cp's OEIS Frontend

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.

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A141459 a(n) = Product_{p-1 divides n} p, where p is an odd prime.

Original entry on oeis.org

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

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Author

Paul Curtz, Aug 08 2008

Keywords

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

Examples

			The denominators of 1, 0, -1/3, 0, 7/15, 0, -31/21, 0, 127/15, 0, -2555/33, 0, 1414477/1365, ...
		

Crossrefs

Programs

  • 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 *)
  • 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

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

Extensions

1 prepended and offset set to 0 by Peter Luschny, May 17 2015
New name from Peter Luschny, Nov 22 2015

A226156 a(n) = BS(n) * W(n) where BS = Sum_{k=0..n} ((-1)^k*k!/(k+1)) S(n, k) and S(n, k) the Stirling subset numbers A048993(n, k). W(n) = Product_{ p primes <= n+1 such that p divides n+1 or p-1 divides n } = A225481(n).

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 35, 0, -3617, 0, 43867, 0, -1222277, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -84802531453387, 0, 90219075042845, 0, -26315271553053477373, 0, 38089920879940267
Offset: 0

Views

Author

Peter Luschny, May 30 2013

Keywords

Comments

a(n)/A225481(n) is a representation of the Bernoulli numbers. This is case m = 1 of the scaled generalized Bernoulli numbers defined as Sum_{k=0..n} ((-1)^k*k!/(k+1)) S_{m}(n,k) where S_{m}(n,k) are generalized Stirling subset numbers. A225481(n) can be seen as an analog of the Clausen numbers A141056(n). Reduced to lowest terms a(n)/A225481(n) becomes A027641(n)/A027642(n).

Examples

			The numerators of 1/1, -1/2, 1/6, 0/2, -1/30, 0/6, 1/42, 0/2, -1/30, 0/10, 5/66, 0/6, -691/2730, 0/14, 35/30, 0/2, -3617/510, 0/6, 43867/798, ... (the denominators are A225481(n)).
		

Crossrefs

Programs

  • Mathematica
    BS[n_] := Sum[((-1)^k*k!/(k + 1)) StirlingS2[n, k], {k, 0, n}];
    W[n_] := Product[If[Divisible[n + 1, p] || Divisible[n, p - 1], p, 1], {p, Prime /@ Range[PrimePi[n + 1]]}];
    a[n_] := BS[n] W[n];
    Table[a[n], {n, 0, 38}] (* Jean-François Alcover, Jul 08 2019 *)
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :   # -- The Eulerian numbers --
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m) + \
               (m*k+1)*EulerianNumber(n-1, k, m)
    @CachedFunction
    def SF_BS(n, m):   # -- The scaled Stirling-Frobenius Bernoulli numbers --
        return add(add(EulerianNumber(n, j, m)*binomial(j, n - k) \
               for j in (0..n))/((-m)^k*(k+1)) for k in (0..n))
    def A226156(n):    # -- The numerators of SF_BS(n, 1) relative to A225481 --
        C = mul(filter(lambda p: ((n+1)%p == 0) or (n%(p-1) == 0), primes(n+2)))
        return C*SF_BS(n, 1)
    [A226156(n) for n in (0..25)]
Showing 1-2 of 2 results.