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%I #13 Mar 11 2024 08:28:33
%S 1,3,14,5,22,7,116,32,38,11,46,13,212,58,62,17,618,19,2748,25,86,23,
%T 94,219,404,106,162,29,118,31,4476,128,134,309,142,37,596,1374,158,41,
%U 166,43,692,115,418,47,1698,119,788,202,206,53,214,81,242,226,916,59
%N a(n) is the least k such that A341857(k) = psi(k^2)/k = n, or 0 if no such k exists.
%C Conjecture: a(n) > 0 for all n.
%C If there exists k such that psi(k^2) | k*n and k*n+1 is prime, then A341857(k*(k*n+1)) = n, so a(n) > 0.
%C For odd n > 1, if a(n) > 0, then a(2n) <= a(n)/2. Proof:
%C Suppose k is the least integer such that A341857(k) = psi(k^2)/k = n is odd. Note that if m > 1 is odd or m is divisible by 8, then A341857(m) is even. So we must have k = 2t or k = 4t with odd t > 1.
%C If k = 2t, then n = psi(4t^2)/(2t) = psi(t^2)/(2t), so A341857(t) = psi(t^2)/t = 2n. This gives a(2n) <= t = k/2.
%C if k = 4t, then n = psi(16t^2)/(4t) = lcm(4, psi(t^2))/(4t). There are two cases: a) if psi(t^2) is divisible by 4, then psi(t^2)/(4t) = n, so A341857(2t) = psi(4t^2)/(2t) = psi(t^2)/(2t) = 2n. This gives a(2n) <= 2t = k/2. b) if psi(t^2) is not divisible by 4, then psi(t^2)/(2t) = n, so A341857(2t) = psi(4t^2)/(2t) = psi(t^2)/(2t) = n. This gives a(n) <= 2t = k/2, contradicting with minimality of k. QED.
%C The smallest odd n such that a(2n) < a(n)/2 is n = 71, where a(71) = 10236 and a(142) = 2276.
%H Jianing Song, <a href="/A341860/b341860.txt">Table of n, a(n) for n = 1..500</a>
%e A341857(116) = psi(116^2)/116 = 812/116 = 7, and there is no k < 116 such that psi(k^2)/k = 7, so a(7) = 116.
%t A341860[n_] := Block[{k = 0}, While[CarmichaelLambda[++k^2]/k != n]; k];
%t Array[A341860, 100] (* _Paolo Xausa_, Mar 11 2024 *)
%o (PARI) a(n) = for(k=1, oo, if(A341857(k)==n, return(k))) \\ Suppose that a(n) > 0. See A341857 for its program
%Y Cf. A341857.
%K nonn
%O 1,2
%A _Jianing Song_, Feb 21 2021
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