A379752 Number of pairs (d, k/d), d | k, d < k/d, such that gcd(d, k/d) > 1 and neither rad(d) | k/d nor rad(k/d) | d, where k is in A375055.
1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 2, 1, 3, 1, 1, 1, 1, 3, 1, 1, 2, 1, 2, 2, 1, 3, 2, 1, 2, 2, 1, 2, 1, 4, 1, 1, 1, 3, 1, 2, 1, 1, 3, 1, 3, 1, 2, 1, 3, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 3, 3, 1, 1, 1, 3, 1, 4, 2, 2, 1
Offset: 1
Examples
Let s(n) = A375055(n). a(1) = 1 since s(1) = 60 = 6 * 10 = (2*3) * (2*5). a(2) = 1 since s(2) = 84 = 6 * 14 = (2*3) * (2*7). a(3) = 1 since s(3) = 90 = 6 * 15 = (2*3) * (3*5). a(4) = 2 since s(4) = 120 = 6*20 = 10*12. a(17) = 3 since s(17) = 240 = 6*40 = 10*24 = 12*20. a(51) = 4 since s(51) = 480 = 6*80 = 10*48 = 12*40 = 20*24. a(117) = 5 since s(117) = 840 = 6*140 = 10*84 = 12*70 = 14*60 = 20*42 = 28*30, etc.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A375055.
Programs
-
Mathematica
nn = 120; rad[x_] := Times @@ FactorInteger[x][[All, 1]]; s = Select[Range[nn], PrimeOmega[#] > PrimeNu[#] > 2 & ]; Table[k = s[[n]]; Count[Transpose@ {#, k/#} &@ #[[2 ;; Ceiling[Length[#]/2]]] &@ Divisors[k], _?(And[1 < GCD @@ {##}, Nor[Divisible[#2, rad[#1]], Divisible[#1, rad[#2]] ] ] & @@ # &)], {n, Length[s]}]
Comments