A337421 -id:A337421 - cp's OEIS frontend

A337419 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1-(k+4)x+sqrt(1+2(k-4)x+((k+4)x)^2)) / (2 * (1+2(k-4)x+((k+4)*x)^2))).

1, 1, 2, 1, 1, 6, 1, 0, -5, 20, 1, -1, -14, -41, 70, 1, -2, -21, -48, -125, 252, 1, -3, -26, -7, 198, 131, 924, 1, -4, -29, 76, 739, 2080, 3301, 3432, 1, -5, -30, 195, 1222, 1629, 1780, 15625, 12870, 1, -6, -29, 344, 1395, -3772, -26859, -57120, 16115, 48620

Offset: 0

Seiichi Manyama, Aug 27 2020

			Square array begins:
    1,    1,    1,    1,     1,      1, ...
    2,    1,    0,   -1,    -2,     -3, ...
    6,   -5,  -14,  -21,   -26,    -29, ...
   20,  -41,  -48,   -7,    76,    195, ...
   70, -125,  198,  739,  1222,   1395, ...
  252,  131, 2080, 1629, -3772, -14873, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..4 give A000984, A337393, A337421, A337422, A337396.
Main diagonal gives A337420.
Cf. A337389, A337464.

T[n_, k_] := Sum[If[k == 0, Boole[n == j], (-k)^(n - j)] * Binomial[2*j, j] * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 27 2020 *)

{T(n, k) = sum(j=0, n, (-k)^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}

T(n,k) = Sum_{j=0..n} (-k)^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).

T(0,k) = 1, T(1,k) = 2-k and n * (2*n-1) * (4*n-5) * T(n,k) = (4*n-3) * (-4*(k-4)*n^2+6*(k-4)*n-k+6) * T(n-1,k) - (k+4)^2 * (n-1) * (2*n-3) * (4*n-1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 28 2020

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

Seiichi Manyama, Aug 25 2020

Seiichi Manyama, Table of n, a(n) for n = 0..939

Column k=2 of A337389.

Cf. A337370, A337421.

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 *)

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))))

{a(n) = sum(k=0, n, 2^(n-k)*binomial(2*k, k)*binomial(2*n, 2*k))}

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 / ( (1-4x+36x^2) * (1-6x+sqrt(1-4x+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

Seiichi Manyama, Aug 28 2020

sign

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Column k=2 of A337464.

Cf. A337370, A337421.

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 *)

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)))))

{a(n) = sum(k=0, n, (-2)^(n-k)*binomial(2*k, k)*binomial(2*n+1, 2*k))}

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

cp's OEIS Frontend

A337419 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1-(k+4)x+sqrt(1+2(k-4)x+((k+4)x)^2)) / (2 * (1+2(k-4)x+((k+4)*x)^2))).

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

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

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cp's OEIS Frontend

A337419 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1-(k+4)*x+sqrt(1+2*(k-4)*x+((k+4)*x)^2)) / (2 * (1+2*(k-4)*x+((k+4)*x)^2))).

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

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

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A337419 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1-(k+4)x+sqrt(1+2(k-4)x+((k+4)x)^2)) / (2 * (1+2(k-4)x+((k+4)*x)^2))).

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

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