A160752 a(n) is the number of sets of (distinct, not necessarily consecutive) positive divisors of n where each set has all of its elements in arithmetic progression, and where each set contains exactly A067131(n) elements.

1, 1, 1, 3, 1, 1, 1, 6, 3, 6, 1, 1, 1, 6, 1, 10, 1, 2, 1, 15, 6, 6, 1, 2, 3, 6, 6, 1, 1, 4, 1, 15, 6, 6, 6, 2, 1, 6, 6, 1, 1, 2, 1, 15, 3, 6, 1, 3, 3, 15, 6, 15, 1, 3, 6, 2, 6, 6, 1, 1, 1, 6, 15, 21, 6, 3, 1, 15, 6, 28, 1, 4, 1, 6, 2, 15, 6, 2, 1, 2, 10, 6, 1, 2, 6, 6, 6, 28, 1, 10, 1, 15, 6, 6, 6, 4, 1, 15

Offset: 1

Author

Leroy Quet, May 25 2009

Keywords

• nonn

Comments

If A067131(n) = 2, then a(n) = d(n)*(d(n)-1)/2, where d(n) is the number of divisors of n.

a(p) = 1 for all primes p.

Examples

			The divisors of 18 are 1,2,3,6,9,18. There are 2 sets of these divisors, (1,2,3) and (3,6,9), that have their terms in arithmetic progression and that each have the maximal number (3) of such divisors of 18. So a(18) = 2.
		

Links

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

Crossrefs

Cf. A067131, A319354.

Programs

PARI

A160752(n) = if(1==n,n,my(d=divisors(n),m=1,counts=vector(#d)); 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++); counts[c]++; m = max(m,c))); (counts[m])); \\ Antti Karttunen, Sep 21 2018

Formula

a(n) = A071178(A319354(n)). - Antti Karttunen, Sep 21 2018

Extensions

Extended by Ray Chandler, Jun 15 2009