A086028 -id:A086028 - cp's OEIS frontend

A085440 a(n) = Sum_{i=1..n} binomial(i+1,2)^5.

1, 244, 8020, 108020, 867395, 4951496, 22161864, 82628040, 267156165, 770440540, 2022773116, 4909947484, 11150268935, 23913084560, 48796284560, 95322158736, 179163294729, 325374464580, 572984364580, 981394464580, 1639143014731, 2675722491224, 4277290592600

Offset: 1

André F. Labossière, Jun 30 2003

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 = 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 (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Column k=5 of A334781.

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

[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n )/Factorial(11): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!, {n,1,50}] (* G. C. Greubel, Nov 22 2017 *)

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

a(n) = (113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!.

G.f.: x*(x^8+232*x^7+5158*x^6+27664*x^5+47290*x^4+27664*x^3+5158*x^2+232*x+1) / (x-1)^12. - Colin Barker, May 02 2014

Formula edited by Colin Barker, May 02 2014

A085441 a(n) = Sum_{i=1..n} binomial(i+1,2)^6.

Original entry on oeis.org

1, 730, 47386, 1047386, 12438011, 98204132, 580094436, 2756876772, 11060642397, 38741283022, 121395233038, 346594833742, 914464085783, 2254559726408, 5240543726408, 11568062614344, 24395756421273, 49397866465794, 96443747465794, 182209868465794

Offset: 1

André F. Labossière, Jul 07 2003

			a(5) = C(7,3)*[191*106 + 450*(18*C(14,10) + 3851*C(13,10) + 61839*C(12,10) + 225352*C(11,10) + 225352*C(10,10))]/10010 = 12438011.

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).

Column k=6 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030, A234253.

[(n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12): n in [1..30]]; // G. C. Greubel, Nov 22 2017

f:= sum(binomial(1+i,2)^6,i=1..n):
seq(f, n=1..30); # Robert Israel, Nov 22 2017

Table[Sum[Binomial[i+1,2]^6,{i,n}],{n,20}] (* or *) LinearRecurrence[ {14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1},{1,730,47386,1047386,12438011, 98204132,580094436, 2756876772,11060642397, 38741283022,121395233038, 346594833742, 914464085783, 2254559726408},20] (* Harvey P. Dale, Jun 05 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^6), ", ")) \\ G. C. Greubel, Nov 22 2017

G.f.: x*(x^10 +716*x^9 +37257*x^8 +450048*x^7 +1822014*x^6 +2864328*x^5 +1822014*x^4 +450048*x^3 +37257*x^2 +716*x +1) / (x -1)^14. - Colin Barker, May 02 2014

a(n) = (n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12). - G. C. Greubel, Nov 22 2017

A086022 a(n) = Sum_{i=1..n} C(i+2,3)^4.

Original entry on oeis.org

1, 257, 10257, 170257, 1670882, 11505378, 61292514, 268652514, 1009853139, 3352413139, 10042998755, 27598188771, 70457539396, 168802499396, 382616259396, 825980472132, 1707628231653, 3396588391653, 6525595601653, 12150082161653, 21987344308134, 38769279231910

Offset: 1

André F. Labossière, Jul 11 2003

			a(8) = C(11,4)*[-41*2793 + 350*(47*C(16,9) + 1749*C(15,9) + 9292*C(14,9) + 9292*C(13,9) + 1749*C(12,9) + 47*C(11,9))]/15015 = 268652514 .

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. A086020, A086021, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127, A024166, A085438, A085439, A085440, A085441, A085442.

[(n/12972960)*(-8856 +60060*n^2 +165165*n^3 +841841*n^4 +2462460*n^5 +3709420*n^6 +3243240*n^7 +1756755*n^8 +600600*n^9 +126490*n^10 +15015*n^11 +770*n^12): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Accumulate[Binomial[Range[3,30],3]^4] (* Harvey P. Dale, Oct 09 2016 *)

for(n=1,30, print1((n/12972960)*(-8856 + 60060*n^2 + 165165*n^3 + 841841*n^4 + 2462460*n^5 + 3709420*n^6 + 3243240*n^7 + 1756755*n^8 + 600600*n^9 + 126490*n^10 + 15015*n^11 + 770*n^12), ", ")) \\ G. C. Greubel, Nov 22 2017

