|
%I #64 Feb 17 2021 12:30:12
%S 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,
%T 3,1,435,1,7161,1,255,1,3,1,959595,1,3,1,6765,1,903,1,345,1,141,1,
%U 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
%N a(n) = Product_{p-1 divides n} p, where p is an odd prime.
%C Previous name was: A027760(n)/2 for n>=1, a(0) = 1.
%C Conjecture: a(n) = denominator of integral_{0..1}(log(1-1/x)^n) dx. - _Jean-François Alcover_, Feb 01 2013
%C 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
%C From _Peter Luschny_, Nov 22 2015: (Start)
%C 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).
%C a(n) is the odd part of A141056(n).
%C a(n) is squarefree, by the von Staudt-Clausen theorem. (End)
%C 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
%H Robert Israel, <a href="/A141459/b141459.txt">Table of n, a(n) for n = 0..10000</a>
%H Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/GeneralizedBernoulliNumbers">Generalized Bernoulli Numbers and Polynomials</a>
%F a(2*n+1) = 1. a(2*n)= A001897(n).
%F a(n) = denominator(0^n + Sum_{j=1..n} zeta(1-j)*(2^j-2)*j*C(n,j)). - _Peter Luschny_, May 17 2015
%F 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
%F 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
%e The denominators of 1, 0, -1/3, 0, 7/15, 0, -31/21, 0, 127/15, 0, -2555/33, 0, 1414477/1365, ...
%p Bfun := (s,z) -> `if`(s=0,1,-s*z^s*Zeta(0,1-s,1/z): # generalized Bernoulli function
%p Bpoly := (m,n,z) -> add(Bfun(j,m)*binomial(n,j)*(z-1)^(n-j),j=0..n): # generalized Bernoulli polynomials
%p seq(Bpoly(2,n,1),n=0..50): denom([%]);
%p # which simplifies to:
%p a := n -> 0^n+add(Zeta(1-j)*(2^j-2)*j*binomial(n,j), j=1..n):
%p seq(denom(a(n)), n=0..50); # _Peter Luschny_, May 17 2015
%p # Alternatively:
%p with(numtheory):
%p ClausenOdd := proc(n) local S, m;
%p S := map(i -> i + 1, divisors(n));
%p S := select(isprime, S) minus {2};
%p mul(m, m = S) end: seq(ClausenOdd(n), n=0..72); # _Peter Luschny_, Nov 22 2015
%p # Alternatively:
%p N:= 1000: # to get a(0) to a(N)
%p V:= Array(0..N, 1):
%p for p in select(isprime, [seq(i,i=3..N+1,2)]) do
%p R:=[seq(j,j=p-1..N, p-1)]:
%p V[R]:= V[R] * p;
%p od:
%p convert(V,list); # _Robert Israel_, Nov 22 2015
%t 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 *)
%t Table[Times @@ Select[Divisors@ n + 1, PrimeQ@ # && OddQ@ # &] + Boole[n == 0], {n, 0, 72}] (* _Michael De Vlieger_, Apr 30 2017 *)
%o (PARI)
%o A141056(n) =
%o {
%o p = 1;
%o if (n > 0,
%o fordiv(n, d,
%o r = d + 1;
%o if (isprime(r) & r>2, p = p*r)
%o )
%o );
%o return(p)
%o }
%o for(n=0, 72, print1(A141056(n), ", ")); \\ _Peter Luschny_, Nov 22 2015
%o (Sage)
%o def A141459_list(size):
%o f = x / sum(x^(n*2+1)/factorial(n*2+1) for n in (0..2*size))
%o t = taylor(f, x, 0, size)
%o return [(factorial(n)*s).denominator() for n,s in enumerate (t.list())]
%o print(A141459_list(72)) # _Peter Luschny_, Jul 05 2016
%Y Cf. A027760, A141056, A141459, A157779, A157818, A160014, A226157, A239275, A340556.
%K nonn
%O 0,3
%A _Paul Curtz_, Aug 08 2008
%E 1 prepended and offset set to 0 by _Peter Luschny_, May 17 2015
%E New name from _Peter Luschny_, Nov 22 2015
|
