A367698 The smallest divisor d of n such that n/d is an exponentially odious number (A270428).
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
maxOdious[e_] := Module[{k = e}, While[EvenQ[DigitCount[k, 2, 1]], k--]; k]; f[p_, e_] := p^(e - maxOdious[e]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
s(n) = {my(k = n); while(!(hammingweight(k)%2), k--); n-k; } a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i, 1]^s(f[i, 2])); }
Formula
a(n) = n/A366905(n).
Multiplicative with a(p^e) = p^(e-s(e)), where s(e) = max({k=1..e, k odious}).
a(n) >= 1, with equality if and only if n is an exponentially odious number (A270428).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} f(1/p) = 1.25857819194624249136..., where f(x) = (1-x)*(1+Sum_{k>=1} x^s(k)), s(k) is defined above for k >= 1, and s(0) = 0.
Comments