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A337390
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Expansion of sqrt((1-2*x+sqrt(1-12*x+4*x^2)) / (2 * (1-12*x+4*x^2))).
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4
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1, 4, 34, 328, 3334, 34904, 372436, 4027216, 43976774, 483860632, 5355697084, 59569288816, 665238165916, 7454247891952, 83769667651816, 943744775565728, 10655369806377542, 120535523282756632, 1365840013196530348, 15500428304345011504, 176148760580561346484
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OFFSET
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0,2
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LINKS
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FORMULA
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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
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)
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MATHEMATICA
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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 *)
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PROG
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(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))}
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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