

A127931


Numbers n such that 13 divides 11*n + 2^n.


3



1, 2, 6, 9, 23, 29, 70, 72, 103, 112, 128, 147, 157, 158, 162, 165, 179, 185, 226, 228, 259, 268, 284, 303, 313, 314, 318, 321, 335, 341, 382, 384, 415, 424, 440, 459, 469, 470, 474, 477, 491, 497, 538, 540, 571, 580, 596, 615, 625, 626, 630, 633, 647, 653
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OFFSET

1,2


COMMENTS

Sequence is infinite: starting with 13th term, a(13)=157, a(i)=a(i12)+156. In general, for p and p2 both prime, starting with pth term, a(i(p1))+p(p1). This particular sequence corresponds to the case p=13.
First differences have period 12.  Charles R Greathouse IV, Oct 11 2013


LINKS

G. C. Greubel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,1,1).


MATHEMATICA

Select[Range[700], Divisible[11#+2^#, 13]&] (* or *) LinearRecurrence[ {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {1, 2, 6, 9, 23, 29, 70, 72, 103, 112, 128, 147, 157}, 60] (* Harvey P. Dale, Sep 03 2016 *)


PROG

(PARI) isok(n) = ((11*n + 2^n) % 13) == 0; \\ Michel Marcus, Oct 11 2013


CROSSREFS

Cf. A125957.
Sequence in context: A047161 A318193 A048083 * A077237 A056879 A006746
Adjacent sequences: A127928 A127929 A127930 * A127932 A127933 A127934


KEYWORD

nonn,easy


AUTHOR

Zak Seidov, Feb 07 2007, Feb 09 2007


STATUS

approved



