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A383614 The unique sequence such that Sum_{d|n} d*a(d)^(n/d) = sigma(n)^2 for every n.

%I A383614 #16 May 19 2025 22:38:52
%S A383614 1,4,5,4,7,-10,9,-44,-23,-197,13,-845,15,-2340,-701,-9164,19,-31578,
%T A383614 21,-124979,-11355,-381326,25,-1778580,-3323,-5162265,-212899,
%U A383614 -21915630,31,-70256029,33,-311369996,-4439583,-1010580635,-129393,-4135827284,39,-14467258386
%N A383614 The unique sequence such that Sum_{d|n} d*a(d)^(n/d) = sigma(n)^2 for every n.
%C A383614 Replace the sequence A072861 on the right-hand side of the equation with any integer sequence. It can be proved that the resulting sequence {a(n)} contain only integer terms if and only if for any prime p and positive integer n such that val(n, p) = k, p^k divides s(n) - s(n/p). The simplest sequence satisfying this property is A000203, and the resulting sequence {a(n)} is the constant sequence of 1's.
%C A383614 In 2025 China Team Selection Test, Test 4, Day 1, Problem 1, this sequence gives {z(n)} when {x(n)} and {y(n)} are constant sequences of 1's.
%H A383614 Art of Problem Solving, <a href="https://artofproblemsolving.com/community/c6h3537503p34399841">2025 China Team Selection Test, Test 4, Day 1, Problem 1</a>
%F A383614 For prime p, a(p) = p + 2.
%e A383614 For n = 1, the equation gives a(1) = sigma(1)^2 = 1;
%e A383614 For n = 6, the equation gives 1*1^6 + 2*4^3 + 3*5^2 + 6*a(6) = sigma(6)^2 = 144, so a(6) = -10.
%o A383614 (PARI) lista(nn) = {my(v=vector(nn)); v[1] = 1; for(n=2, nn, s = sigma(n)^2; fordiv(n, d, s -= d*v[d]^(n/d)); v[n]=s/n); v}
%Y A383614 Cf. A000203, A072861.
%K A383614 sign,easy
%O A383614 1,2
%A A383614 _Yifan Xie_, May 02 2025