G.f.: x*(1+x)*(x^8 +242*x^7 +6508*x^6 +43174*x^5 +84950*x^4 +43174*x^3 +6508*x^2 +242*x + 1) / (x-1)^14 . - R. J. Mathar, Dec 22 2013
(n-1)^4*a(n) +(-2*n^4 -4*n^3 -30*n^2 -28*n -17)*a(n-1) +(n+2)^4*a(n-2)=0. - R. J. Mathar, Dec 22 2013
a(n) = C(n+3,4)*[-41*F3(n) +350*(47*C(n+8,9) + 1749*C(n+7,9) + 9292*C(n+6,9) + 9292*C(n+5,9) + 1749*C(n+4,9) + 47*C(n+3,9))]/15015, where F3(n) = -C(3,0)*C(n+3,0) + C(4,1)*C(n+3,1) - C(5,2)*C(n+3,2) + C(6,3)*C(n+3,3). The value of F3(n), (n=0..8) is: 1, 35, 119, 273, 517, 871, 1355, 1989, 2793, ... - Yahia Kahloune, Dec 23 2013
a(n) = (n/12972960)*(-8856 + 60060*n^2 + 165165*n^3 + 841841*n^4 + 2462460*n^5 + 3709420*n^6 + 3243240*n^7 + 1756755*n^8 + 600600*n^9 + 126490*n^10 + 15015*n^11 + 770*n^12). - G. C. Greubel, Nov 22 2017

A087107 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of tetrahedral numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 3p-2, where a(i,p) satisfies Sum_{i=1..n} C(i+2,3)^p = 4 C(n+3,4) * Sum_{i=1..3p-2} a(i,p) C(n-1,i-1)/(i+3).

Original entry on oeis.org

1, 1, 3, 3, 1, 1, 15, 69, 147, 162, 90, 20, 1, 63, 873, 5191, 16620, 31560, 36750, 25830, 10080, 1680, 1, 255, 9489, 130767, 919602, 3832650, 10238000, 18244380, 21990360, 17745000, 9198000, 2772000, 369600, 1, 1023, 97953, 2903071, 40317780

Offset: 1

André F. Labossière, Aug 11 2003

Let s_n denote the sequence (1, 4^n, 10^n, 20^n, ...) regarded as an infinite column vector, where 1, 4, 10, 20, ... is the sequence of tetrahedral numbers A000292. It appears that the n-th row of this table is determined by the matrix product P^(-1)s_n, where P denotes Pascal's triangle A007318. - Peter Bala, Nov 26 2017
From Peter Bala, Mar 11 2018: (Start)
The observation above is correct.
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+3,3)^p of degree 3*p in terms of falling factorials: C(x+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the tetrahedral numbers is also given by Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 3..3*p} A299041(p,k)*C(n+3,k+1) for p >= 1. (End)

			Row 3 contains 1,15,69,147,162,90,20, so Sum_{i=1..n} C(i+2,3)^3 = 4 * C(n+3,4) * [ a(1,3)/4 + a(2,3)*C(n-1,1)/5 + a(3,3)*C(n-1,2)/6 + ... + a(7,3)*C(n-1,6)/10 ] = 4 * C(n+3,4) * [ 1/4 + 15*C(n-1,1)/5 + 69*C(n-1,2)/6 + 147*C(n-1,3)/7 + 162*C(n-1,4)/8 + 90*C(n-1,5)/9 + 20*C(n-1,6)/10 ]. Cf. A086021 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
n=0 | 1
n=1 | 1  3   3    1
n=2 | 1 15  69  147   162    90    20
n=3 | 1 63 873 5191 16620 31560 36750 25830 10080 1680
...
Row 2: C(i+3,3)^2 = C(i,0) + 15*C(i,1) + 69*C(i,2) + 147*C(i,3) + 162*C(i,4) + 90*C(i,5) + 20*C(i,6). Hence, Sum_{i = 0..n-1} C(i+3,3)^2 =  C(n,1) + 15*C(n,2) + 69*C(n,3) + 147*C(n,4) + 162*C(n,5) + 90*C(n,6) + 20*C(n,7). (End)

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

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A087110, A174266, A299041.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+3, 3)^n, i= 0..k), k = 0..3*n), n = 0..8); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 4, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 3, 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, 3*p - 2}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

