This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A281817 #16 Jul 05 2017 02:09:09 %S A281817 0,1,1,4,19,116,871,7764,80179,941812,12403711,181056404,2901669739, %T A281817 50656307508,956922611191,19449063226324,423206168046499, %U A281817 9816562636678004,241805428075379311,6303793707327637524,173401707643671303259 %N A281817 a(n) = 2*Sum_{k odd} k!*Stirling2(n,k)/(k + 1). %C A281817 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 A281817 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 A281817 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 A281817 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 A281817 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 A281817 For prime p, a(p) = 1 (mod p). Conjecture: for prime p, a(2*p) = 1 (mod p). %p A281817 seq(add((2*k+1)!*Stirling2(n,2*k+1)/(k + 1), k = 0..floor((n-1)/2)), n = 0..20); %Y A281817 Cf. A000629, A008277, A027641, A027642, A027760. %K A281817 nonn,easy %O A281817 0,4 %A A281817 _Peter Bala_, Jan 31 2017