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A341860
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a(n) is the least k such that A341857(k) = psi(k^2)/k = n, or 0 if no such k exists.
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2
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1, 3, 14, 5, 22, 7, 116, 32, 38, 11, 46, 13, 212, 58, 62, 17, 618, 19, 2748, 25, 86, 23, 94, 219, 404, 106, 162, 29, 118, 31, 4476, 128, 134, 309, 142, 37, 596, 1374, 158, 41, 166, 43, 692, 115, 418, 47, 1698, 119, 788, 202, 206, 53, 214, 81, 242, 226, 916, 59
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) > 0 for all n.
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.
For odd n > 1, if a(n) > 0, then a(2n) <= a(n)/2. Proof:
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.
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.
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.
The smallest odd n such that a(2n) < a(n)/2 is n = 71, where a(71) = 10236 and a(142) = 2276.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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A341860[n_] := Block[{k = 0}, While[CarmichaelLambda[++k^2]/k != n]; k];
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PROG
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(PARI) a(n) = for(k=1, oo, if(A341857(k)==n, return(k))) \\ Suppose that a(n) > 0. See A341857 for its program
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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