A212160 Numbers that are congruent to {2, 10} mod 13.

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

Offset: 0

Author

Wolfdieter Lang, May 09 2012

Keywords

• nonn
• easy

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).

Examples

			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.
		

Links

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

Crossrefs

Cf. A047219 (case p=5).

Programs

Magma

[1/4*(26*n-3*(-1)^n+11): n in [0..60]]; // Vincenzo Librandi, May 24 2012

Maple

A212160:=n->(26*n-3*(-1)^n+11)/4; seq(A212160(n), n=0..100); # Wesley Ivan Hurt, Feb 26 2014

Mathematica

Table[1/4*(26*n-3*(-1)^n+11),{n,0,60}] (* Vincenzo Librandi, May 24 2012 *)

PARI

a(n) = (26*n - 3*(-1)^n + 11)/4 \\ David Lovler, Aug 09 2022

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)/((1-x)*(1-x^2)).

E.g.f.: ((26*x + 11)*exp(x) - 3*exp(-x))/4. - David Lovler, Aug 09 2022