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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.
3
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).
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. 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
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 20 2021
STATUS
approved