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A360119
Number of divisors of n which are not also differences between consecutive divisors, minus the number of differences between consecutive divisors of n which are not also divisors of n. Here the differences are counted with repetition if they occur more than once.
3
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 4, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 5, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 6, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 6, 1, 1, 1, 3, 1, 3, 1, 1, 1
OFFSET
1,6
COMMENTS
Because the algorithm for computing this sequence (see the PARI program) starts with s set to the number of divisors, and s is decremented at most once on each iteration in the loop over the first differences of the divisors, and because there is one less difference than there are divisors, it implies that a(n) >= 1 for all n.
Note that if a(n) = 1, then A088722(n) = 0, but not vice versa, i.e., the positions of 1's in this sequence is just a subsequence of A088725. See A360129 for the exceptions.
LINKS
FORMULA
a(n) = A060764(n) - A360118(n).
a(n) <= A000005(n).
PROG
(PARI) A360119(n) = { my(d=divisors(n), erot=vecsort(vector(#d-1, k, d[k+1] - d[k])), s=#d); for(i=1, #erot, if(n%erot[i], s--, if(1==i || erot[i]!=erot[i-1], s--))); (s); };
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 20 2023
STATUS
approved