A380056 E.g.f. (exp(x) - 1)/cos(2*x).

1, 1, 13, 25, 441, 1261, 30213, 115025, 3529201, 16792021, 629401213, 3593565625, 159175688361, 1060279600381, 54189700721013, 412526870321825, 23894940183997921, 204641610743378341, 13248060325188261613, 126065945039257743625, 9020317522757414377881, 94419130586604915837901

Offset: 1

Author

Paul D. Hanna, Jan 28 2025

Keywords

• nonn

Comments

Conjecture: a(n) == 1 (mod 4) for n >= 1.

Conjecture: a(4*n) is divisible by 5 for n >= 1.

Examples

			E.g.f.: A(x) = x + x^2/2! + 13*x^3/3! + 25*x^4/4! + 441*x^5/5! + 1261*x^6/6! + 30213*x^7/7! + 115025*x^8/8! + 3529201*x^9/9! + 16792021*x^10/10! + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..400

Crossrefs

Cf. A380053, A000364, A380555.

Programs

PARI

{a(n) = my(A,X = x + x*O(x^n)); n!*polcoef( (exp(X) - 1)/cos(2*X), n)}
for(n=1,25,print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies the following formulas.

(1) A(x) = (exp(x) - 1)/cos(2*x).

(2) A(F(x)) = x where F(x) = log(1 + x*cos(2*F(x))) equals the e.g.f. of A380555.

(3.a) a(2*n+1) = Sum_{k=0..n} binomial(2*n+1, 2*k+1) * A000364(n-k) * 4^(n-k) for n >= 0.

(3.b) a(2*n) = Sum_{k=1..n} binomial(2*n, 2*k) * A000364(n-k) * 4^(n-k) for n >= 1.

a(n) ~ (exp(Pi/4) - 1 + (exp(-Pi/4) - 1)*(-1)^n) * 2^(2*n + 3/2) * n^(n + 1/2) / (sqrt(Pi) * exp(n) * Pi^n). - Vaclav Kotesovec, Jan 29 2025