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a(n) = Sum{i=1..A343901(n)} n (mod A343902(n, i)^2).
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%I #6 May 07 2021 09:22:04

%S 3,0,6,2,10,8,1,3,5,3,31,7,24,23,26,2,23,22,46,6,8,30,33,37,30,32,7,

%T 10,69,12,48,18,91,47,125,42,98,19,63,67,53,71,111,3,5,26,5,57,61,64,

%U 12,7,64,71,133,14,8,16,214,85,156,79,164,87,92,91,95,120

%N a(n) = Sum{i=1..A343901(n)} n (mod A343902(n, i)^2).

%C A theorem of Pomerance says that if a(n) = 0, then A000010(x) = A000010(y) has only the solution y = x (cf. Pomerance, 1974).

%H Carl Pomerance, <a href="https://doi.org/10.1090/S0002-9939-1974-0340161-0">On Carmichael's conjecture</a>, Proceedings of the American Mathematical Society 43 (1974), 297-298.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Carmichael%27s_totient_function_conjecture">Carmichael's totient function conjecture</a>

%o (PARI) row_a343902(n) = my(e=eulerphi(n), v=[]); forprime(p=1, e, if(e%(p-1)==0, v=concat(v, [p]))); v

%o a(n) = my(v=row_a343902(n)); sum(i=1, #v, n%v[i]^2)

%Y Cf. A000010, A343901, A343902.

%K nonn

%O 3,1

%A _Felix Fröhlich_, May 03 2021