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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A092964 Numbers > 1 in A051168, with a(0) = 1.

Original entry on oeis.org

1, 2, 2, 2, 3, 2, 3, 5, 5, 3, 3, 7, 8, 7, 3, 4, 9, 14, 14, 9, 4, 4, 12, 20, 25, 20, 12, 4, 5, 15, 30, 42, 42, 30, 15, 5, 5, 18, 40, 66, 75, 66, 40, 18, 5, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 6, 26, 70, 143, 212, 245, 212, 143, 70, 26, 6, 7, 30, 91, 200, 333, 429, 429, 333, 200
Offset: 0

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Author

Thomas O. Hoffbauer (Thomas.Hoffbauer(AT)cibamberg.de), Apr 20 2004

Keywords

Comments

Original definition: Triangle read by rows in which row n gives the number of equal configurations under cyclic shift.
T(n,k) = A051168(n+3,k+1), if 0Michael Somos, Jul 17 2004
From Paul Weisenhorn, Dec 21 2010 (Start):
T(n,k) is the number of classes that have (k+1) ordered sums of (n+4) with (k+1) positive integers, that can be transformed into each other by a cyclic permutation.
m has 2^(m-1) ordered sums; for each sum one remove the first part z(1) and add 1 to the next z(1) parts to get a new ordered sum until a period is reached. T(n,k)=a(m) with m=(n+4)*(n+3)/2+k+1 gives for m the number of periods with length (n+4).
The numbers m=n*(n+3)/2 with 1<=n have one period with length (n+1).
The numbers m=n*(n+3)/2+2 with 1<=n have one period with length (n+2).
The triangular numbers n*(n+1)/2 with 1<=n have one period [(n+(n-1)+...+2+1)] with length 1. (End)

Examples

			As triangle, starts:
  1;
  2,2;
  2,3,2;
  3,5,5,3;
  3,7,8,7,3;
  4,9,14,14,9,4;
  4,12,20,25,20,12,4; ...
From _Paul Weisenhorn_, Dec 21 2010: (Start)
T(2,2)=3 classes with 3 ordered sums of 6; [(1+1+4),(1+4+1),(4+1+1)]; [(1+2+3),(2+3+1),(3+1+2)]; [(1+3+2),(3+2+1),(2+1+3)].
T(2,2)=a(m)=3 periods with length 6 for m=6*5/2+3=18 [(5+5+4+3+1),(6+5+4+2+1),(6+5+3+2+1+1),(6+4+3+2+2+1),(5+4+3+3+2+1),(5+4+4+3+2)]; [(5+5+3+3+2),(6+4+4+3+1),(5+5+4+2+1+1),(6+5+3+2+2),(6+4+3+3+1+1),(5+4+4+2+2+1)]; [(5+5+3+2+2+1),(6+4+3+3+2),(5+4+4+3+1+1),(5+5+4+2+2),(6+5+3+3+1),(6+4+4+2+1+1)]. (End)
		

Crossrefs

Row sums give A093210. Essentially the same as A051168. See A185158 for another version.

Programs

  • Mathematica
    T[n_, k_] := DivisorSum[GCD[k + 1, n + 4], Binomial[(n + 4)/#, (k + 1)/#] * MoebiusMu[#] & ]/(n + 4); Table[T[n, k], {n, 0, 12}, {k, 1, n + 1}] // Flatten (* Jean-François Alcover, Dec 02 2015 *)
  • PARI
    T(n,k)=local(A,ps,c); n+=3; k++; if(k<1||k>=n-1, 0, A=x*O(x^n) + y*O(y^n); ps=1-x-y+A; for(m=1,n,for(i=0,m,c=polcoeff(polcoeff(ps,i,x),m-i, y); if(m==n&i==k,break(2),ps*=(1-y^(m-i)*x^i+A)^c)));-c) /* Michael Somos, Jul 17 2004 */

Extensions

Edited with better definition by Omar E. Pol, Jan 05 2009

A060477 Number of orbits of length n in map whose periodic points are A000051.

Original entry on oeis.org

3, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335, 630, 1161, 2182, 4080, 7710, 14532, 27594, 52377, 99858, 190557, 364722, 698870, 1342176, 2580795, 4971008, 9586395, 18512790, 35790267, 69273666, 134215680, 260300986, 505286415, 981706806, 1908866960, 3714566310
Offset: 1

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Author

Keywords

Examples

			a(3)=2 since the 3rd term of A000051 is 9 and the first term is 3.
		

Crossrefs

Cf. A000051.
Cf. A001037, A059966 (both nearly identical to this sequence).
Cf. A093210.

Programs

  • PARI
    a000051(n) = 2^n+1;
    a(n) = (1/n)*sumdiv(n, d, moebius(d)*a000051(n/d)); \\ Michel Marcus, Sep 11 2017
    
  • Python
    from sympy import mobius, divisors
    def A060477(n): return sum(mobius(n//d)*(2**d+1) for d in divisors(n,generator=True))//n # Chai Wah Wu, Feb 03 2022

Formula

a(n) = (1/n)* Sum_{d|n} mu(d)*A000051(n/d).

Extensions

A048578 replaced by A000051 in name and formula by Michel Marcus, Sep 11 2017
Showing 1-2 of 2 results.