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A141459
a(n) = Product_{p-1 divides n} p, where p is an odd prime.
15
1, 1, 3, 1, 15, 1, 21, 1, 15, 1, 33, 1, 1365, 1, 3, 1, 255, 1, 399, 1, 165, 1, 69, 1, 1365, 1, 3, 1, 435, 1, 7161, 1, 255, 1, 3, 1, 959595, 1, 3, 1, 6765, 1, 903, 1, 345, 1, 141, 1, 23205, 1, 33, 1, 795, 1, 399, 1, 435, 1, 177, 1, 28393365, 1, 3, 1, 255, 1, 32361, 1, 15, 1, 2343, 1, 70050435
OFFSET
0,3
COMMENTS
Previous name was: A027760(n)/2 for n>=1, a(0) = 1.
Conjecture: a(n) = denominator of integral_{0..1}(log(1-1/x)^n) dx. - Jean-François Alcover, Feb 01 2013
Define the generalized Bernoulli function as B(s,z) = -s*z^s*HurwitzZeta(1-s,1/z) for Re(1/z) > 0 and B(0,z) = 1 for all z; further the generalized Bernoulli polynomials as Bp(m,n,z) = Sum_{j=0..n} B(j,m)*C(n,j)*(z-1)^(n-j) then the a(n) are denominators of Bp(2,n,1), i. e. of the generalized Bernoulli numbers in the case m=2. The numerators of these numbers are A157779(n). - Peter Luschny, May 17 2015
From Peter Luschny, Nov 22 2015: (Start)
a(n) are the denominators of the centralized Bernoulli polynomials 2^n*Bernoulli(n, x/2+1/2) evaluated at x=1. The numerators are A239275(n).
a(n) is the odd part of A141056(n).
a(n) is squarefree, by the von Staudt-Clausen theorem. (End)
Apparently a(n) = denominator(Sum_{k=0..n-1}(-1)^k*E2(n-1, k+1)/binomial(2*n-1, k+1)) where E2(n, k) denotes the second-order Eulerian numbers A340556. - Peter Luschny, Feb 17 2021
FORMULA
a(2*n+1) = 1. a(2*n)= A001897(n).
a(n) = denominator(0^n + Sum_{j=1..n} zeta(1-j)*(2^j-2)*j*C(n,j)). - Peter Luschny, May 17 2015
Let P(x)= Sum_{n>=0} x^(2*n+1)/(2*n+1)! then a(n) = denominator( n! [x^n] x/P(x) ). - Peter Luschny, Jul 05 2016
a(n) = A157818(n)/4^n. See a comment under A157817, also for other Bernoulli numbers B[4,1] and B[4,3] with this denominator. - Wolfdieter Lang, Apr 28 2017
EXAMPLE
The denominators of 1, 0, -1/3, 0, 7/15, 0, -31/21, 0, 127/15, 0, -2555/33, 0, 1414477/1365, ...
MAPLE
Bfun := (s, z) -> `if`(s=0, 1, -s*z^s*Zeta(0, 1-s, 1/z): # generalized Bernoulli function
Bpoly := (m, n, z) -> add(Bfun(j, m)*binomial(n, j)*(z-1)^(n-j), j=0..n): # generalized Bernoulli polynomials
seq(Bpoly(2, n, 1), n=0..50): denom([%]);
# which simplifies to:
a := n -> 0^n+add(Zeta(1-j)*(2^j-2)*j*binomial(n, j), j=1..n):
seq(denom(a(n)), n=0..50); # Peter Luschny, May 17 2015
# Alternatively:
with(numtheory):
ClausenOdd := proc(n) local S, m;
S := map(i -> i + 1, divisors(n));
S := select(isprime, S) minus {2};
mul(m, m = S) end: seq(ClausenOdd(n), n=0..72); # Peter Luschny, Nov 22 2015
# Alternatively:
N:= 1000: # to get a(0) to a(N)
V:= Array(0..N, 1):
for p in select(isprime, [seq(i, i=3..N+1, 2)]) do
R:=[seq(j, j=p-1..N, p-1)]:
V[R]:= V[R] * p;
od:
convert(V, list); # Robert Israel, Nov 22 2015
MATHEMATICA
a[n_] := If[OddQ[n], 1, Denominator[-2*(2^(n - 1) - 1)*BernoulliB[n]]]; Table[a[n], {n, 0, 72}] (* Jean-François Alcover, Jan 30 2013 *)
Table[Times @@ Select[Divisors@ n + 1, PrimeQ@ # && OddQ@ # &] + Boole[n == 0], {n, 0, 72}] (* Michael De Vlieger, Apr 30 2017 *)
PROG
(PARI)
A141056(n) =
{
p = 1;
if (n > 0,
fordiv(n, d,
r = d + 1;
if (isprime(r) & r>2, p = p*r)
)
);
return(p)
}
for(n=0, 72, print1(A141056(n), ", ")); \\ Peter Luschny, Nov 22 2015
(Sage)
def A141459_list(size):
f = x / sum(x^(n*2+1)/factorial(n*2+1) for n in (0..2*size))
t = taylor(f, x, 0, size)
return [(factorial(n)*s).denominator() for n, s in enumerate (t.list())]
print(A141459_list(72)) # Peter Luschny, Jul 05 2016
KEYWORD
nonn
AUTHOR
Paul Curtz, Aug 08 2008
EXTENSIONS
1 prepended and offset set to 0 by Peter Luschny, May 17 2015
New name from Peter Luschny, Nov 22 2015
STATUS
approved