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

1,24

Comments

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

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)

Example

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.

Prog

(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); };

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.

Sequence in context: A066087 A294927 A080224 * A261488 A341353 A010105

Adjacent sequences: A341505 A341506 A341507 * A341509 A341510 A341511

Keyword

nonn

Author

Antti Karttunen, Feb 20 2021

Status

approved