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A124924
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Primes p such that p^2 divides A124923((3p-1)/2) = ((3p-1)/2)^(3(p-1)/2) + 1.
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2
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OFFSET
| 1,1
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COMMENTS
| p divides A124923((3p-1)/2) for primes p in A003628. Hence this sequence is a subsequence of A003628.
Also, primes p such that (-1/8)^((p-1)/2) == -1 + 9p/2 (mod p^2).
No other terms below 10^11.
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EXAMPLE
| 5 is in this sequence because A124923((3*5-1)/2) = A124923(7) = 7^8 + 1 = 117650 is divisible by 5^2 = 25.
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MATHEMATICA
| Do[ p = Prime[n]; m = (3p-1)/2; f = PowerMod[ m, m-1, p^2 ] + 1; If[ IntegerQ[ f/p^2 ], Print[p] ], {n, 2, 10000} ]
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CROSSREFS
| Cf. A124923, A003628.
Sequence in context: A187894 A159261 A117077 * A124878 A085554 A067135
Adjacent sequences: A124921 A124922 A124923 * A124925 A124926 A124927
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KEYWORD
| hard,more,nonn
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AUTHOR
| Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 12 2006
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EXTENSIONS
| Edited by Max Alekseyev (maxale(AT)gmail.com), Jan 28 2012
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