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Number of solutions 0<k<n to the equation sigma(n) = sigma(k) + sigma(n-k), where sigma is the sum of divisors function.
3

%I #7 Feb 20 2023 12:26:33

%S 0,0,2,0,0,0,0,2,2,2,0,0,0,0,2,0,0,0,0,2,2,0,0,0,0,0,0,0,0,2,0,4,2,0,

%T 0,0,0,0,2,2,0,0,0,0,0,0,0,0,2,0,2,0,0,0,4,2,4,0,0,0,0,2,4,0,0,0,0,0,

%U 4,2,0,0,0,0,2,0,0,0,0,0,0,0,0,0,2,0,4,2,0,2,0,2,2,2,0,2,0,0,2,0,0,0,0,2,4

%N Number of solutions 0<k<n to the equation sigma(n) = sigma(k) + sigma(n-k), where sigma is the sum of divisors function.

%C The number of solutions is always even because k=n/2 cannot be a solution for even n.

%H Antti Karttunen, <a href="/A110177/b110177.txt">Table of n, a(n) for n = 1..16384</a>

%t a[n_] := Select[Range[n-1], DivisorSigma[1, n]==DivisorSigma[1, n-# ]+DivisorSigma[1, # ]&]; Table[Length[a[n]], {n, 150}]

%o (PARI) A110177(n) = { my(x=sigma(n)); sum(k=1,n-1,(x == (sigma(k)+sigma(n-k)))); }; \\ _Antti Karttunen_, Feb 20 2023

%Y Cf. A110176 (least k such that sigma(n)=sigma(k)+sigma(n-k)).

%Y Cf. also A110174.

%K nonn

%O 1,3

%A _T. D. Noe_, Jul 15 2005