A341508 a(n) = 0 if n is nonabundant, otherwise a(n) is the number of abundant divisors of the last abundant number in the iteration x -> A003961(x) (starting from x=n) before a nonabundant number is reached.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 5, 0, 0, 0, 0, 0, 1, 0, 3, 0, 0, 0, 3, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 5, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Antti Karttunen, Feb 20 2021

Keywords

• nonn

Comments

Question: Is a(A336389(n)) = 1 for all n >= 2? Note that all the terms of A047802 are obviously primitively abundant (in A091191).

Examples

			Starting from n = 120 = 2^3 * 3 * 5, the number of its abundant divisors is A080224(120) = 7. Then we apply a prime shift (A003961) to obtain the next number, 3^3 * 5 * 7 = 945, which has A080224(945) = 1 abundant divisors (as 945 is a term of A091191). The next prime shift gives 5^3 * 7* 11 = 9625, which has zero abundant divisors (as it is nonabundant, in A263837), so A080224(9625) = 0, and a(120) = 1, the last nonzero value encountered.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences computed from indices in prime factorization
• Index entries for sequences related to sigma(n)

Crossrefs

Cf. A003961, A047802, A091191, A336389, A336835, A337468.

Cf. A263837 (positions of zeros), A005101 (and of nonzeros).

Differs from A080224 for the first time at n=120, with a(120) = 1, while A080224(120) = 7.

Programs

PARI

A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A080224(n) = sumdiv(n, d, sigma(d)>2*d);
A341508(n) = { my(t, u=0); while((t=A080224(n))>0, u=t; n = A003961(n)); (u); };