cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Nikita D. Gogin

Nikita D. Gogin's wiki page.

Nikita D. Gogin has authored 3 sequences.

A329913 The fifth moments of the squared binomial coefficients; a(n) = Sum_{m=0..n} m^5*binomial(n, m)^2.

Original entry on oeis.org

0, 1, 36, 540, 6080, 56250, 455112, 3342192, 22809600, 146988270, 904475000, 5358254616, 30750385536, 171773279860, 937514244240, 5014575000000, 26351064760320, 136319273714070, 695429503781400, 3503580441563400, 17452918098000000, 86055711108818220
Offset: 0

Author

Nikita D. Gogin, Nov 24 2019

Keywords

References

  • H. W. Gould, Combinatorial Identities, 1972. (See formulas 3.77, 3.78, and 3.79 on page 31.)

Programs

  • Magma
    [(&+[Binomial(n,k)^2*k^5: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jun 23 2022
    
  • Maple
    seq( binomial(2*n,n)*n^4*(n^3 + 3*n^2 - 3*n - 5)/((16*n-8)*(2*n-3)),n=0..30); # Robert Israel, Jan 26 2020
  • Mathematica
    Table[Sum[m^5*(Binomial[n, m])^2, {m, 0, n}], {n, 21}]
  • PARI
    a(n) = sum(k=0, n, k^5*binomial(n, k)^2); \\ Michel Marcus, Nov 24 2019
    
  • SageMath
    [n^4*(n+1)*(n^3+3*n^2-3*n-5)/(8*(2*n-1)*(2*n-3))*catalan_number(n) for n in (0..30)] # G. C. Greubel, Jun 23 2022

Formula

a(n) = binomial(2*n,n)*n^4*(n^3 + 3*n^2 - 3*n - 5)/(8*(2*n-1)*(2*n-3)).
G.f.: x*(1 + 14*x - 54*x^2 + 404*x^3 - 1544*x^4 + 2880*x^5 - 2160*x^6)/(1-4*x)^(11/2). - Stefano Spezia, Jan 03 2020
(-12960 + 8640*n)*a(n) + (7200 - 13680*n)*a(n + 1) + (3920 + 9056*n)*a(n + 2) + (-4184 - 3160*n)*a(n + 3) + (1404 + 620*n)*a(n + 4) + (-584 - 110*n)*a(n + 5) + (14 + 10*n)*a(n + 6) + (n + 6)*a(n + 7) = 0. - Robert Israel, Jan 26 2020

A329444 The sixth moments of the squared binomial coefficients; a(n) = Sum_{m=0..n} m^6*binomial(n, m)^2.

Original entry on oeis.org

0, 1, 68, 1314, 18080, 197350, 1836792, 15233316, 115776768, 821760390, 5520171800, 35438827996, 219038609088, 1310833221724, 7629754810160, 43348888067400, 241117582878720, 1316197491501510, 7065439665315480, 37362065079691500, 194909773207512000, 1004374157379474420
Offset: 0

Author

Nikita D. Gogin, Nov 16 2019

Keywords

References

  • H. W. Gould, Combinatorial Identities, 1972. (See formulas 3.77, 3.78, and 3.79 on page 31.)

Programs

  • Magma
    [(&+[Binomial(n,k)^2*k^6: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jun 23 2022
    
  • Mathematica
    Table[Sum[m^6*(Binomial[n, m])^2, {m, 0, n}], {n, 21}]
  • PARI
    a(n) = sum(m=0, n, m^6*binomial(n, m)^2); \\ Jinyuan Wang, Nov 23 2019
    
  • SageMath
    [n^3*(n+1)*(n^6+3*n^5-13*n^4-15*n^3+30*n^2+8*n-2)*catalan_number(n)/(8*(2*n-1)*(2*n-3)*(2*n-5)) for n in (0..30)] # G. C. Greubel, Jun 23 2022

Formula

a(n) = binomial(2*n, n) * n^3*(n^6 + 3*n^5 - 13*n^4 - 15*n^3 + 30*n^2 + 8*n - 2)/(8*(2*n-1)*(2*n-3)*(2*n-5)).
G.f.: x*(1 + 42*x - 168*x^2 + 1648*x^3 - 7608*x^4 + 18144*x^5 - 19376*x^6 - 1440*x^7 + 14400*x^8)/((1-4*x)^6*sqrt(1-4*x)). - G. C. Greubel, Jun 23 2022

A329521 The sixth moments of the alternated squared binomial coefficients; a(n) = Sum_{m=0..n} (-1)^m*m^6*binomial(n, m)^2.

Original entry on oeis.org

0, -1, 60, -162, -5280, 20250, 128520, -569380, -1854720, 9338490, 20097000, -113704668, -181621440, 1142905764, 1447926480, -10042461000, -10529925120, 79859881530, 71384175720, -587933314540, -457825368000, 4070529226764
Offset: 0

Author

Nikita D. Gogin, Nov 15 2019

Keywords

References

  • H. W. Gould, Combinatorial Identities, 1972.

Crossrefs

Programs

  • Magma
    [&+[(-1)^m*m^6*(Binomial(n,m))^2:m in [0..n]]:n in [0..21]]; // Marius A. Burtea, Nov 15 2019
    
  • Mathematica
    Table[Sum[(-1)^m*m^6*(Binomial[n, m])^2, {m, 0, n}], {n, 21}]
  • PARI
    a(n) = sum(m=0, n, (-1)^m*m^6*binomial(n , m)^2); \\ Michel Marcus, Nov 15 2019

Formula

a(n) = (-1)^((n+2)/2)*binomial(n, n/2)*(n^3*(n+1)*(3n-1)/4), if n is even,
a(n) = (-1)^((n-1)/2)*binomial(n,((n+1)/2))*(n^2*( n+1)*(n^3+n^2-9n+3)/8), if n is odd.
G.f.: x*(-1 + 60*x - 188*x^2 - 3720*x^3 + 15752*x^4 + 8400*x^5 - 90928*x^6 + 79680*x^7 + 42112*x^8 - 69120*x^9 + 17408*x^10)/(1+4*x^2)^(13/2). - Stefano Spezia, Nov 15 2019