A067131 Number of elements in the largest set of divisors of n which are in arithmetic progression.

1, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 4, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 4, 2, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 6, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 4, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 3, 2

Offset: 1

Author

Amarnath Murthy, Jan 09 2002

Keywords

• easy
• nonn

Examples

			a(12) = 4 as the divisors of 12 are {1,2,3,4,6,12} and the maximal subset in arithmetic progression is {1,2,3,4}. a(15) = 3; the maximal set is {1,3,5}.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A067132, A091009, A160752, A319354.

Programs

Mathematica

lap[s_] := Module[{}, l=Length[s]; If[l<2, Return[l]]; val=2; For[i=1, ival, val=k]]]; val]; lap/@Divisors/@Range[1, 200]

PARI

A067131(n) = { my(d=divisors(n),m=1); for(i=1,(#d-1), for(j=(i+1),#d,my(c=1,k=d[j],s=(d[j]-d[i])); while(!(n%k), k+=s; c++); m = max(m,c))); (m); }; \\ Antti Karttunen, Sep 21 2018

Formula

a(n) = A061395(A319354(n)). - Antti Karttunen, Sep 21 2018

Extensions

Edited by Dean Hickerson, Jan 15 2002