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

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+4, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+3, i-2*k)^(p-1) ].
From Peter Bala, Nov 26 2017: (Start)
Conjectural formula for table entries: T(n,k) = Sum_{j = 0..k} (-1)^(k+j)*binomial(k,j)*binomial(j+3,3)^n.
Conjecturally, the n-th row polynomial R(n,x) = 1/(1 + x)*Sum_{i >= 0} binomial(i+3,3)^n *(x/(1 + x))^n. (End)
From Peter Bala, Mar 11 2018: (Start)
The conjectures above are correct.
The following remarks assume the row and column indices start at 0.
T(n+1,k) = C(k+3,3)*T(n,k) + 3*C(k+2,3)*T(n,k-1) + 3*C(k+1,3)*T(n,k-2) + C(k,3)*T(n,k-3) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 3*n.
Sum_{k = 0..3*n} T(n,k)*binomial(x,k) = (binomial(x+3,3))^n.
x^3*R(n,x) = (1 + x)^3 * the n-th row polynomial of A299041.
R(n+1,x) = 1/3!*(1 + x)^3*(d/dx)^3 (x^3*R(n,x)).
(1 - x)^(3*n)*R(n,x/(1 - x)) gives the n-th row polynomial of A174266.
R(n,x) = (1 + x)^3 o (1 + x)^3 o ... o (1 + x)^3 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^3 o ... o x^3 (n factors) is the n-th row polynomial of A299041. (End)

Edited by Dean Hickerson, Aug 16 2003

A087111 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,7). The p-th row (p>=1) contains a(i,p) for i=1 to 7p-6, where a(i,p) satisfies Sum_{i=1..n} C(i+6,7)^p = 8 C(n+7,8) * Sum_{i=1..7p-6} a(i,p) C(n-1,i-1)/(i+7).

Original entry on oeis.org

1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 63, 1169, 10703, 58821, 214123, 545629, 1004307, 1356194, 1347318, 974862, 500346, 172788, 36036, 3432, 1, 511, 45633, 1589567, 29302889, 333924087, 2577462937, 14287393351, 59159005164, 188008120188

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+7,7)^p of degree 7*p in terms of falling factorials: C(x+7,7)^p = Sum_{k = 0..7*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+7,7)^p = Sum_{k = 0..7*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,63,1169,...,3432, so Sum_{i=1..n} C(i+6,7)^3 = 8 * C(n+7,8) * [ a(1,3)/8 + a(2,3)*C(n-1,1)/9 + a(3,3)*C(n-1,2)/10 + ... + a(15,3)*C(n-1,14)/22 ] = 8 * C(n+7,8) * [ 1/8 + 63*C(n-1,1)/9 + 1169*C(n-1,2)/10 + ... + 3432*C(n-1,14)/22 ]. Cf. A086030 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattebed
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A027555, A086029, A086030, A087127.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+7, 7)^n, i = 0..k), k = 0..7*n), n = 0..4); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 8, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 7, 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, 7*p - 6}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 8, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 7, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 7*p-6, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+8, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+7, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+7,7)^n. Equivalently, let v_n denote the sequence (1, 8^n, 36^n, 120^n, ...) regarded as an infinite column vector, where 1, 8, 36, 120, ... is the sequence binomial(n+7,7) - see A000580. 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.
Recurrence: T(n+1,k) = Sum_{i = 0..7} C(7,i)*C(k+7-i,7)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 7*n.
n-th row polynomial R(n,x) = (1 + x)^7 o (1 + x)^7 o ... o (1 + x)^7 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/7!*(1 + x)^7 * (d/dx)^7(x^7*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+7,7)^n*x^i/(1 + x)^(i+1).
(End)

Edited by Dean Hickerson, Aug 16 2003

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

Original entry on oeis.org

1, 1, 36, 225, 400, 225, 36, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1

Offset: 1

Yahia Kahloune, Feb 05 2014

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,6,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 6 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 06 2020

			For example :
  T(n,0) = 1;
  T(n,1) = 7^n - (6*n+1);
  T(n,2) = 28^n - (6*n+1)*7^n + C(6*n+1,2);
  T(n,3) = 84^n - (6*n+1)*28^n + C(6*n+1,2)*7^n + C(6*n+1,3);
  T(n,4) = 210^n - (6*n+1)*84^n + C(6*n+1,2)*28^n - C(6*n+1,3)*7^n + C(6*n+1,4).
Triangle T(n,k) begins:
 1;
 1, 36, 225, 400, 225, 36, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1;
 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1;
 1, 16776, 16689816, 3656408776, 286691702976, 10255094095176, 192698692565176, 2080037792142216, 13690633212385551, 57229721552316976, 156200093827061616, 283397584598631216, 345271537321293856, 283397584598631216, 156200093827061616, 57229721552316976,13690633212385551, 2080037792142216, 192698692565176, 10255094095176, 286691702976, 3656408776, 16689816, 16776, 1;
...
Example:
Sum_{i=1..n} C(5+i,6)^2 = A086027(n) = C(n+6,13) + 36*C(n+7,13) + 225*C(n+8,13) + 400*C(n+9,13) + 225*C(n+10,13) + 36*C(n+11,13) + C(n+12,13).
binomial(n,6)^2 = C(n,12) + 36*C(n+1,12) + 225*C(n+2,12) + 400*C(n+3,12) + 225*C(n+4,12) + 36*C(n+5,12) + C(n+6,12).

