A385043 The sum of the unitary divisors of n whose exponents in their prime factorizations are all powers of 2 (A138302).
1, 3, 4, 5, 6, 12, 8, 1, 10, 18, 12, 20, 14, 24, 24, 17, 18, 30, 20, 30, 32, 36, 24, 4, 26, 42, 1, 40, 30, 72, 32, 1, 48, 54, 48, 50, 38, 60, 56, 6, 42, 96, 44, 60, 60, 72, 48, 68, 50, 78, 72, 70, 54, 3, 72, 8, 80, 90, 60, 120, 62, 96, 80, 1, 84, 144, 68, 90, 96
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
The unitary analog of A353900.
The sum of unitary divisors of n that are: A092261 (squarefree), A192066 (odd), A358346 (exponentially odd), A358347 (square), A360720 (powerful), A371242 (cubefree), A380396 (cube), A383763 (exponentially squarefree), this sequence (exponentially 2^n), A385045 (5-rough), A385046 (3-smooth), A385047 (power of 2), A385048 (cubefull), A385049 (biquadratefree).
Programs
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Mathematica
f[p_, e_] := If[e == 2^IntegerExponent[e, 2], p^e + 1, 1]; a[ 1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
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PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1<
Formula
Multiplicative with a(p^e) = p^(A209229(e)) + 1.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1/(p*(p+1)) + Sum_{k>=2} (1/p^(2^k)-1/p^(2^k-1))) = 1.21427559551509410114... .
Comments