A379592 Number of coreful divisor pairs (d, k/d), d | k, d < k/d, such that only one divisor divides the other, where k is in A320966.
1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 4, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 4, 1, 3, 3, 1, 1, 1, 1, 2, 2, 3, 1, 1, 2, 2, 1, 5, 3, 1, 3, 1, 1, 1, 4, 1, 1, 1, 1, 2, 1, 2, 2, 3, 1, 1, 3, 1, 1, 2, 4, 1, 2, 5, 1, 1, 1, 4, 1, 1, 2, 5, 1, 1
Offset: 1
Keywords
Examples
Let s(n) = A320966(n). a(1) = 1 since s(1) = 8 = 2*4. a(2) = 1 since s(2) = 16 = 2*8. a(3) = 1 since s(3) = 27 = 3*9. a(4) = 2 since s(4) = 32 = 2*16 = 4*8. a(10) = 3 since s(10) = 128 = 2*64 = 4*32 = 8*16. a(23) = 4 since s(23) = 512 = 2*256 = 4*128 = 8*64 = 16*32. a(181) = 7 since s(181) = 20736 = 6*3456 = 12*1728 = 18*1152 = 24*864 = 36*576 = 48*432 = 72*288, etc.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
nn = 5400; rad[x_] := Times @@ FactorInteger[x][[All, 1]]; s = Union@ Select[Flatten@ Table[a^2*b^3, {b, Surd[nn, 3]}, {a, Sqrt[nn/b^3]}], Length@ Select[FactorInteger[#][[All, -1]], # > 2 &] > 0 &]; Table[k = s[[n]]; Count[Transpose@ {#, k/#} &@ #[[2 ;; Ceiling[Length[#]/2]]] &@ Divisors[k], _?(And[rad[#1] == rad[#2], Xor[Divisible[#2, #1], Divisible[#1, #2]]] & @@ # &)], {n, Length[s]}]
Comments