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A337157 a(n) is the smallest m such that A054024(m) = prime(n), where A054024(m) is A000203(m) mod m, or -1 if there is no such m.

%I A337157 #19 Jan 29 2021 14:21:36
%S A337157 20,4,-1,8,21,27,39,36,57,115,32,155,63,50,129,235,265,371,305,201,98,
%T A337157 365,237,171,245,291,485,309,325,327,128,189,279,917,1507,1529,242,
%U A337157 785,489,835,865,1211,385,605,579,324,338,2321,669,1115,687,1165,399,1936
%N A337157 a(n) is the smallest m such that A054024(m) = prime(n), where A054024(m) is A000203(m) mod m, or -1 if there is no such m.
%C A337157 From _Bernard Schott_, Jan 28 2021: (Start)
%C A337157 a(n) > 0 when n <> 3.
%C A337157 Proof: When n = 1, 2, 4 then a(n) = 20, 4, 8; then, following Goldbach conjecture when p = prime(n) >=  11 with p-1 = p1 + p2, then sigma(p1*p2) = 1+p1+p2+p1*p2 == 1+p1+p2 = p (mod p1*p2); hence, for each n>= 5, a(n) exists and a(n) <= p1*p2 (see examples).
%C A337157 Now, when n = 3, no k is known such that sigma(k) == 5 (mod k)? (End)
%H A337157 Michel Marcus, <a href="/A337157/b337157.txt">Table of n, a(n) for n = 1..4000</a>
%e A337157 For prime(5) = 11, 11-1=3+7 with 3*7 = 21, so a(5) <= 21 and a(5) = 21.
%e A337157 For prime(6) = 13, 13-1=5+7 with 5*7 = 35, so a(6) <= 35 but sigma(27) = 40 == 13 (mod 27), hence a(6)=27.
%o A337157 (PARI) f(n) = sigma(n) % n; \\ A054024
%o A337157 a(n) = if (n==3, return (-1)); my(k=1, p=prime(n)); while (f(k) != p, k++); k;
%Y A337157 Cf. A000203, A054024, A308150.
%K A337157 sign
%O A337157 1,1
%A A337157 _Michel Marcus_, Jan 28 2021