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.

A112028 a(n) = Sum_{k=0..n} binomial(n+k,k)^3.

Original entry on oeis.org

1, 9, 244, 9065, 389376, 18188478, 897376152, 46011772521, 2427553965160, 130930630643384, 7186614533569296, 400132290102421214, 22543708920891189136, 1282873288801683197250, 73628947696550668509744, 4257138240245923453355625, 247733479854085081062353400
Offset: 0

Views

Author

N. J. A. Sloane, Nov 28 2005

Keywords

References

  • M.J. Coster: Supercongruences, [Thesis] Univ. of Leiden, the Netherlands, 1988.

Links

Crossrefs

Cf. A001700, A112029, A219562, A219563, A219564.
A375178 is an essentially identical sequence.

Programs

  • Maple
    y2 := hypergeom([2/3, 2/3],[4/3],-x^2/27)*x^(1/3);
h := hypergeom([1/4, 1/4],[1],64*x)^2;
H := (243+247*x)*x*diff(h,x,x) + (463*x+234)*diff(h,x) + (48-24/x)*h;
ogf := y2*Int(Int(y2*H,x)/(8*x*(x^2+27)*y2^2),x);
# Check ogf by computing a series expansion:
SER := proc(a, x) series(a, x, 20) end:
INT := proc(a, x) int(SER(a, x), x) end:
SER(eval(ogf, Int = INT), x); # Mark van Hoeij, Apr 04 2013
  • Mathematica
    Table[Sum[Binomial[n+k,k]^3,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Sep 24 2012 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n+k, k)^3); \\ Michel Marcus, Mar 09 2016

Formula

a(n) ~ 2^(6*n+3)/(7*(Pi*n)^(3/2)). - Vaclav Kotesovec, Nov 23 2012
Recurrence: 3*(3*n-1)*(3*n+1)*(15799*n^5 - 103177*n^4 + 265789*n^3 - 336367*n^2 + 208000*n - 49852)*n^3*a(n) = 24*(2*n-3)^3*(3*n-4)*(3*n-2)*(15799*n^5 - 24182*n^4 + 11071*n^3 - 72*n^2 - 1080*n + 192)*a(n-1) - (n-1)^2*(15799*n^5 - 103177*n^4 + 265789*n^3 - 336367*n^2 + 208000*n - 49852)*n^3*a(n-2) + 8*(n-2)^2*(2*n-3)^3*(15799*n^5 - 24182*n^4 + 11071*n^3 - 72*n^2 - 1080*n + 192)*a(n-3). - Vaclav Kotesovec, Nov 23 2012
O.g.f. can be expressed in terms of hypergeometric functions (see Maple program). - Mark van Hoeij, Apr 01 2013
From Peter Bala, Mar 29 2023: (Start)
The supercongruence a(p-1) == 1 (mod p^5) appears to hold for all primes p >= 7 (checked up to p = 199). Coster, Theorem 4, proves that a(p-1) == 1 (mod p^3) for primes p >= 5.
For r >= 2, the supercongruence a(p^r - 1) == a(p^(r-1) - 1) (mod p^(3*r+3)) may hold for all primes p >= 7. (End)

A375179 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k)^3.

Original entry on oeis.org

0, 1, 63, 3160, 154175, 7623126, 385867944, 20012582304, 1061140695615, 57352579248250, 3150527165580938, 175455318597459264, 9885323482971453608, 562487110567071566800, 32279596239888192636000, 1866149091367606758188160, 108583784076103277424118335, 6354063452012073380091209250
Offset: 0

Views

Author

Peter Bala, Aug 04 2024

Keywords

Comments

Compare with the identity Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k) = (1/2) * binomial(2*n, n) = (1/2) * A000984(n) for n >= 1.
The central binomial coefficients satisfy the supercongruence (1/2) * binomial(2*p, p) == 1 (mod p^3) for all primes p >= 5 (Wolstenholme's theorem).
We conjecture that for the present sequence the stronger supercongruence a(p) == 1 (mod p^5) holds for all primes p >= 7.
Further, we conjecture that for r >= 2 and prime p >= 5, a(p^r) == a(p^(r-1)) (mod p^(3*r+3)).
More generally, for a positive integer m, define a sequence {b_m(n) : n >= 0} by setting b_m(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k)^(2*m+1). We conjecture that the supercongruence b_m(p) == 1 (mod p^(2*m+3)) holds for all primes p >= 2*m + 5, and for r >= 2, the supercongruence b_m(p^r) == b_m(p^(r-1)) (mod p^(3*r+2*m+1)) also holds for all primes p >= 2*m + 5.

Examples

			Examples of supercongruences:
a(7) - a(1) = 20012582304 - 1 = (7^5)*389*3061 == 0 (mod 7^5);
a(17) - a(1) = 6354063452012073380091209250 - 1 = (17^5)*6007*15725921* 47373254231 == 0 (mod 17^5).

Links

Crossrefs

Cf. A000984, A375178, A375180.

Programs

  • Maple
    seq(add( (-1)^(n+k+1)*binomial(2*n, k)^3, k = 0..n-1), n = 0..20);

Formula

a(n) ~ 2^(6*n-1) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Aug 08 2024

A375180 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(3*n, k)^3.

Original entry on oeis.org

0, 1, 215, 45928, 10362231, 2450260001, 600869373182, 151570671244560, 39096342054496887, 10267275084850974619, 2736324289110748127715, 738255282011665067114400, 201254884472471159485086750, 55352399437924814524429123488, 15341068552569688728602977821596
Offset: 0

Views

Author

Peter Bala, Aug 05 2024

Keywords

Comments

Compare with the identity Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(3*n, k) = (1/3) * binomial(3*n, n) = (1/3) * A005809(n) for n >= 1.
The binomial coefficients satisfy the supercongruence (1/3) * binomial(3*p, p) == 1 (mod p^3) for all primes p >= 5 (Meštrović, Equation 35).
We conjecture that for the present sequence the stronger supercongruence a(p) == 1 (mod p^5) holds for all primes p >= 7.
Further, we conjecture that for r >= 2 and prime p >= 5, a(p^r) == a(p^(r-1)) (mod p^(3*r+3)).
More generally, for a positive integer m, define a sequence {b_m(n) : n >= 0} by setting b_m(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(3*n, k)^(2*m+1). We conjecture that the supercongruence b_m(p) == 1 (mod p^(2*m+3)) holds for all primes p >= 2*m + 5, and for r >= 2, the supercongruence b_m(p^r) == b_m(p^(r-1)) (mod p^(3*r+2*m+1)) also holds for all primes p >= 2*m + 5.

Examples

			Examples of supercongruences:
a(7) - a(1) = 151570671244560 - 1 = (7^5)*379*2269*10487 == 0 (mod 7^5);
a(13) - a(1) = 55352399437924814524429123488 - 1 = (13^5)*149080105032749915900459 == 0 (mod 13^5).

Links

Crossrefs

Cf. A005809, A375178, A375179.

Programs

  • Maple
    seq(add( (-1)^(n+k+1)*binomial(3*n, k)^3, k = 0..n-1), n = 0..20);

Formula

a(n) ~ 3^(9*n - 1/2) / (Pi^(3/2) * n^(3/2) * 2^(6*n+3)). - Vaclav Kotesovec, Aug 08 2024
Showing 1-3 of 3 results.

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