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.

A255051 a(1)=1, a(n+1) = a(n)/gcd(a(n),n) if this GCD is > 1, else a(n+1) = a(n) + n + 1.

Original entry on oeis.org

1, 3, 6, 2, 1, 7, 14, 2, 1, 11, 22, 2, 1, 15, 30, 2, 1, 19, 38, 2, 1, 23, 46, 2, 1, 27, 54, 2, 1, 31, 62, 2, 1, 35, 70, 2, 1, 39, 78, 2, 1, 43, 86, 2, 1, 47, 94, 2, 1, 51, 102, 2, 1, 55, 110, 2, 1, 59, 118, 2, 1, 63, 126, 2, 1, 67, 134, 2, 1, 71, 142, 2, 1
Offset: 1

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Author

M. F. Hasler, Feb 15 2015

Keywords

Comments

A somehow "trivial" variant of A133058 and A255140, both of which have very similar definitions, but enter 4-periodic loops only at later indices.
There could be two motivated values for an initial term: either a(0)=0 which would yield a(1)=1 and the following values via the recursion formula, or a(0)=2 according to the general formula for a(4k).

Examples

			a(2) = a(1)+2 = 3, a(3) = a(2)+3 = 6, a(4) = a(3)/3 = 2, a(5) = a(4)/2 = 1;
a(6) = a(5)+6 = 7, a(7) = a(6)+7 = 14, a(8) = a(7)/7 = 2, a(9) = a(8)/2 = 1; ...
		

Crossrefs

Programs

  • Magma
    &cat [[1, 4*n+3, 8*n+6, 2]: n in [0..20]]; // Bruno Berselli, Feb 16 2015
  • Mathematica
    Table[(2 (3 + (-1)^n) - (2 - 3 n + n (-1)^n) (1 - (-1)^((n - 1) n/2)))/4, {n, 1, 80}] (* Bruno Berselli, Feb 16 2015 *)
    nxt[{n_,a_}]:={n+1,If[GCD[a,n]>1,a/GCD[a,n],a+n+1]}; Transpose[ NestList[ nxt, {1,1},80]][[2]] (* or *) LinearRecurrence[{0,0,0,2,0,0,0,-1},{1,3,6,2,1,7,14,2},80] (* Harvey P. Dale, Oct 13 2015 *)
  • PARI
    (A255051_upto(N)=vector(N, n, if(gcd(N, n-1)>1, N\=gcd(N, n-1), N+=n)))(99) \\ simplified by M. F. Hasler, Jan 11 2020
    
  • PARI
    A255051(n)=if(n%4>1,if(bittest(n,0),n*2,n+1),2-bittest(n,0)) \\ M. F. Hasler, Feb 18 2015
    

Formula

a(4k+1) = 1, a(4k+2) = 4k+3, a(4k+3) = 2*a(4k+2) = 8k+6, a(4k) = 2.
G.f.: x*(1 + 3*x + 6*x^2 + 2*x^3 - x^4 + x^5 + 2*x^6 - 2*x^7)/((1 - x)^2*(1 + x)^2*(1 + x^2)^2). - Bruno Berselli, Feb 16 2015
a(n) = ( 2*(3 + (-1)^n) - (2 - 3*n + n*(-1)^n)*(1 - (-1)^((n-1)*n/2)) )/4. - Bruno Berselli, Feb 16 2015