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
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
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, May 09 2012
STATUS
approved