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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A276549 Number of primitive (aperiodic) reversible string structures with n beads using an infinite alphabet.

Original entry on oeis.org

1, 1, 3, 9, 31, 112, 467, 2141, 10739, 58454, 340389, 2110093, 13830234, 95475087, 691543059, 5240282987, 41432986587, 341040306207, 2916376237349, 25862097428262, 237434959190586, 2253358056942644, 22076003468637449, 222979436688500085, 2319295172178428701
Offset: 1

Views

Author

Andrew Howroyd, Apr 09 2017

Keywords

Crossrefs

Row sums of A276544.
Cf. A103293.

Programs

  • Mathematica
    b[n_] := SeriesCoefficient[Exp[(Exp[2*x] - 3)/2 + Exp[x]], {x, 0, n}]*n!;
    c[n_] := If[n == 0, 1, (BellB[n - 1] + If[Mod[n, 2] == 1, b[(n - 1)/2], Sum[Binomial[n/2 - 1, k]*b[k], {k, 0, n/2 - 1}]])/2];
    a[n_] := DivisorSum[n, MoebiusMu[n/#] c[# + 1]&];
    Array[a, 25] (* Jean-François Alcover, Jun 16 2017, using Alois P. Heinz's code for A103293 *)

Formula

a(n) = Sum_{d|n} mu(n/d) * A103293(d+1).

A327612 Number of length n reversible string structures that are not palindromic using any number of colors.

Original entry on oeis.org

0, 1, 2, 9, 27, 112, 453, 2137, 10691, 58435, 340187, 2110016, 13829358, 95474679, 691538954, 5240280999, 41432965441, 341040295916, 2916376121375, 25862097370783, 237434958512487, 2253358056604465, 22076003464423853, 222979436686398848, 2319295172150784296
Offset: 1

Views

Author

Andrew Howroyd, Sep 18 2019

Keywords

Crossrefs

Row sums of A309748(n > 1).

Programs

  • PARI
    \\ Ach is A304972 as square matrix.
    Ach(n)={my(M=matrix(n, n, i, k, i>=k)); for(i=3, n, for(k=2, n, M[i, k]=k*M[i-2, k] + M[i-2, k-1] + if(k>2, M[i-2, k-2]))); M}
    seq(n)={my(A=Ach(n)); vector(n, i, sum(k=1, n, (A[i,k] + stirling(i, k, 2))/2 - stirling((i+1)\2, k, 2)))}

Formula

a(n) = A103293(n + 1) - A188164(n).

A376620 Number of equational laws for magmas involving n operations, up to relabeling and symmetry.

Original entry on oeis.org

2, 5, 41, 364, 4294, 57882, 888440, 15120105, 281942218, 5698630860, 123850400282, 2875187314622, 70909556575040, 1849319825544900, 50801676938400207, 1464954360561398340, 44213852151914127210, 1392971702129279452950, 45705100441643456206404, 1558551328538087579977710
Offset: 0

Views

Author

Terence Tao, Sep 30 2024

Keywords

Comments

Is always at least A289679(n+2)/2 (with equality when n is odd), and at most A289679(n+2).
If one does not invoke symmetry, the sequence becomes A289679(n+2).
For a Python script to enumerate the laws (which also deletes trivial laws w=w) see the links.

Examples

			For n=0 the distinct laws are x=x and x=y.
For n=1 the distinct laws are x=x*x, x=x*y, x=y*x, x=y*y, and x=y*z.  (x*y=z, for instance, is a relabeling of x=y*z after applying symmetry.)
		

Crossrefs

Programs

  • PARI
    \\ All functions that are needed
    a110(n) = sum(k=0, n, stirling(n,k,2)); \\ Bell
    a108(n) = binomial(2*n,n)/(n+1); \\ Catalan
    a289679(n) = a108(n-1)*a110(n);
    Ach(n,k)= my(s=n<2 && n>=0 && n==k); if(n<=1, s, k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2) + s);
    a103293(n) = if(n<3, 1, sum(k=0, n-1, stirling(n-1,k,2) + Ach(n-1,k))/2);
    a376620(n) = if(n%2==0,(a289679(n+2) + a108(n/2) * (2*a103293(n+3) - a110(n+2)))/2, a289679(n+2)/2); \\ Hugo Pfoertner, Sep 30 2024
    
  • Python
    from functools import lru_cache
    from sympy.functions.combinatorial.numbers import stirling, bell, catalan
    def A376620(n):
        if n&1:
            return catalan(n+1)*bell(n+2)>>1
        else:
            @lru_cache(maxsize=None)
            def ach(n,k): return (n==k) if n<2 else k*ach(n-2,k)+ach(n-2,k-1)+ach(n-2,k-2)
            return catalan(n+1)*bell(n+2)+catalan(n>>1)*((sum(stirling(n+2,k,kind=2)+ach(n+2,k)>>1 for k in range(n+3))<<1)-bell(n+2))>>1 # Chai Wah Wu, Oct 15 2024

Formula

For odd n, a(n) = A289679(n+2)/2.
For even n, a(n) = (A289679(n+2) + A000108(n/2) * (2*A103293(n+3) - A000110(n+2)))/2.

Extensions

a(7) and beyond from Michael S. Branicky, Sep 30 2024 using formulas
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