|
Search: id:A081728
|
|
|
| A081728 |
|
Length of periods of Euler numbers modulo prime(n). |
|
+0
1
|
|
| 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, 50, 102, 106, 54, 56, 126, 130, 68, 138, 74, 150, 78, 162, 166, 86, 178, 90, 190, 96, 98, 198, 210, 222, 226, 114, 116, 238, 120, 250, 128, 262, 134, 270, 138, 140, 282, 146
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
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 (noe(AT)sspectra.com), Mar 16 2007
|
|
FORMULA
|
a(n)=prime(n)-1 if prime(n) == 2 or 3 (mod 4)
|
|
EXAMPLE
|
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.
|
|
MATHEMATICA
|
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] (*Chandler*)
|
|
CROSSREFS
|
Cf. A000364, A045326, A080148.
Adjacent sequences: A081725 A081726 A081727 this_sequence A081729 A081730 A081731
Sequence in context: A034805 A051765 A077063 this_sequence A080460 A077017 A127404
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 06 2003
|
|
EXTENSIONS
|
More terms from John W. Layman (layman(AT)math.vt.edu), Jul 29 2005
Extended by Ray Chandler (rayjchandler(AT)sbcglobal.net), Mar 15 2007
|
|
|
Search completed in 0.002 seconds
|
