

A212160


Numbers congruent to 2 or 10 modulo 13.


6



2, 10, 15, 23, 28, 36, 41, 49, 54, 62, 67, 75, 80, 88, 93, 101, 106, 114, 119, 127, 132, 140, 145, 153, 158, 166, 171, 179, 184, 192, 197, 205, 210, 218, 223, 231, 236, 244, 249, 257, 262, 270, 275, 283, 288, 296, 301, 309, 314, 322, 327, 335, 340, 348
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OFFSET

0,1


COMMENTS

A001844(N) = N^2 + (N+1)^2 = 4*A000217(N) + 1 is divisible by 13 if and only if N=a(n), n>=0. For the proof it suffices to show that only N=2 and N=10 from {0,1,..,12} satisfy A001844(N)== 0 (mod 13). Note that only primes of the form p= 4*k+1 (A002144) can be divisors of A001844 (see a Wolfdieter Lang comment there giving the reference).
Partial sums of the sequence [2,5,8,5,8,5,8,5,8,...] (see the o.g.f., and subtract 2 to see the 5,8 periodicity).


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000


FORMULA

Bisection: a(2*n) = 13*n + 2, a(2*n+1) = 13*n + 10, n>=0.
O.g.f.: (2 + 8*x + 3*x^2)/((1x)*(1x^2)).
a(n) = (1/4)*(26*n  3*(1)^n + 11).  Paolo P. Lava, May 16 2012


EXAMPLE

Divisibility of A001844 by 13:
n=0: A001844(2) = 13 == 0 (mod 13).
n=3: A001844(23) = 1105 = 85*13 == 0 (mod 13).
However, 8^2 + 9^2 = 145 == 2 (mod 13) is not divisible by 13 because 8 is not a member of the present sequence.


MAPLE

A212160:=n>(26*n3*(1)^n+11)/4; seq(A212160(n), n=0..100); # Wesley Ivan Hurt, Feb 26 2014


MATHEMATICA

Table[1/4*(26*n3*(1)^n+11), {n, 0, 60}] (* Vincenzo Librandi, May 24 2012 *)


PROG

(MAGMA) [1/4*(26*n3*(1)^n+11): n in [0..60]]; // Vincenzo Librandi, May 24 2012


CROSSREFS

Cf. A047219 (case p=5).
Sequence in context: A299378 A031222 A272041 * A134861 A063610 A181474
Adjacent sequences: A212157 A212158 A212159 * A212161 A212162 A212163


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, May 09 2012


STATUS

approved



