A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)) ).
1, 5, 51, 587, 7123, 89055, 1135005, 14660805, 191253843, 2513963567, 33244446601, 441772827105, 5894323986301, 78912561223553, 1059543126891027, 14261959492731387, 192392702881384275, 2600355510685245087, 35206018016510388345, 477377227987055971905
Offset: 0
Examples
G.f.: A(x) = 1 + 5*x + 51*x^2 + 587*x^3 + 7123*x^4 + 89055*x^5 +... where A(x)^2 = (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)). Explicitly, A(x)^2 = 1 + 10*x + 127*x^2 + 1684*x^3 + 22717*x^4 + 309214*x^5 + 4231675*x^6 + 58117672*x^7 + 800173945*x^8 +...+ A245923(n)*x^n +...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..870
Programs
-
Maple
A245926 := n -> sqrt(add(binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2, k=0..2*n)); seq(A245926(n), n=0..20); # Peter Luschny, Aug 17 2014
-
Mathematica
CoefficientList[Series[Sqrt[(1 - x + Sqrt[1 - 14*x + x^2])/(2*(1 - 14*x + x^2))], {x,0,50}], x] (* G. C. Greubel, Jan 29 2017 *) a[n_] := (-1)^n Hypergeometric2F1[-n, n + 1/2, 1, 4]; Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 16 2018 *) -
PARI
/* From definition: */ {a(n)=polcoeff( sqrt( (1-x + sqrt(1-14*x+x^2 +x*O(x^n))) / (2*(1-14*x+x^2 +x*O(x^n))) ), n)} for(n=0, 20, print1(a(n), ", ")) -
PARI
/* From formula for a(n)^2: */ {a(n)=sqrtint(sum(k=0, 2*n, sum(j=0, 4*n-2*k, (-1)^(j+k)*binomial(4*n-k,j+k)^2*binomial(j+k, k)^2)))} for(n=0, 20, print1(a(n), ", ")) -
PARI
/* From formula for a(n)^2: */ {a(n) = sqrtint( sum(k=0, 2*n, binomial(2*k, k)^2*binomial(2*n+k, 2*n-k)*(-1)^k) )} for(n=0, 20, print1(a(n), ", "))
Formula
a(n)^2 = Sum_{k=0..2*n} Sum_{j=0..4*n-2*k} (-1)^(j+k) * C(4*n-k,j+k)^2 * C(j+k,k)^2.
a(n) ~ (3*sqrt(3)-5) * (7+4*sqrt(3))^(n+1) / (4*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 16 2014
a(n)^2 = C(4*n,2*n)*hyper4F3([-2*n,-2*n,-2*n,-2*n+1/2],[1,-4*n,-4*n],4). - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2. - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} (-1)^k * C(2*k, k)^2 * C(2*n+k, 2*n-k). - Paul D. Hanna, Aug 17 2014
From Peter Bala, Mar 14 2018: (Start)
a(n) = (-1)^n*P(2*n,sqrt(-3)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = (1/C(2*n,n))*Sum_{k = 0..n} (-1)^(n+k)*C(n,k)*C(n+k,k)* C(2*n+2*k,n+k) = Sum_{k = 0..n} (-1)^(n+k)*C(2*k,k)*C(n,k) *C(2*n+2*k,2*n)/C(n+k,n). In general, P(2*n,sqrt(1+4*x)) = (1/C(2*n,n))*Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k) *x^k.
a(n) = Sum_{k = 0..2*n} C(2*n,k)^2 * u^(n-k), where u = (1 - sqrt(-3))/2 is a primitive sixth root of unity.
a(n) = (-1)^n*Sum_{k = 0..2*n} C(2*n,k)*C(2*n+k,k)*u^(2*k).
(End)
a(n) = (-1)^n*hypergeom([-n, n + 1/2], [1], 4). - Peter Luschny, Mar 16 2018
a(0) = 1, a(1) = 5 and n * (2*n-1) * (4*n-5) * a(n) = (4*n-3) * (28*n^2-42*n+9) * a(n-1) - (n-1) * (2*n-3) * (4*n-1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 28 2020
From Peter Bala, May 03 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{i = 0..k} Sum_{j = 0..n-k-i} C(n, k)*C(k, i)*C(2*n, j)*C(3*n-k-i-j-1, n-k-i-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 >= 5 and positive integers n and k. (End)
Comments