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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A181147 a(n) = Sum_{k=0..n-1} (105k+44)*C(2k,k)^2*T(k)*(-1)^(n-1-k)/(2n*C(2n,n)), where T(k) (k=0,1,2,...) are central trinomial coefficients given by A002426.

Original entry on oeis.org

11, 23, 224, 1747, 16754, 162392, 1651206, 17126327, 181182446, 1943132842, 21080299228, 230802972664, 2546569337336, 28280754214358, 315824396838386, 3544003431783795, 39936833763112790, 451718158386620678
Offset: 1

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Author

Zhi-Wei Sun, Jan 24 2011

Keywords

Comments

On Jan 22 2011, Zhi-Wei Sun conjectured that a(n) is a positive integer for every n=1,2,3,... Let p > 3 be a prime. He also conjectured that a(p) == 5 + 6*(p/3)*(2-3^(p-1)) (mod p^2). Another conjecture of his states that Sum_{k=0..p-1} (-1)^k*binomial(2k,k)^2*T(k) is congruent to b(p) modulo p^2, where b(p)=0 if (p/15)=-1, b(p) = 4x^2-2p if p == 1,4 (mod 15) and p = x^2+15y^2 with x,y integers, and b(p) = 20x^2-2p if p == 2,8 (mod 15) and p=5x^2+3y^2 with x,y integers.

Examples

			For n=2 we have a(2) = (44*1^2*T(0)(-1) + (105+44)*2^2*T(1))/(2*2*binomial(4,2)) = 23.
		

Crossrefs

Programs

  • Maple
    T:= n-> `if`(n=0, 1, coeff ((x^2+x+1)^n, x, n)):
    a:= n-> add ((105*k+44) *binomial (2*k, k)^2 *T(k)*(-1)^(n-1-k),
                 k=0..n-1)/ (2*n*binomial (2*n,  n)):
    seq (a(n), n=1..30);
  • Mathematica
    T[k_]:=If[k>0,Coefficient[(x^2+x+1)^k,x^k],1]
    A[n_]:=Sum[(105k+44)Binomial[2k,k]^2*T[k](-1)^(n-1-k),{k,0,n-1}]/(2n*Binomial[2n,n])
    Table[A[n],{n,1,50}]

A179100 a(n) = (1/n) * Sum_{k=0..n-1} (8k+5) T_k^2, where T_0, T_1, ... are central trinomial coefficients given by A002426.

Original entry on oeis.org

5, 9, 69, 407, 2997, 22005, 169389, 1325889, 10573677, 85386881, 697013325, 5739021051, 47599593941, 397234035333, 3332690347437, 28089543969855, 237711099004461, 2018856328439841, 17200553934626253, 146966002696538271
Offset: 1

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Author

Zhi-Wei Sun, Jun 29 2010

Keywords

Comments

On Jun 17 2010, Zhi-Wei Sun conjectured that a(n) is an integer for every n=1,2,3,... and that a(p) == 3(p/3) (mod p) for any prime p, where (p/3) is the Legendre symbol. He also observed that Sum_{k=0..n-1} (2k+1) T_k*3^{n-1-k} = n * Sum_{k=0..n-1} C(n-1,k)*(-1)^(n-1-k)*(k+1)*C(2k,k).

Examples

			For n=3 we have a(3) = (5*T_0^2 + 13*T_1^2 + 21*T_2^2)/3 = (5 + 13 + 21*9)/3 = 69.
		

Crossrefs

Programs

  • Mathematica
    TT[n_]:=Sum[Binomial[n,2k]Binomial[2k,k],{k,0,Floor[n/2]}] SS[n_]:=Sum[(8k+5)*TT[k]^2,{k,0,n-1}]/n Table[SS[n],{n,1,50}]

A179537 a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(n-k,k)^2*(-16)^k.

Original entry on oeis.org

1, 1, -63, -575, 6913, 224001, 420801, -69020223, -918270975, 14596918273, 511845045697, 336721812417, -198449271643391, -2498857696947455, 51614254703660481, 1666776235855331265, -1588877076116525055
Offset: 0

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Author

Zhi-Wei Sun, Jul 18 2010

Keywords

Comments

On July 17, 2010 Zhi-Wei Sun introduced this sequence and made the following conjecture: If p is a prime with (p/7)=1 and p=x^2+7y^2 with x,y integers, then sum_{k=0}^{p-1}(-1)^k*a(k)=4x^2-2p (mod p^2); if p is a prime with (p/7)=-1, then sum_{k=0}^{p-1}(-1)^k*a(k)=0 (mod p^2). He also conjectured that sum_{k=0}^{n-1}(42k+37)(-1)^k*a(k)=0 (mod n) for all n=1,2,3,... and that sum_{k=0}^{p-1}(42k+37)(-1)^k*a(k)=p(21(p/7)+16) (mod p^2) for any prime p.

Examples

			For n=2 we have a(2)=1+2^2*(-16)=-63.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[Binomial[n,k]^2Binomial[n-k,k]^2*(-16)^k,{k,0,n}] Table[a[n],{n,0,25}]
Previous Showing 11-13 of 13 results.