A324539 Number of divisors d of n such that A276086(d) = (n/d).

0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Antti Karttunen, Mar 10 2019

Keywords

• nonn

Comments

a(n) tells how many times n occurs in A324580. See also comments in A324541.

Question: Where is the next term larger than one in this sequence after a(2250) = 2 and a(5402250) = 2 ? Are there terms larger than 2 ?

Links

• Antti Karttunen, Table of n, a(n) for n = 1..100000
• Index entries for sequences related to primorial base
• Index entries for sequences related to primorial numbers

Crossrefs

Cf. A276086, A324540 (positions of zeros), A324541 (nonzeros), A324580.

Programs

PARI

A276086(n) = { my(i=0,m=1,pr=1,nextpr); while((n>0),i=i+1; nextpr = prime(i)*pr; if((n%nextpr),m*=(prime(i)^((n%nextpr)/pr));n-=(n%nextpr));pr=nextpr); m; };
A324539(n) = sumdiv(n,d,(d==A276086(n/d)));

Formula

a(n) = Sum_{d|n} [d == A276086(n/d)], where [ ] is the Iverson bracket.