%I #8 Jan 02 2026 09:30:43
%S 1,1,1,1,1,1,1,1,1,1,1,1,3,2,5,3,7,4,9,5,1,3,1,5,3,7,1,9,5,11,1,2,5,1,
%T 7,4,1,5,11,2,1,5,3,7,1,9,5,11,3,13,1,3,7,4,9,1,11,6,13,7,1,7,1,1,5,
%U 11,1,13,7,5,1,4,9,5,11,6,13,1,15,8,1,9,5,11,3,13,7,15,1,17
%N a(n) is the denominator of the harmonic mean of the digits of n.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/HarmonicMean.html">Harmonic Mean</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Harmonic_mean">Harmonic mean</a>.
%F a(A011540(n)) = 1.
%F a(A392051(n)) = 1.
%e a(43) = denominator(2*4*3/(4 + 3)) = denominator(24/7) = 7.
%t a[n_]:=Denominator[HarmonicMean[IntegerDigits[n]]]; Array[a,90,0]
%o (Python)
%o from fractions import Fraction
%o def a(n):
%o d = list(map(int, str(n)))
%o return 1 if 0 in d else (len(d)/sum(Fraction(1, di) for di in d)).denominator
%o print([a(n) for n in range(90)]) # _Michael S. Branicky_, Dec 29 2025
%Y Cf. A011540, A371384, A392049 (numerator), A392051.
%K nonn,base,easy,frac
%O 0,13
%A _Stefano Spezia_, Dec 29 2025