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A006517 Numbers k such that k divides 2^k + 2.
(Formerly M1719)
21
1, 2, 6, 66, 946, 8646, 180246, 199606, 265826, 383846, 1234806, 3757426, 9880278, 14304466, 23612226, 27052806, 43091686, 63265474, 66154726, 69410706, 81517766, 106047766, 129773526, 130520566, 149497986, 184416166, 279383126 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
All terms greater than 1 are even. If an odd term n>1 exists then n = m*2^k + 1 for some k>=1 and odd m. Then n divides 2^(m*2^k) + 1 and so does every prime factor p of n, implying that 2^(k+1) divides the multiplicative order of 2 modulo p and thus p-1. Therefore n = m*2^k + 1 is the product of prime factors of the form t*2^(k+1) + 1, implying that n-1 is divisible by 2^(k+1), a contradiction. - Max Alekseyev, Mar 16 2009
The sequence is infinite. In fact, its intersection with A055685 (given by A219037) is infinite (see Li et al. link). - Max Alekseyev, Oct 11 2012
All terms greater than 6 have at least three distinct prime factors. - Robert Israel, Aug 21 2014
REFERENCES
R. Honsberger, Mathematical Gems, M.A.A., 1973, p. 142.
W. Sierpiński, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #18
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Kin Y. Li et al., Solution to Problem 323, Mathematical Excalibur 14(2), 2009, p. 3.
MATHEMATICA
Do[ If[ PowerMod[ 2, n, n ] + 2 == n, Print[n]], {n, 2, 1500000000, 4} ]
Join[{1}, Select[Range[28*10^7], PowerMod[2, #, #]==#-2&]] (* Harvey P. Dale, Aug 13 2018 *)
PROG
(PARI) is_A006517(n)=!(Mod(2, n)^n+2) \\ M. F. Hasler, Oct 08 2012
CROSSREFS
Sequence in context: A082619 A046399 A082617 * A217630 A091458 A335934
KEYWORD
nonn,nice
AUTHOR
EXTENSIONS
Corrected and extended by Joe K. Crump (joecr(AT)carolina.rr.com), Sep 12 2000 and Robert G. Wilson v, Sep 13 2000
STATUS
approved

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Last modified March 19 04:58 EDT 2024. Contains 370952 sequences. (Running on oeis4.)