# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a337390 Showing 1-1 of 1 %I A337390 #33 May 05 2022 08:07:38 %S A337390 1,4,34,328,3334,34904,372436,4027216,43976774,483860632,5355697084, %T A337390 59569288816,665238165916,7454247891952,83769667651816, %U A337390 943744775565728,10655369806377542,120535523282756632,1365840013196530348,15500428304345011504,176148760580561346484 %N A337390 Expansion of sqrt((1-2*x+sqrt(1-12*x+4*x^2)) / (2 * (1-12*x+4*x^2))). %H A337390 Seiichi Manyama, Table of n, a(n) for n = 0..939 %F A337390 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*k,k) * binomial(2*n,2*k). %F A337390 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 %F A337390 a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(2*n + 1/2) / sqrt(Pi*n). - _Vaclav Kotesovec_, Aug 31 2020 %F A337390 From _Peter Bala_, May 02 2022: (Start) %F A337390 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). %F A337390 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. %F A337390 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) %t A337390 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 *) %o A337390 (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)))) %o A337390 (PARI) {a(n) = sum(k=0, n, 2^(n-k)*binomial(2*k, k)*binomial(2*n, 2*k))} %Y A337390 Column k=2 of A337389. %Y A337390 Cf. A337370, A337421. %K A337390 nonn,easy %O A337390 0,2 %A A337390 _Seiichi Manyama_, Aug 25 2020 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE