|
Search: id:A124240
|
|
| |
|
| 1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 40, 42, 48, 54, 60, 64, 72, 80, 84, 96, 100, 108, 120, 126, 128, 144, 156, 160, 162, 168, 180, 192, 200, 216, 220, 240, 252, 256, 272, 288, 294, 300, 312, 320, 324, 336, 342, 360, 378, 384, 400, 420, 432, 440, 468, 480
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
A124239[n] = Sum[ Sum[ (2k-1)^m, {m,1,n} ], {k,1,n} ] = n + Sum[ (2k-1)((2k-1)^n-1) / (2(k-1)), {k,2,n} ]. If k is in the sequence then 2k is also in the sequence, but if 2m is in the sequence m isn't necessary a term of the sequence. It appears that a(n) almost coincides with A068563[n] Numbers n such that 2^n (mod n) = 4^n (mod n). The first term that is different is A068563[27] = 136. The terms of A068563[n] that are not the terms of a(n) are listed in A124241[n] = {136,408,620,680,820,...}.
It appears that this sequence (except for the initial 1) is the same as the sequence of numbers n such that p-1 divides n for all primes p that divide n. - Leroy Quet, Jun 27 2008. Maximilian Hasler and Edwin Clark have confirmed that this is true for n <= 2500. Is there a proof?
|
|
EXAMPLE
|
A124239[n] begins {1, 14, 197, 3704, 90309, 2704470, 95856025, 3921108576, ...}.
Thus a(0) = 1 because 1 divides A124239[1] = 1.
a(1) = 2 because 2 divides A124239[2] = 14.
a(3) = 4 because 4 divides A124239[4] = 3704, but 3 does not divide A124239[3] = 197.
|
|
MATHEMATICA
|
Do[f=n + Sum[ (2k-1)((2k-1)^n-1) / (2(k-1)), {k, 2, n} ]; If[IntegerQ[f/n], Print[n]], {n, 1, 900}]
|
|
CROSSREFS
|
Cf. A124239, A124241, A068563.
Sequence in context: A092903 A005153 A068563 this_sequence A068997 A067712 A060765
Adjacent sequences: A124237 A124238 A124239 this_sequence A124241 A124242 A124243
|
|
KEYWORD
|
nonn,new
|
|
AUTHOR
|
Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 22 2006
|
|
|
Search completed in 0.002 seconds
|