G. C. Greubel, Table of n, a(n) for the first 25 rows, flattened
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. 12.

Columns k=2..6 are A151651, A151652, A151653, A151654, A151655.
Row sums are A248814.
Similar triangles for e=1..5: A173018 (or A008292), A154283, A174266, A236463, A237202.
Sum_{i=1..n} binomial(5+i,6)^p for p=1..3 gives: A000580, A086027, A086028.
Cf. A087127, A086023, A086024, A086025, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 6, p_] := Sum[(-1)^i*Binomial[6*p+1, i]*Binomial[k-i, 6]^p /. k -> 6+i, {i, 0, k-6}]; row[p_] := Table[b[k, 6, p], {k, 6, 6*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(6*n+1, i)*binomial(k+6-i, 6)^n)} \\ Andrew Howroyd, May 06 2020

Sum_{i=1..n} binomial(5+i,6)^p = Sum{k=0..6*(p-1)} T(p,k) * binomial(n+6+k, 6*p+1).

binomial(n,6)^p = Sum_{k=0..6*(p-1)} T(p,k) * binomial(n+k, 6*p).

Edited by Andrew Howroyd, May 06 2020

A087108 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,4). The p-th row (p>=1) contains a(i,p) for i=1 to 4p-3, where a(i,p) satisfies Sum_{i=1..n} C(i+3,4)^p = 5 C(n+4,5) * Sum_{i=1..4p-3} a(i,p) C(n-1,i-1)/(i+4).

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 124, 3126, 33124, 191251, 681000, 1596120, 2543520, 2780820, 2058000, 987000, 277200, 34650, 1, 624, 49376, 1350624, 18308751, 146500500, 763418870, 2749648020, 7101675070, 13440210000

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+4,4)^p of degree 4*p in terms of falling factorials: C(x+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,24,176,...,70, so Sum_{i=1..n} C(i+3,4)^3 = 5 * C(n+4,5) * [ a(1,3)/5 + a(2,3)*C(n-1,1)/6 + a(3,3)*C(n-1,2)/7 + ... + a(9,3)*C(n-1,8)/13 ] = 5 * C(n+4,5) * [ 1/5 + 24*C(n-1,1)/6 + 176*C(n-1,2)/7 + ... + 70*C(n-1,8)/13 ]. Cf. A086024 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
  n = 0 | 1
  n = 1 | 1   4    6     4      1
  n = 2 | 1  24  176   624   1251   1500    1070  420  70
  n = 3 | 1 124 3126 33124 191251 681000 1596120 ...
  ...
Row 2: C(i+4,4)^2 = C(i,0) + 24*C(i,1) + 176*C(i,2) + 624*C(i,3) + 1251*C(i,4) + 1500*C(i,5) + 1070*C(i,6) + 420*C(i,7) + 70*C(i,8). Hence, Sum_{i = 0..n-1} C(i+4,4)^2 =  C(n,1) + 24*C(n,2) + 176*C(n,3) + 624*C(n,4) + 1251*C(n,5) + 1500*C(n,6) + 1070*C(n,7) + 420*C(n,8) + 70*C(n,9) .(End)

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

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A236463.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+4, 4)^n, i = 0..k), k = 0..4*n), n = 0..6); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 5, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 4, 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, 4*p - 3}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 5, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 4, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 4*p-3, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+5, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+4, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+4,4)^n. Equivalently, let v_n denote the sequence (1, 5^n, 15^n, 35^n, ...) regarded as an infinite column vector, where 1, 5, 15, 35, ... is the sequence binomial(n+4,4) - see A000332. 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.
Recurrence: T(n+1,k) = C(k+4,4)*T(n,k) + 4*C(k+3,4)*T(n,k-1) + 6*C(k+2,4)*T(n,k-2) + 4*C(k+1,4)*T(n,k-3) + C(k,4)*T(n,k-4) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 4*n.
n-th row polynomial R(n,x) = (1 + x)^4 o (1 + x)^4 o ... o (1 + x)^4 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n,x) = Sum_{i >= 0} binomial(i+4,4)^n*x^i/(1 + x)^(i+1).
R(n+1,x) = 1/4! * (1 + x)^4 * (d/dx)^4(x^4*R(n,x)).
(1 - x)^(4*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A236463. (End)

Edited by Dean Hickerson, Aug 16 2003

A087109 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,5). The p-th row (p>=1) contains a(i,p) for i=1 to 5p-4, where a(i,p) satisfies Sum_{i=1..n} C(i+4,5)^p = 6 C(n+5,6) * Sum_{i=1..5p-4} a(i,p) C(n-1,i-1)/(i+5).

