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A328026
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Number of divisible pairs of consecutive divisors of n.
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11
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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
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OFFSET
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1,4
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COMMENTS
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The number m = 2^n, n >= 0, is the smallest for which a(m) = n. - Marius A. Burtea, Nov 20 2019
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LINKS
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FORMULA
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a(p^k) = k for any prime number p and k >= 0. - Rémy Sigrist, Oct 05 2019
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EXAMPLE
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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.
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MATHEMATICA
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Table[Length[Split[Divisors[n], !Divisible[#2, #1]&]]-1, {n, 100}]
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PROG
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(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
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CROSSREFS
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Positions of 0's and 2's are A328028.
Positions of terms > 2 are A328189.
Successive pairs of consecutive divisors are counted by A129308.
Cf. A000005, A027750, A033676, A060680, A060681, A060682, A060683, A193829, A328023, A328024, A328194.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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