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.

A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2*(k+4)*x+((k-4)*x)^2) * (1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) )).

Original entry on oeis.org

1, 1, 6, 1, 7, 30, 1, 8, 51, 140, 1, 9, 74, 393, 630, 1, 10, 99, 736, 3139, 2772, 1, 11, 126, 1175, 7606, 25653, 12012, 1, 12, 155, 1716, 14499, 80464, 212941, 51480, 1, 13, 186, 2365, 24310, 183195, 864772, 1787607, 218790, 1, 14, 219, 3128, 37555, 352716, 2351805, 9400192, 15134931, 923780
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2020

Keywords

Examples

			Square array begins:
     1,     1,     1,      1,      1,      1, ...
     6,     7,     8,      9,     10,     11, ...
    30,    51,    74,     99,    126,    155, ...
   140,   393,   736,   1175,   1716,   2365, ...
   630,  3139,  7606,  14499,  24310,  37555, ...
  2772, 25653, 80464, 183195, 352716, 610897, ...
		

Crossrefs

Columns k=0..5 give A002457, A273055, A337370, A245927, A002458, A243947.
Main diagonal gives A337387.

Programs

  • Mathematica
    T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n + 1, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    {T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n+1, 2*j))}

Formula

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n+1,2*j).
T(0,k) = 1, T(1,k) = k+6 and n * (2*n+1) * (4*n-3) * T(n,k) = (4*n-1) * (4*(k+4)*n^2-2*(k+4)*n-k-2) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-1) * (4*n+1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 29 2020
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 3/2) / sqrt(8*k*Pi*n). - Vaclav Kotesovec, Aug 31 2020

A337390 Expansion of sqrt((1-2*x+sqrt(1-12*x+4*x^2)) / (2 * (1-12*x+4*x^2))).

Original entry on oeis.org

1, 4, 34, 328, 3334, 34904, 372436, 4027216, 43976774, 483860632, 5355697084, 59569288816, 665238165916, 7454247891952, 83769667651816, 943744775565728, 10655369806377542, 120535523282756632, 1365840013196530348, 15500428304345011504, 176148760580561346484
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2020

Keywords

Crossrefs

Column k=2 of A337389.

Programs

  • Mathematica
    a[n_] := Sum[2^(n - k) * Binomial[2*k, k] * Binomial[2*n, 2*k], {k, 0, n}]; Array[a, 21, 0] (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    N=40; x='x+O('x^N); Vec(sqrt((1-2*x+sqrt(1-12*x+4*x^2))/(2*(1-12*x+4*x^2))))
    
  • PARI
    {a(n) = sum(k=0, n, 2^(n-k)*binomial(2*k, k)*binomial(2*n, 2*k))}

Formula

a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*k,k) * binomial(2*n,2*k).
a(0) = 1, a(1) = 4 and n * (2*n-1) * (4*n-5) * a(n) = (4*n-3) * (24*n^2-36*n+8) * a(n-1) - 4 * (n-1) * (2*n-3) * (4*n-1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 28 2020
a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(2*n + 1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 31 2020
From Peter Bala, May 02 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{j = 0..n-2*k} binomial(n,k)*binomial(2*n,j)*binomial(3*n-2*k-j-1,n-2*k-j).
If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 3 and positive integers n and k.
It appears that a(n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(2*n-k)*binomial(2*k,k)^2* binomial(2*n+k,2*k). Compare with the pair of identities: binomial(2*n,n) = Sum_{k = 0..n} 2^(n-2*k)*binomial(2*k,k)*binomial(n,2*k) and binomial(2*n,n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(4*n-2*k)*binomial(2*k,k)^2*binomial(2*n+k,2*k). (End)

A337466 Expansion of sqrt(2 / ( (1-4*x+36*x^2) * (1-6*x+sqrt(1-4*x+36*x^2)) )).

Original entry on oeis.org

1, 4, -6, -120, -266, 2520, 17380, -13104, -599130, -1853544, 12391116, 108252144, 6439356, -3577917200, -14043012984, 65962248352, 730407220998, 602517029400, -22507424996420, -108316306187600, 347406564086868, 5073542740156752, 7904100039294456, -143838603813578400
Offset: 0

Views

Author

Seiichi Manyama, Aug 28 2020

Keywords

Crossrefs

Column k=2 of A337464.

Programs

  • Mathematica
    a[n_] := Sum[(-2)^(n-k) * Binomial[2*k, k] * Binomial[2*n+1, 2*k], {k, 0, n}]; Array[a, 24, 0] (* Amiram Eldar, Apr 29 2021 *)
    CoefficientList[Series[Sqrt[2/((1-4x+36x^2)(1-6x+Sqrt[1-4x+36x^2]))],{x,0,40}],x] (* Harvey P. Dale, Sep 07 2023 *)
  • PARI
    N=40; x='x+O('x^N); Vec(sqrt(2/((1-4*x+36*x^2)*(1-6*x+sqrt(1-4*x+36*x^2)))))
    
  • PARI
    {a(n) = sum(k=0, n, (-2)^(n-k)*binomial(2*k, k)*binomial(2*n+1, 2*k))}

Formula

a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(2*k,k) * binomial(2*n+1,2*k).
a(0) = 1, a(1) = 4 and n * (2*n+1) * (4*n-3) * a(n) = (4*n-1) * (8*n^2-4*n) * a(n-1) - 36 * (n-1) * (2*n-1) * (4*n+1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 29 2020
Showing 1-3 of 3 results.