

A319233


Numbers k such that k^2 + 1 divides 2^k + 4.


1



0, 1, 8, 28, 32, 128, 2048, 8192, 23948, 131072, 524288, 8388608, 536870912, 2147483648, 137438953472
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OFFSET

1,3


COMMENTS

This sequence corresponds to numbers k such that k^2 + 1 divides 2^k + 2^m where m = 2 (A247220 (m = 0), A319216 (m = 1)).
a(16) > 10^12.  Hiroaki Yamanouchi, Sep 17 2018


LINKS

Table of n, a(n) for n=1..15.


EXAMPLE

32 = 2^5 is a term since (2^(2^5) + 2^2)/((2^5)^2 + 1) = 2^22  2^12 + 2^2.


PROG

(PARI) isok(n)=Mod(2, n^2+1)^n==4;


CROSSREFS

Cf. A034785, A247220, A319216.
Sequence in context: A201105 A184614 A006377 * A323198 A121739 A166729
Adjacent sequences: A319230 A319231 A319232 * A319234 A319235 A319236


KEYWORD

nonn,more


AUTHOR

Altug Alkan, Sep 14 2018


EXTENSIONS

a(15) from Hiroaki Yamanouchi, Sep 17 2018


STATUS

approved



