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.

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A361716 a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)^2*binomial(n-1,k).

Original entry on oeis.org

0, 1, -3, -8, 45, 126, -840, -2400, 17325, 50050, -378378, -1100736, 8576568, 25069968, -199536480, -585307008, 4732755885, 13919870250, -113936715750, -335813478000, 2775498395670, 8194328596740, -68263497731520, -201822515032320
Offset: 0

Views

Author

Peter Bala, Mar 23 2023

Keywords

Comments

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integers n and k.

Links

Crossrefs

Cf. A006480, A245086, A361710, A361711.

Programs

  • Maple
    seq(add((-1)^k*binomial(n,k)^2*binomial(n-1,k), k = 0..n-1), n = 0..20);
  • Mathematica
    A361716[n_]:=Sum[(-1)^k*Binomial[n,k]^2Binomial[n-1,k],{k,0,n-1}];Array[A361716,30,0] (* Paolo Xausa, Oct 06 2023 *)
  • PARI
    a(n) = sum(k = 0, n-1, (-1)^k*binomial(n,k)^2*binomial(n-1,k)); \\ Michel Marcus, Mar 26 2023
  • Python
    from math import comb
def A361716(n): return (sum(comb(n,k)**3*k if k&1 else -comb(n,k)**3*k for k in range(n+1)))//(n if n&1 else -n) if n else 0 # Chai Wah Wu, Mar 27 2023

Formula

a(n) = Sum_{k = 0..n} (-1)^(n+k) * (k/n) * binomial(n,k)^3.
a(2*n) = (-1)^n * (1/2) * (3*n)!/n!^3 for n >= 1; a(2*n+1) = (-1)^n * (3*n+1)/(2*n+1) * (3n)!/n!^3.
a(2*n) = (1/2)*A245086(2*n) = (1/2)*(-1)^n*A006480(n) for n >= 1.
a(2*n+1) = A361710(2*n+1) = A361711(2*n+1).
a(n) = hypergeom([1 - n, - n, - n], [1, 1], 1) for n >= 1.
P-recursive: n^2*(n-1)*(6*n^2-20*n+17)*a(n) = -( 6*(3*n^2-6*n+2)*(n-1)*a(n-1) + (3*n-6)*(3*n-5)*(3*n-4)*(6*n^2-8*n+3)*a(n-2) ).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(3*n,n-k)*binomial(n+k,k)^2 for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n+k,n)*binomial(2*n-k-1,n). - Peter Bala, Sep 13 2023

A361710 a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n-1,k)^2.

Original entry on oeis.org

0, 1, -1, -8, 15, 126, -280, -2400, 5775, 50050, -126126, -1100736, 2858856, 25069968, -66512160, -585307008, 1577585295, 13919870250, -37978905250, -335813478000, 925166131890, 8194328596740, -22754499243840, -201822515032320, 564121960420200, 5009403008531376
Offset: 0

Views

Author

Peter Bala, Mar 21 2023

Keywords

Comments

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integers n and k.
Compare with A005258(n-1) = Sum_{k = 0..n-1} (-1)^k*binomial(-n,k)*binomial(n-1,k)^2.

Links

Crossrefs

Cf. A005258, A006480, A245086, A361711, A361716.

Programs

  • Maple
    seq( add((-1)^k*binomial(n,k)*binomial(n-1,k)^2, k = 0..n-1), n = 0..25);
  • PARI
    a(n) = sum(k = 0, n-1, (-1)^k*binomial(n,k)*binomial(n-1,k)^2); \\ Michel Marcus, Mar 26 2023

Formula

a(n) = (1/n^2) * Sum_{k = 0..n} (-1)^(n+k) * k^2 * binomial(n,k)^3 for n >= 1.
a(n) = (1/(3*n)) * Sum_{k = 0..n} (-1)^(n+k+1) * (n - 3*k) * binomial(n,k)^3 for n >= 1.
a(2*n) = (-1)^n * (1/6) * (3*n)!/n!^3 for n >= 1; a(2*n+1) = (-1)^n * (3*n+1)/(2*n+1) * (3*n)!/n!^3.
a(2*n) = (1/3)*A361716(2*n); a(2*n+1) = A361711(2*n+1) = A361716(2*n+1).
a(2*n) = (1/6)*A245086(2*n) = (1/6)*(-1)^n*A006480(n) for n >= 1.
a(n) = hypergeom([-n, 1 - n, 1 - n], [1, 1], 1);
P-recursive: n^2*(n - 1)*(6*n^2 - 16*n + 11)*a(n) = - 6*(n - 1)*(3*n^2 - 6*n + 2)*a(n-1) - (3*n - 4)*(3*n - 5)*(3*n - 6)*(6*n^2 - 4*n + 1)*a(n-2) with a(0) = 0 and a(1) = 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n+k-1,k)*binomial(2*n-k-1,n). - Peter Bala, Jul 01 2023
Showing 1-2 of 2 results.

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