|
%I #16 Jul 05 2017 02:09:09
%S 0,1,1,4,19,116,871,7764,80179,941812,12403711,181056404,2901669739,
%T 50656307508,956922611191,19449063226324,423206168046499,
%U 9816562636678004,241805428075379311,6303793707327637524,173401707643671303259
%N a(n) = 2*Sum_{k odd} k!*Stirling2(n,k)/(k + 1).
%C 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.
%C 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.
%C 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).
%H Bai-Ni Guo, István Mező, Feng Qi, <a href="https://arxiv.org/abs/1402.2340">An explicit formula for Bernoulli polynomials in terms of r-Stirling numbers of the second kind</a>, arxiv:1402.2340v1 [math.CO], 2014.
%F 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).
%F For prime p, a(p) = 1 (mod p). Conjecture: for prime p, a(2*p) = 1 (mod p).
%p seq(add((2*k+1)!*Stirling2(n,2*k+1)/(k + 1), k = 0..floor((n-1)/2)), n = 0..20);
%Y Cf. A000629, A008277, A027641, A027642, A027760.
%K nonn,easy
%O 0,4
%A _Peter Bala_, Jan 31 2017
|
