A345058 Number of distinct k > 0 for which A011772(k) = n.

1, 2, 1, 3, 2, 2, 2, 7, 4, 2, 2, 4, 4, 3, 1, 5, 4, 4, 3, 7, 2, 3, 2, 7, 6, 4, 3, 3, 4, 7, 4, 13, 3, 4, 2, 4, 6, 3, 4, 7, 6, 3, 4, 7, 1, 5, 2, 7, 11, 6, 2, 7, 4, 5, 2, 11, 3, 4, 2, 7, 8, 5, 5, 7, 2, 3, 4, 4, 5, 3, 4, 13, 9, 6, 6, 9, 2, 4, 4, 8, 15, 6, 2, 11, 3, 3, 3, 7, 6, 3, 1, 5, 4, 7, 3, 13, 10, 11, 7, 6, 6, 5, 4, 8, 2

Offset: 1

Author

Antti Karttunen, Jun 07 2021

Keywords

• nonn

Comments

Question: Are there any numbers n other than 1, 3, 15, 45, 91 for which a(n) = 1?

Examples

			A011772 obtains the value 6 only as A011772(7)=6 and A011772(36)=6, therefore a(6) = 2.
		

Crossrefs

Cf. A000217, A002024, A011772, A066561, A345056, A345057.

Programs

PARI

A011772(n) = { if(n==1, return(1)); my(f=factor(if(n%2, n, 2*n)), step=vecmax(vector(#f~, i, f[i, 1]^f[i, 2]))); forstep(m=step, 2*n, step, if(m*(m-1)/2%n==0, return(m-1)); if(m*(m+1)/2%n==0, return(m))); }; \\ From A011772
A345058(n) = { my(x=A011772(n), y=binomial(x+1,2)); sum(i=1,y,(A011772(i)==x)); };

Formula

a(n) = Sum_{i=1..A000217(n)} [A011772(i) = A011772(n)], where [ ] is the Iverson bracket.