 19269250*n^5 + 15996695*n^6 + 8678670*n^7 + 3138135*n^8 + 750750*n^9 + 114205*n^10 + 10010*n^11 + 385*n^12): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(n/69189120)*(13824 + 960960*n^2 + 5885880*n^3 + 14370356*n^4 + 19269250*n^5 + 15996695*n^6 + 8678670*n^7 + 3138135*n^8 + 750750*n^9 + 114205*n^10 + 10010*n^11 + 385*n^12), {n,1,30}] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1},{1,126,3501,46376,389376,2389752,11650752,47587752,168875127,534401002,1537404003,4080706128,10109274128,23590546128},30] (* Harvey P. Dale, Feb 18 2024 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+3, 4)^3), ", ")) \\ G. C. Greubel, Nov 22 2017

a(n) = ( C(n+4, 5)/1001 )*( 1001 +20020*C(n-1, 1) +125840*C(n-1, 2) +390390*C(n-1, 3) +695695*C(n-1, 4) +750750*C(n-1, 5) +486850*C(n-1, 6) +175175*C(n-1, 7) +26950*C(n-1, 8) ).
G.f.: x*(1 +112*x +1828*x^2 +8464*x^3 +13840*x^4 +8464*x^5 +1828*x^6 +112*x^7 +x^8)/(x-1)^14 . - R. J. Mathar, Dec 22 2013
-(n-1)^3*a(n) +2*(n+1)*(n^2+2*n+13)*a(n-1) -(n+3)^3*a(n-2)=0. - R. J. Mathar, Dec 22 2013
a(n) = (n/69189120)*(13824 + 960960*n^2 + 5885880*n^3 + 14370356*n^4 + 19269250*n^5 + 15996695*n^6 + 8678670*n^7 + 3138135*n^8 + 750750*n^9 + 114205*n^10 + 10010*n^11 + 385*n^12). - G. C. Greubel, Nov 22 2017

A086029 a(n) = Sum_{i=1..n} C(i+6,7)^2.

Original entry on oeis.org

1, 65, 1361, 15761, 124661, 751925, 3696581, 15475205, 56884430, 187758030, 565982734, 1578749710, 4117700254, 10127050654, 23648089054, 52733344990, 112835299639, 232623278455, 463695768455, 896396608455, 1684993889355, 3086944610955, 5522978819355

Offset: 1

André F. Labossière, Jul 11 2003

			a(3) = C(10,8)*(-1*C(10,0) + 8*C(10,1) - 36*C(10,2) + 120*C(10,3) - 330*C(10,4) + 792*C(10,5) - 1716*C(10,6) + 3432*C(10,7))/6435 = 1361.

G. C. Greubel, Table of n, a(n) for n = 1..5000
John Engbers and Christopher Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (16, -120, 560, -1820, 4368, -8008, 11440, -12870, 11440, -8008, 4368, -1820, 560, -120, 16, -1).

Cf. A087127, A024166, A085438-A085442, A086020-A086030, A234253.

[(1/108972864000)*n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7+n)*(7 + 2*n)*(60 + 64974*n + 126245*n^2 + 82467*n^3 + 23408*n^4 + 3003*n^5 + 143*n^6): n in [1..30]]; // G. C. Greubel, Nov 22 2017

A086029:=n->add(binomial(i+6, 7)^2, i=1..n); seq(A086029(n), n=1..30); # Wesley Ivan Hurt, Dec 22 2013

Table[Sum[Binomial[i + 6, 7]^2, {i, n}], {n, 30}] (* Wesley Ivan Hurt, Dec 22 2013 *)
LinearRecurrence[{16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1},{1,65,1361,15761,124661,751925, 3696581, 15475205, 56884430,187758030,565982734,1578749710,4117700254, 10127050654,23648089054,52733344990},40] (* Harvey P. Dale, Apr 25 2016 *)

a(n) = sum(i=1, n, binomial(i+6, 7)^2); \\ Michel Marcus, Dec 22 2013

Vec(x*(x+1)*(x^6+48*x^5+393*x^4+832*x^3+393*x^2+48*x+1)/(x-1)^16 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = C(n+7,8)*(-C(n+7,0) + 8*C(n+7,1) - 36*C(n+7,2) + 120*C(n+3,7) - 330*C(n+7,4) + 792*C(n+7,5) - 1716*C(n+7,6) + 3432*C(n+7,7))/6435. - Yahia Kahloune, Dec 22 2013
(n-1)^2*a(n) +(-2*n^2-10*n-37)*a(n-1) +(n+6)^2*a(n-2)=0. - R. J. Mathar, Dec 22 2013
G.f.: x*(x+1)*(x^6 +48*x^5 +393*x^4 +832*x^3 +393*x^2 +48*x +1)/(1-x)^16. - Colin Barker, May 02 2014
a(n) = (1/108972864000)*n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7+n)*(7 + 2*n)*(60 + 64974*n + 126245*n^2 + 82467*n^3 + 23408*n^4 + 3003*n^5 + 143*n^6). - G. C. Greubel, Nov 22 2017

More terms from Michel Marcus, Dec 22 2013

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A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

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A000540 Sum of 6th powers: 0^6 + 1^6 + 2^6 + ... + n^6.

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A085438 a(n) = Sum_{i=1..n} binomial(i+1,2)^3.

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A085442 a(n) = Sum_{i=1..n} binomial(i+1,2)^7.

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A086030 a(n) = Sum_{i=1..n} C(i+6,7)^3.

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