A024166 -id:A024166 - cp's OEIS frontend

Previous Showing 21-30 of 51 results. Next

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

A101094 a(n) = n(n+1)(n+2)(n+3)(1+3*n+n^2)/120.

Original entry on oeis.org

1, 11, 57, 203, 574, 1386, 2982, 5874, 10791, 18733, 31031, 49413, 76076, 113764, 165852, 236436, 330429, 453663, 612997, 816431, 1073226, 1394030, 1791010, 2277990, 2870595, 3586401, 4445091, 5468617, 6681368, 8110344, 9785336

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Partial sums of A024166. Third partial sums of cubes (A000578).

Antidiagonal sums of the array A213564. - Clark Kimberling, Jun 18 2012

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
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.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000537, A101097, A101102, A101104.

[n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 : n in [1..35]]; // Vincenzo Librandi, Apr 23 2015

Table[n*(n + 1)*(n + 2)*(n + 3)*(1 + 3*n + n^2)/120, {n, 31}] (* Michael De Vlieger, Apr 20 2015 *)

[n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 for n in range(1,32)] # Danny Rorabaugh, Apr 20 2015

This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k)+(k-1)*k/6)/((k+3)!/6) at k=3. - Alexander R. Povolotsky, May 17 2008
G.f.: -x*(1+4*x+x^2) / (x-1)^7. - R. J. Mathar, Dec 06 2011
Sum_{n>0} 1/a(n) = (8/3)*(25-9*sqrt(5)*Pi*tan(sqrt(5)*Pi/2)). - Enrique Pérez Herrero, Dec 02 2014
a(k) = MagicNKZ(3,k,7) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n. (Cf. A101104.) - Danny Rorabaugh, Apr 23 2015

Edited by Ralf Stephan, Dec 16 2004

A101097 a(n) = n(n+1)(n+2)(n+3)(n+4)(2 + 4n + n^2)/840.

Original entry on oeis.org

1, 12, 69, 272, 846, 2232, 5214, 11088, 21879, 40612, 71643, 121056, 197132, 310896, 476748, 713184, 1043613, 1497276, 2110273, 2926704, 3999930, 5393960, 7184970, 9462960, 12333555, 15919956, 20365047, 25833664, 32515032, 40625376, 50410712

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Fourth partial sums of cubes (A000578). Partial sums of A101094.

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. 15.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [broken link]
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000217, A101102, A101094, A024166, A000537.

A000217:=func; [&+[A000217(k)^2*A000217(n-k+1): k in [1..n]]: n in [1..40]]; // Bruno Berselli, Sep 04 2013

Mathematica

Table[Binomial[n+4,5]*(2+4*n+n^2)/7, {n,0,50}] (* G. C. Greubel, Feb 17 2017 *)

PARI

{A101097(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(2+4*n+n^2)/840} \\ R. J. Mathar, Dec 06 2011

Sage

[binomial(n+4,5)*(2+4*n+n^2)/7 for n in (1..40)] # G. C. Greubel, Dec 01 2018

Formula

a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) for k=4. - Alexander R. Povolotsky, May 17 2008

G.f.: x*(1 + 4*x + x^2)/(1-x)^8. - R. J. Mathar, Jun 13 2008

a(n) = Sum_{k=1..n} A000217(k)^2*A000217(n-k+1). - Bruno Berselli, Sep 04 2013

E.g.f.: x*(840 + 4200*x + 5040*x^2 + 2240*x^3 + 427*x^4 + 35*x^5 + x^6) *exp(x)/840. - G. C. Greubel, Dec 01 2018

Extensions

Edited by Ralf Stephan, Dec 16 2004

A101102 Fifth partial sums of cubes (A000578).

Original entry on oeis.org

1, 13, 82, 354, 1200, 3432, 8646, 19734, 41613, 82225, 153868, 274924, 472056, 782952, 1259700, 1972884, 3016497, 4513773, 6624046, 9550750, 13550680, 18944640, 26129610, 35592570, 47926125, 63846081, 84211128, 110044792, 142559824, 183185200, 233595912

Offset: 1

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Keywords

easy
nonn

Links

Colin Barker, 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. 15.

Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Archive link]

Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]

Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Crossrefs

Partial sums of A101097.

Cf. A000537, A024166, A101094.

Programs

Magma

[Binomial(n+5,6)*(3*n^2+15*n+10)/28: n in [1..30]]; // G. C. Greubel, Dec 01 2018

Mathematica

Table[Binomial[n+5,6]*(3*n^2+15*n+10)/28, {n,1,30}] (* G. C. Greubel, Dec 01 2018 *) Nest[Accumulate,Range[40]^3,5] (* Harvey P. Dale, Feb 06 2023 *)

PARI

a(n)=sum(t=1,n,sum(s=1,t,sum(l=1,s,sum(j=1,l, 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

PARI

Vec(-x*(x^2+4*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Apr 23 2015

PARI

a(n) = binomial(n+5,6)*(3*n^2+15*n+10)/28 \\ Charles R Greathouse IV, Apr 23 2015

Sage

[binomial(n+5,6)*(3*n^2+15*n+10)/28 for n in (1..30)] # G. C. Greubel, Dec 01 2018

Formula

a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(10 + 3*n*(n+5))/20160.

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=5. - Alexander R. Povolotsky, May 17 2008

G.f.: x*(x^2+4*x+1) / (1-x)^9. - Colin Barker, Apr 23 2015

Sum_{n>=1} 1/a(n) = -162*sqrt(21/5)*Pi*tan(sqrt(35/3)*Pi/2) - 136269/100. - Amiram Eldar, Jan 26 2022

Extensions

Edited by Ralf Stephan, Dec 16 2004

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 3*p-2, where a(i,p) satisfies Sum_{i=1..n} C(i+2,3)^p = 4 * C(n+3,4) * Sum_{i=1..3*p-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

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

André F. Labossière, Aug 11 2003

Keywords

easy
nonn
tabf

Comments

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)

Examples

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)

Links

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.

Crossrefs

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.

Programs

Maple

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

Mathematica

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 *)

PARI

{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

Formula

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)

Extensions

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 7*p-6, where a(i,p) satisfies Sum_{i=1..n} C(i+6,7)^p = 8 * C(n+7,8) * Sum_{i=1..7*p-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

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

André F. Labossière, Aug 11 2003

Keywords

easy
nonn
tabf

Comments

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)

Examples

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.

Links

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.

Crossrefs

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.

Programs

Maple

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

Mathematica

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 *)

PARI

{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

Formula

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)

Extensions

Edited by Dean Hickerson, Aug 16 2003

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 4*p-3, where a(i,p) satisfies Sum_{i=1..n} C(i+3,4)^p = 5 * C(n+4,5) * Sum_{i=1..4*p-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

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

André F. Labossière, Aug 11 2003

Keywords

easy
nonn
tabf

Comments

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)

Examples

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)

Links

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.

Crossrefs

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.

Programs

Maple

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

Mathematica

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 *)

PARI

{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

Formula

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)

Extensions

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 5*p-4, where a(i,p) satisfies Sum_{i=1..n} C(i+4,5)^p = 6 * C(n+5,6) * Sum_{i=1..5*p-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

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

André F. Labossière, Aug 11 2003

Keywords

easy
nonn
tabf

Comments

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)

Examples

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)

Links

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.

Crossrefs

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.

Programs

Maple

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

Mathematica

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 *)

PARI

{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

Formula

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)

Extensions

Edited by Dean Hickerson, Aug 16 2003

Previous Showing 21-30 of 51 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.
Content is available under The OEIS End-User License Agreement.

cp's OEIS Frontend

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

Views

Author

Keywords

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A101094 a(n) = n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A101097 a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(2 + 4*n + n^2)/840.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

Extensions

A101102 Fifth partial sums of cubes (A000578).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

PARI

Sage

Formula

A101094 a(n) = n(n+1)(n+2)(n+3)(1+3*n+n^2)/120.

A101097 a(n) = n(n+1)(n+2)(n+3)(n+4)(2 + 4n + n^2)/840.