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.

Showing 1-3 of 3 results.

A361883 a(n) = (1/n) * Sum_{k = 0..n} (n+2*k) * binomial(n+k-1,k)^3.

Original entry on oeis.org

4, 98, 3550, 150722, 6993504, 343542572, 17560824138, 924397069250, 49770307114528, 2728028537409848, 151717661909940724, 8539838104822762220, 485583352521437530000, 27850592121190001279928, 1609345458428168657866050
Offset: 1

Views

Author

Peter Bala, Mar 28 2023

Keywords

Comments

Compare with the closed form evaluation of the binomial sums (1/n) * Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k) * binomial(n+k-1,k) = binomial(2*n,n) and (1/n) * Sum_{k = 0..n} (n + 2*k) * binomial(n+k-1,k)^2 = binomial(2*n,n)^2.
The central binomial coefficients u(n) := binomial(2*n,n) = A000984(n) satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for positive integers n and r and all primes p >= 5. We conjecture that the present sequence satisfies the same congruences.
More generally, for m >= 3, the sequences {b_m(n) : n >= 1} and {c_m(n) : n >= 1} defined by b_m(n) = (1/n) * Sum_{k = 0..n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may satisfy the same congruences.

Crossrefs

Cf. A000984, A002894, A361884, A361885, A361886.

Programs

  • Maple
    seq( (1/n)*add((n + 2*k) * binomial(n+k-1,k)^3, k = 0..n), n = 1..20);
  • Mathematica
    Table[Sum[(3*n - 2*k) * Binomial[2*n-k-1, n-1]^3, {k,0,n}]/n, {n,1,20}] (* Vaclav Kotesovec, Mar 29 2023 *)
  • PARI
    a(n) = (1/n) * sum(k = 0, n, (n+2*k) * binomial(n+k-1,k)^3); \\ Michel Marcus, Mar 30 2023

Formula

a(n) ~ 3 * 2^(6*n) / (7 * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2023

A361885 a(n) = (1/n) * Sum_{k = 0..2*n} (n+2*k) * binomial(n+k-1,k)^3.

Original entry on oeis.org

9, 979, 165816, 33372819, 7380882509, 1732912534168, 424032181044264, 106952563532680339, 27609695174536836075, 7259294757681340436979, 1937215339689731617386000, 523352118643145676922317336, 142854011885066484369862826496, 39337931825265398967484384872560
Offset: 0

Views

Author

Peter Bala, Mar 28 2023

Keywords

Comments

Compare with the closed form evaluation of the binomial sums (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k) = binomial(3*n,n) and (1/n) * Sum_{k = 0..2*n} (n + 2*k) * binomial(n+k-1,k)^2 = binomial(3*n,n)^2.
The binomial coefficients u(n) := binomial(3*n,n) = A005809(n) satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for positive integers n and r and all primes p >= 5. We conjecture that the present sequence satisfies the same congruences.
More generally, for m >= 3, the sequences {b_m(n) : n >= 1} and {c_m(n) : n >= 1} defined by b_m(n) = (1/n) * Sum_{k = 0..2*n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may satisfy the same congruences.

Crossrefs

Cf. A005809, A188662, A361883, A361884, A361886.

Programs

  • Maple
    seq( (1/n)*add((n + 2*k) * binomial(n+k-1,k)^3, k = 0..2*n), n = 1..20);
  • Mathematica
    Table[Sum[(n+2*k) * Binomial[n+k-1,k]^3, {k,0,2*n}]/n, {n,1,20}] (* Vaclav Kotesovec, Mar 29 2023 *)
  • PARI
    a(n) = (1/n) * sum(k = 0, 2*n,  (n+2*k) * binomial(n+k-1,k)^3); \\ Michel Marcus, Mar 30 2023

Formula

a(n) ~ 5 * 3^(9*n + 3/2) / (19 * Pi^(3/2) * n^(3/2) * 2^(6*n + 3)). - Vaclav Kotesovec, Mar 29 2023

A361886 a(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n+2*k) * binomial(n+k-1,k)^3.

Original entry on oeis.org

3, 435, 79464, 16551315, 3732732003, 887492378136, 219081875199120, 55618197870142611, 14429522546341842225, 3808899907812064500435, 1019705941257612879722400, 276212555234100323977483800, 75563424471884688135891640224
Offset: 1

Views

Author

Peter Bala, Mar 28 2023

Keywords

Comments

Compare with the closed form evaluation of the binomial sum (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k) = binomial(3*n,n).
The binomial coefficients u(n) := binomial(3*n,n) = A005809(n) satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for positive integers n and r and all primes p >= 5. We conjecture that the present sequence satisfies the same congruences.
More generally, for m >= 3, the sequences {b_m(n) : n >= 1} and {c_m(n) : n >= 1} defined by b_m(n) = (1/n) * Sum_{k = 0..2*n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may satisfy the same congruences.

Crossrefs

Cf. A005809, A361883, A361884, A361885.

Programs

  • Maple
    seq( (1/n)*add( (-1)^k * (n + 2*k) * binomial(n+k-1,k)^3, k = 0..2*n), n = 1..20);
  • Mathematica
    Table[Sum[(-1)^k * (n+2*k) * Binomial[n+k-1,k]^3, {k,0,2*n}]/n, {n,1,20}] (* Vaclav Kotesovec, Mar 29 2023 *)
  • PARI
    a(n) = (1/n) * sum(k = 0, 2*n, (-1)^k * (n+2*k) * binomial(n+k-1,k)^3); \\ Michel Marcus, Mar 30 2023

Formula

a(n) ~ 3^(9*n + 3/2) / (7 * Pi^(3/2) * n^(3/2) * 2^(6*n + 3)). - Vaclav Kotesovec, Mar 29 2023
Showing 1-3 of 3 results.

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