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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A287706 a(n) = denominator(Bernoulli(n, x/2) - Bernoulli(n, x)).

Original entry on oeis.org

1, 2, 4, 8, 16, 96, 64, 384, 256, 2560, 1024, 6144, 4096, 860160, 81920, 98304, 65536, 1966080, 1310720, 55050240, 7340032, 346030080, 20971520, 125829120, 16777216, 9160359936, 469762048, 134217728, 268435456, 8053063680, 1073741824, 496068722688, 330712481792
Offset: 0

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Author

Peter Luschny, Aug 08 2017

Keywords

Crossrefs

Programs

  • Maple
    a := n -> denom(bernoulli(n, x/2)-bernoulli(n, x)): seq(a(i), i=0..32);

A290645 a(n) = denominator(Bernoulli(n, x) - Bernoulli(n, 1/2)).

Original entry on oeis.org

1, 2, 4, 2, 16, 6, 64, 6, 768, 10, 1024, 6, 4096, 210, 245760, 6, 196608, 30, 1310720, 210, 22020096, 330, 62914560, 30, 251658240, 546, 1409286144, 14, 268435456, 30, 1073741824, 462, 992137445376, 3570, 1803886264320, 6, 68719476736, 51870, 375208342978560
Offset: 0

Views

Author

Peter Luschny, Aug 08 2017

Keywords

Crossrefs

Programs

  • Maple
    a := n -> denom(bernoulli(n, x)-bernoulli(n, 1/2)): seq(a(i), i=0..38);

A338025 a(n) = LCM(k*j_1!*...*j_k! : j_1,...,j_k>=1, j_1+...+j_k=n, k=1,...,n)/n!.

Original entry on oeis.org

1, 1, 2, 1, 6, 2, 6, 3, 10, 2, 6, 2, 210, 30, 12, 3, 30, 10, 210, 42, 330, 30, 60, 30, 546, 42, 28, 2, 60, 4, 924, 231, 3570, 210, 6, 2, 51870, 2730, 420, 42, 2310, 330, 4620, 210, 9660, 420, 420, 210, 6630, 1326, 1716, 66, 660, 220, 1596, 114, 1740, 60, 60, 12, 1861860, 60060
Offset: 1

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Author

Harald Hofstätter, Oct 07 2020

Keywords

Comments

For each prime p >= 2, the exponent of p in a(n) is the largest integer t such that p^t is less than or equal to the sum of digits of n in base p.
n!*a(n) is the smallest common denominator of the n-th degree coefficients of the Baker-Campbell-Hausdorff series.

Crossrefs

Cf. A007947 (squarefree kernel), A195441.

Programs

  • Julia
    using Primes
    A338025(n::Int) =
        prod([p^(floor(Int, log(p, sum(digits(n, base=p)))))
      for p in 2:n if isprime(p)])
    println([A338025(n) for n = 1:50])
    
  • Maple
    A338025 := n->mul(map(p->p^(ilog[p](add(i, i=convert(n, base, p)))), select(isprime, [seq(p, p=2..n)]))):
    seq(A338025(n), n=1..50);
  • PARI
    a(n) = {my(v = matrix(primepi(n), 2, i, j, my(p=prime(i)); if (j==1, p, logint(sumdigits(n, p), p)))); factorback(v);} \\ Michel Marcus, Oct 08 2020

Formula

A007947(a(n)) = A195441(n-1).

A341109 a(n) = denominator(p(n, x)) / (n!*denominator(bernoulli(n, x))), where p(n, x) = Sum_{k=0..n} E2(n, k)*binomial(x + k, 2*n) / Product_{j=1..n} (j - x) and E2(n, k) are the second-order Eulerian numbers A201637.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 96, 192, 1152, 768, 1536, 3072, 18432, 36864, 221184, 147456, 884736, 1769472, 10616832, 21233664, 637009920, 424673280, 2548039680, 5096079360, 152882380800, 61152952320, 366917713920, 81537269760, 163074539520, 326149079040, 1956894474240
Offset: 0

Views

Author

Peter Luschny, Feb 06 2021

Keywords

Comments

The challenge is to characterize the sequence purely arithmetically, i.e., without reference to the Eulerian numbers or the Bernoulli polynomials.

Crossrefs

Programs

  • Maple
    Epoly := proc(n, x) add(combinat:-eulerian2(n, k)*binomial(x+k, 2*n), k = 0..n) / mul(j-x, j = 1..n): simplify(expand(%)) end:
    seq(denom(Epoly(n, x)) / (n!*denom(bernoulli(n, x))), n = 0..30);
  • Mathematica
    A053657[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k,0,n}],{p, Prime[Range[n]]}];
    A144845[n_] := Denominator[Together[BernoulliB[n, x]]];
    A163176[n_] := A053657[n] / n!;
    Table[(n + 1) A163176[n + 1] / A144845[n], {n, 0, 30}]
  • Sage
    def A341109(n): # uses[A341108, A318256]
        return A341108(n)//A318256(n)
    print([A341109(n) for n in (0..30)])

Formula

a(n) = A053657(n+1)/(n!*A144845(n)).
a(n) = (n+1)*A163176(n+1)/A144845(n).
a(n) = A341108(n)/A318256(n).
a(n) = A341107(n)*A324369(n+1).
a(n) = A341108(n)/A324370(n+1).
a(n) = A341108(n)*A007947(n+1)/A144845(n).
a(n) = A341108(n)*A324369(n+1)/A195441(n).
prime(n) divides a(k) for k >= A036689(n).
2^(n-1) divides exactly a(n) for n >= 2.
Previous Showing 21-24 of 24 results.