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A128677
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Least k>p such that (kp)^3 divides (p-1)^(kp)^2+1 for prime p = A000040(n).
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20
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19, 41, 29, 23, 79, 41617, 20939, 47, 40427, 4093, 4441, 2543, 1033, 659, 2612032921, 394502321, 14958421, 17957, 569, 14747, 12641, 167, 174263, 100493, 285629
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OFFSET
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2,1
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COMMENTS
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For every prime p>2, p^3 divides (p-1)^(p^2)+1 and furthermore p divides all numbers n>1 such that n^3 divides (p-1)^(n^2)+1.
Some further terms: a(28)-a(36) = {857, 3271, 7243979, 509, 263, 43019, 38921, 2683, 312055091}. a(38)-a(43) = {7499, 88588425539, 9689, 359, 1087, 383}. a(45)-a(61) = {931417, 40597, 2111, 2677, 14983, 261061, 1302937, 479, 17935703, 503, 4227137, 39398453, 2153, 1627, 1109, 28663, 1699}. a(63)-a(69) = {1229, 1867, 78877, 500861, 1987, 62683, 2777}. a(71)-a(75) = {275884327, 719, 44041, 3122698559, 15161}. a(77)-a(80) = {907927, 202471, 5788837, 16361}.
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LINKS
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FORMULA
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a(n) = smallest prime divisor of (p-1)^(p^2)+1 other than p, where p=A000040(n).
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EXAMPLE
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MATHEMATICA
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a[n_] := Module[{p, k}, p = Prime[n]; k = p + 1;
While[! Divisible[(p - 1)^(k p)^2 + 1, (k p)^3], k++]; k];
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CROSSREFS
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KEYWORD
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hard,more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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