A379448 a(n) is the number of ones in the binary expansion of n^(n^n).
1, 1, 30, 1, 3683, 36635, 1156050, 1, 614037343, 11609679812, 493508438640
Offset: 1
Programs
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Mathematica
Table[DigitCount[n^n^n,2,1],{n,9}] (* James C. McMahon, Dec 26 2024 *) -
PARI
a379448(n) = hammingweight(n^n^n)
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PARI
a(n)=hammingweight((n>>valuation(n,2))^n^n) \\ Charles R Greathouse IV, Dec 26 2024
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Python
def A379448(n): return 1 if n.bit_count()==1 else (n**n**n).bit_count() # Chai Wah Wu, Dec 24 2024
Formula
a(2^k) = 1. - Chai Wah Wu, Dec 24 2024
Probably a(n) = n^n * log(A000265(n))/log(4) + O(n^(n/2)) by analogy with the Law of the Iterated Logarithm. - Charles R Greathouse IV, Dec 27 2024
Extensions
a(11) from Markus Sigg, Dec 27 2024