OFFSET
0,4
COMMENTS
Recall the result Sum_{k = 0..n} (-1)^k*k!*Stirling2(n,k)/(k + 1) = Bernoulli(n) = A027641(n)/A027642(n). We can write this result as Bernoulli(n) = S_1(n) - S_2(n), where S1 = Sum_{k even} k!*Stirling2(n,k)/(k + 1) and S2 = Sum_{k odd} k!* Stirling2(n,k)/(k + 1). Here we record the values of the sums 2*S_2(n), which are easily seen to be integers.
The numbers a(n) are derived from a formula for the numbers Bernoulli(n). Surprisingly, there also appears to be a connection between a(2*n) and Bernoulli(2*n - 2): we conjecture a(2*n) - 1 = integer * the denominator of Bernoulli(2*n - 2) = integer * (Product_{p prime, p - 1 | 2*n - 2} p) (checked up to n = 200). For example, a(14) - 1 = 956922611190 is divisible by 2*3*5*7*13 where 2, 3, 5, 7 and 13 are the primes p such that p - 1 divides 12, while a(18) - 1 = 241805428075379310 is divisible by 2*3*5*17 where 2, 3, 5 and 17 are the primes p such that p - 1 divides 16.
The same result also appears to hold for the integer sequence b(n) := 2*Sum_{k odd} (-1)^((k-1)/2)*k!*Stirling2(n,k)/(k + 1).
LINKS
Bai-Ni Guo, István Mező, Feng Qi, An explicit formula for Bernoulli polynomials in terms of r-Stirling numbers of the second kind, arxiv:1402.2340v1 [math.CO], 2014.
FORMULA
E.g.f.: ( -x - log(2 - exp(x)) )/(exp(x) - 1) = x + x^2/2! + 4*x^3/3! + 19*x^4/4! + 116*x^5/5! + .... (use the first equation on page 3 of Guo et al. with r = 0 and s = 1).
For prime p, a(p) = 1 (mod p). Conjecture: for prime p, a(2*p) = 1 (mod p).
MAPLE
seq(add((2*k+1)!*Stirling2(n, 2*k+1)/(k + 1), k = 0..floor((n-1)/2)), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jan 31 2017
STATUS
approved