A375034 The difference between the maximum odd exponent and the maximum even exponent in the prime factorization of n, where 0 is assigned to each maximum exponent if no such exponent exists.
0, 1, 1, -2, 1, 1, 1, 3, -2, 1, 1, -1, 1, 1, 1, -4, 1, -1, 1, -1, 1, 1, 1, 3, -2, 1, 3, -1, 1, 1, 1, 5, 1, 1, 1, -2, 1, 1, 1, 3, 1, 1, 1, -1, -1, 1, 1, -3, -2, -1, 1, -1, 1, 3, 1, 3, 1, 1, 1, -1, 1, 1, -1, -6, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, -1, -1, 1, 1, 1, -3
Offset: 1
Links
Programs
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Mathematica
a[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, Max[0, Max[Select[e, OddQ]]] - Max[0, Max[Select[e, EvenQ]]]]; a[1] = 0; Array[a, 100]
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PARI
a(n) = {my(e = factor(n)[,2], e1 = select(x -> (x % 2), e), e2 = select(x -> !(x % 2), e)); if(#e1 == 0, 0, vecmax(e1)) - if(#e2 == 0, 0, vecmax(e2));}
Formula
a(n) = 0 if and only if n = 1.
a(n) <= 0 if and only if n is in A368714.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} (-1)^(k+1)*k*d(k) = 0.5741591604302832339078..., where d(k) = Product_{p prime} (1 - 1/(p^(k+1)*(p+1))) - Product_{p prime} (1 - 1/(p^(k-1)*(p+1))) for k >= 2, and d(1) = Product_{p prime} (1 - 1/(p^2*(p+1))).
Comments