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A308935
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a(n) is the smallest m > n such that n^2*(n^2 + 1) divides m^2*(m^2 + 1).
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1
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2, 8, 12, 64, 18, 216, 35, 112, 360, 818, 660, 348, 208, 2744, 693, 4096, 493, 450, 3420, 4832, 1071, 2112, 1242, 13824, 7800, 17576, 1998, 4368, 10133, 1560, 1178, 1280, 3597, 3060, 8582, 46656, 5032, 1292, 29640, 12768, 1189, 14868, 3182, 13112, 36468, 6670
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OFFSET
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1,1
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COMMENTS
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For any n > 0, a(n) exists as n^2*(n^2+1) divides (n^3)^2*((n^3)^2+1).
Tsz Ho Chan proved that a(n) >> n*log(n)^(1/8)/log(log(n))^12.
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LINKS
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FORMULA
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a(n) <= n^3.
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EXAMPLE
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For n = 2:
- hence a(2) = 8.
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MATHEMATICA
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a[n_] := With[{n2 = n^2(n^2+1)}, For[m = n+1, True, m++, If[Divisible[ m^2(m^2+1), n2], Print[n, " ", m]; Return[m]]]];
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PROG
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(PARI) a(n, f = x->x^2*(x^2+1)) = my (fn=f(n)); for (m=n+1, oo, if (f(m)%fn==0, return (m)))
(Python)
n2, m, m2 = n**2*(n**2+1), n+1, ((n+1)**2*((n+1)**2+1)) % (n**2*(n**2+1))
while m2:
m2, m = (m2 + 2*(2*m+1)*(m**2+m+1)) % n2, (m+1) % n2
(Magma) a:=[]; for n in [1..50] do m:=n+1; while not IsIntegral( (m^2*(m^2 + 1))/(n^2*(n^2 + 1) )) do m:=m+1; end while; Append(~a, m); end for; a; // Marius A. Burtea, Dec 20 2019
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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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