A212161 Numbers congruent to 6 or 10 mod 17.

6, 10, 23, 27, 40, 44, 57, 61, 74, 78, 91, 95, 108, 112, 125, 129, 142, 146, 159, 163, 176, 180, 193, 197, 210, 214, 227, 231, 244, 248, 261, 265, 278, 282, 295, 299, 312, 316, 329, 333, 346, 350

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 17 if and only if N = a(n), n >= 0. For the proof it suffices to show that only N=6 and N=10 from {0,1,...,16} satisfy A001844(N) == 0 (mod 17). 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). Note also that if N^2 + (N+1)^2 == 0 (mod p), with any prime p (necessarily from A002144), then also p-1-N satisfies this congruence. This explains why 10 = 17-1-6 is the (incongruent) companion of 6.

Partial sums of the sequence 6,4,13,4,13,4,13,4,13,4,13,... (see the o.g.f., and subtract 6 to see the remaining 4, 13=17-4 periodicity).

Examples

			Divisibility of A001844 by 17:
n=0: A001844(6) = 85 = 5*17 == 0 (mod 17).
n=2: A001844(23) = 1105 = 5*13*17 == 0 (mod 17).
However, 8^2 + 9^2 = 145 == 9 (mod 17) is not divisible by 17 because 8 is not a term of the present sequence.
		

Links

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

Crossrefs

Cf. A047219 (p=5), A212160 (p=13).

Programs

Magma

[1/4*(34*n+9*(-1)^n+15): n in [0..60]]; // Vincenzo Librandi, May 24 2012

Mathematica

Table[1/4*(34*n+9*(-1)^n+15),{n,0,60}] (* Vincenzo Librandi, May 24 2012 *)

PARI

a(n) = (34*n + 9*(-1)^n + 15)/4 \\ David Lovler, Aug 09 2022

Formula

Bisection: a(2*n) = 17*n + 6, a(2*n+1) = 17*n + 10, n >= 0.

O.g.f.: (6 + 4*x + 7*x^2)/((1-x)*(1-x^2)).

E.g.f.: ((34*x + 15)*exp(x) + 9*exp(-x))/4. - David Lovler, Aug 09 2022