A069852 a(n) = Sum_{i=0..2n} B(i)*C(2n+1,i)*5^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial numbers.

6, -74, 1946, -88434, 6154786, -607884394, 80834386026, -13923204233954, 3015393801263666, -801997872697905114, 256982712667627683706, -97641716941862894337874, 43406301788286350509870146, -22319737637152541506923644234

Offset: 1

Author

Benoit Cloitre, May 01 2002

Keywords

• easy
• sign

Comments

Related to those formulas derived from Bernoulli polynomials: Sum_{k>0} sin(k*x)/k^(2n+1) = (-1)^(n+1)/2*x^(2n+1)/(2n+1)!*Sum_{i=0..2n}(2Pi/x)^i*B(i)*C(2n+1,i).

Links

• Robert Israel, Table of n, a(n) for n = 1..232
• Wikipedia, Bernoulli Polynomials

Crossrefs

Cf. A002111, A009843, A069994.

Programs

Maple

seq(5^(2*n+1)*bernoulli(2*n+1,1/5),n=1..14); # (after Peter Bala) Peter Luschny, Mar 08 2015

Mathematica

Table[5^(2n+1) BernoulliB[2n+1, 1/5], {n, 1, 14}] (* Jean-François Alcover, Jun 03 2019, from Maple *)

PARI

for(n=1,25,print1(sum(i=0,2*n,binomial(2*n+1,i)*bernfrac(i)*5^i),","))

Formula

From Peter Bala, Mar 02 2015: (Start)

a(n) = 5^(2*n + 1)*B(2*n + 1,1/5), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A009843 and A069994.

Conjecturally, a(n) = the signed numerator of B(2*n + 1,1/5).

G.f.: t/2*( 3 - 5*sinh(3*t/2)/sinh(5*t/2) ) = 6*t^3/3! - 74*t^5/5! + 1946*t^7/7! - ....

G.f. for signed version of sequence: 3/2 + 3/2*Sum_{n >= 0} { 1/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (5*k + 1)*x)*(1 - (5*k + 4)*x) ) } = 6*x^2 - 74*x^4 + 1946*x^6 + .... (End)