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%I #13 Jul 08 2015 16:55:18
%S 1,2,2,6,10,6,8,18,22,14,30,18,20,42,46,26,58,30,66,70,36,78,82,44,48,
%T 50,102,106,54,56,126,130,68,138,74,150,78,162,166,86,178,90,190,96,
%U 98,198,210,222,226,114,116,238,120,250,128,262,134,270,138,140,282,146
%N Length of periods of Euler numbers modulo prime(n).
%C As proved by Kummer, if the actual signed Euler numbers (A122045) are used, then the period is prime(n)-1 for n>1. - _T. D. Noe_, Mar 16 2007
%F a(n)=prime(n)-1 if prime(n) == 2 or 3 (mod 4)
%e A000364 modulo 5=prime(3) gives : 1,1,0,1,0,1,0,1,0,1,0,... with period (1,0) of length 2, hence a(3)=2.
%t f[n_] := Block[{p = Prime[n], t, d = Divisors[p - 1], dk, k = 1},t = Mod[Table[Abs@EulerE[2i], {i, 2, p}], p];While[dk = d[[k]];Nand @@ Equal @@@ Partition[Partition[t, dk], 2, 1], k++ ];dk];Array[f, 63] (* _Ray Chandler_, Mar 15 2007 *)
%Y Cf. A000364, A045326, A080148.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 06 2003
%E More terms from _John W. Layman_, Jul 29 2005
%E Extended by _Ray Chandler_, Mar 15 2007
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