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 215, 8830, 148480, 1352615, 7665757, 29224020, 78518790, 152794740, 218270220, 229279512, 175227360, 94864770, 34504470, 7567560, 756756, 1, 1295, 191890

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+5,5)^p of degree 5*p in terms of falling factorials: C(x+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,35,370,...,252, so Sum_{i=1..n} C(i+4,5)^3 = 6 * C(n+5,6) * [ a(1,3)/6 + a(2,3)*C(n-1,1)/7 + a(3,3)*C(n-1,2)/8 + ... + a(11,3)*C(n-1,10)/16 ] = 6 * C(n+5,6) * [ 1/6 + 35*C(n-1,1)/7 + 370*C(n-1,2)/8 + ... + 252*C(n-1,10)/16 ]. Cf. A086026 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
1
1  5  10   10    5     1
1 35 370 1920 5835 11253 14240 11830 6230 1890 252
...
Row 2: C(i+5,5)^2 = C(i,0) + 35*C(i,1) + 370*C(i,2) + 1920*C(i,3) + 5835*C(i,4) + 11253*C(i,5) + 14240*C(i,6) + 11830*C(i,7) + 6230*C(i,8) + 1890*C(i,9) + 252*C(i,10). Hence, Sum_{i = 0..n-1} C(i+5,5)^2 = C(n,1) + 35*C(n,2) + 370*C(n,3) + 1920*C(n,4) + 5835*C(n,5) + 11253*C(n,6) + 14240*C(n,7) + 11830*C(n,8) + 6230*C(n,9) + 1890*C(n,10) + 252*C(n,11). (End)

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

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237202.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+5, 5)^n, i = 0..k), k = 0..5*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 6, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 5, 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, 5*p - 4}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 6, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 5, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 5*p-4, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+6, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+5, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+5,5)^n. Equivalently, let v_n denote the sequence (1, 6^n, 21^n, 56^n, ...) regarded as an infinite column vector, where 1, 6, 21, 56, ... is the sequence binomial(n+5,5) - see A000389. 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.
Recurrence: T(n+1,k) = Sum_{i = 0..5} C(5,i)*C(k+5-i,5)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 5*n.
n-th row polynomial R(n,x) = (1 + x)^5 o (1 + x)^5 o ... o (1 + x)^5 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/5!*(1 + x)^5 * (d/dx)^5(x^5*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+5,5)^n*x^i/(1 + x)^(i+1).
(1 - x)^(5*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237202. (End)

Edited by Dean Hickerson, Aug 16 2003

A087110 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,6). The p-th row (p>=1) contains a(i,p) for i=1 to 6p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 C(n+6,7) * Sum_{i=1..6p-5} a(i,p) C(n-1,i-1)/(i+6).

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 342, 21267, 527876, 7020525, 58015362, 324610399, 1297791264, 3839203452, 8595153000, 14760228672, 19560928464, 19987430694

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+6,6)^p of degree 6*p in terms of falling factorials: C(x+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,48,687,...,924, so Sum_{i=1..n} C(i+5,6)^3 = 7 * C(n+6,7) * [ a(1,3)/7 + a(2,3)*C(n-1,1)/8 + a(3,3)*C(n-1,2)/9 + ... + a(13,3)*C(n-1,12)/19 ] = 7 * C(n+6,7) * [ 1/7 + 48*C(n-1,1)/8 + 687*C(n-1,2)/9 + ... + 924*C(n-1,12)/19 ]. Cf. A086028 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237252.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+6, 6)^n, i = 0..k), k = 0..6*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 7, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 6, 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, 6*p - 5}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 7, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 6, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 6*p-5, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+7, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+6, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+6,6)^n. Equivalently, let v_n denote the sequence (1, 7^n, 28^n, 84^n, ...) regarded as an infinite column vector, where 1, 7, 28, 84, ... is the sequence binomial(n+6,6) - see A000579. 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.
Recurrence: T(n+1,k) = Sum_{i = 0..6} C(6,i)*C(k+6-i,6)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 6*n.
n-th row polynomial R(n,x) = (1 + x)^6 o (1 + x)^6 o ... o (1 + x)^6 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).
(1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

Edited by Dean Hickerson, Aug 16 2003

cp's OEIS Frontend

A085440 a(n) = Sum_{i=1..n} binomial(i+1,2)^5.

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A085441 a(n) = Sum_{i=1..n} binomial(i+1,2)^6.

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A086022 a(n) = Sum_{i=1..n} C(i+2,3)^4.

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A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6*n+1,i) * binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

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A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).