login
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.
4
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
1,4
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.
LINKS
FORMULA
a(n) = A071178(A319354(n)). - Antti Karttunen, Sep 21 2018
EXAMPLE
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.
PROG
(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
CROSSREFS
Sequence in context: A016566 A096744 A180051 * A186024 A080002 A291722
KEYWORD
nonn
AUTHOR
Leroy Quet, May 25 2009
EXTENSIONS
Extended by Ray Chandler, Jun 15 2009
STATUS
approved