A027760 -id:A027760 - cp's OEIS frontend

A212197 Numbers k that divide the 3k-th Clausen number.

1, 2, 6, 14, 42, 114, 602, 798, 1806, 5334, 34314, 101346, 229362, 4357878, 9786714, 12198858, 168241542, 185947566, 231778302, 524550894

Offset: 1

Peter Luschny, May 05 2012

The classical Clausen numbers are given in A141056. See A160014 for generalizations. Related sequences are A014117 and A106741.

Thomas Clausen, Theorem. Lehrsatz aus einer Abhandlung ueber die Bernoullischen Zahlen, Astr. Nachr. 17 (22) (1840), 351-352.

Cf. A014117, A027760, A106741, A141056, A160014.

(* This program is not convenient for more than 15 terms *) c[n_] := Sum[Boole[PrimeQ[d+1]]/(d+1), {d, Divisors[n]}] // Denominator; Reap[For[n = 1, n < 10^7, n++, If[Divisible[c[3*n], n], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, May 21 2013 *)

A212197_list(searchlimit) =
{
    for (n=1, searchlimit,
        p = 1;
        fordiv(3*n, d,
            r = d + 1;
            if (isprime(r), p = p*r;)
        );
        if (Mod(p, n) == 0, print1(n, ", "));
    );
}

A281817 a(n) = 2Sum_{k odd} k!Stirling2(n,k)/(k + 1).

Original entry on oeis.org

0, 1, 1, 4, 19, 116, 871, 7764, 80179, 941812, 12403711, 181056404, 2901669739, 50656307508, 956922611191, 19449063226324, 423206168046499, 9816562636678004, 241805428075379311, 6303793707327637524, 173401707643671303259

Offset: 0

Peter Bala, Jan 31 2017

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).

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.

Cf. A000629, A008277, A027641, A027642, A027760.

seq(add((2*k+1)!*Stirling2(n,2*k+1)/(k + 1), k = 0..floor((n-1)/2)), n = 0..20);

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).

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A212197 Numbers k that divide the 3k-th Clausen number.

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A281817 a(n) = 2Sum_{k odd} k!Stirling2(n,k)/(k + 1).

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cp's OEIS Frontend

A212197 Numbers k that divide the 3k-th Clausen number.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

A281817 a(n) = 2*Sum_{k odd} k!*Stirling2(n,k)/(k + 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A281817 a(n) = 2Sum_{k odd} k!Stirling2(n,k)/(k + 1).