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A328026
Number of divisible pairs of consecutive divisors of n.
11
0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 1, 2, 2, 4, 1, 3, 1, 4, 2, 2, 1, 2, 2, 2, 3, 4, 1, 2, 1, 5, 2, 2, 2, 2, 1, 2, 2, 4, 1, 4, 1, 4, 2, 2, 1, 2, 2, 3, 2, 4, 1, 4, 2, 4, 2, 2, 1, 2, 1, 2, 2, 6, 2, 4, 1, 4, 2, 2, 1, 2, 1, 2, 3, 4, 2, 4, 1, 4, 4, 2, 1, 2, 2, 2, 2, 6, 1, 2, 2, 4, 2, 2, 2, 2, 1, 3, 4, 6, 1, 4, 1, 6, 2
OFFSET
1,4
COMMENTS
The number m = 2^n, n >= 0, is the smallest for which a(m) = n. - Marius A. Burtea, Nov 20 2019
LINKS
FORMULA
a(p^k) = k for any prime number p and k >= 0. - Rémy Sigrist, Oct 05 2019
EXAMPLE
The divisors of 500 are {1,2,4,5,10,20,25,50,100,125,250,500}, with consecutive divisible pairs {1,2}, {2,4}, {5,10}, {10,20}, {25,50}, {50,100}, {125,250}, {250,500}, so a(500) = 8.
MATHEMATICA
Table[Length[Split[Divisors[n], !Divisible[#2, #1]&]]-1, {n, 100}]
PROG
(PARI) a(n) = {my(d=divisors(n), nb=0); for (i=2, #d, if ((d[i] % d[i-1]) == 0, nb++)); nb; } \\ Michel Marcus, Oct 05 2019
(Magma) f:=func<n, i|D[i+1] mod D[i] eq 0 where D is Divisors(n) >; g:=func<k| #[i:i in [1..#Divisors(k)-1]| f(k, i)]>; [g(n):n in [1..100]]; // Marius A. Burtea, Nov 20 2019
CROSSREFS
Positions of 1's are A000040.
Positions of 0's and 2's are A328028.
Positions of terms > 2 are A328189.
Successive pairs of consecutive divisors are counted by A129308.
Sequence in context: A076558 A328195 A235875 * A326975 A204893 A251717
KEYWORD
nonn
AUTHOR
Gus Wiseman, Oct 03 2019
EXTENSIONS
Data section extended up to a(105) by Antti Karttunen, Feb 23 2023
STATUS
approved