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

0,1

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).

Links

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

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

Example

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.

Mathematica

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

Prog

(Magma) [1/4*(34*n+9*(-1)^n+15): n in [0..60]]; // Vincenzo Librandi, May 24 2012
(PARI) a(n) = (34*n + 9*(-1)^n + 15)/4 \\ David Lovler, Aug 09 2022

Crossrefs

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

Sequence in context: A166305 A047651 A079426 * A132994 A199885 A024499

Adjacent sequences: A212158 A212159 A212160 * A212162 A212163 A212164

Keyword

nonn,easy

Author

Wolfdieter Lang, May 09 2012

Status

approved