%I #34 Oct 28 2025 22:01:29

%S 6,10,23,27,40,44,57,61,74,78,91,95,108,112,125,129,142,146,159,163,

%T 176,180,193,197,210,214,227,231,244,248,261,265,278,282,295,299,312,

%U 316,329,333,346,350,363,367,380,384,397,401,414,418,431,435,448,452,465

%N Numbers congruent to 6 or 10 mod 17.

%C 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.

%C 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).

%H Vincenzo Librandi, <a href="/A212161/b212161.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

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

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

%F E.g.f.: ((34*x + 15)*exp(x) + 9*exp(-x))/4. - _David Lovler_, Aug 09 2022

%e Divisibility of A001844 by 17:

%e n=0: A001844(6) = 85 = 5*17 == 0 (mod 17).

%e n=2: A001844(23) = 1105 = 5*13*17 == 0 (mod 17).

%e However, 8^2 + 9^2 = 145 == 9 (mod 17) is not divisible by 17 because 8 is not a term of the present sequence.

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

%o (Magma) [1/4*(34*n+9*(-1)^n+15): n in [0..60]]; // _Vincenzo Librandi_, May 24 2012

%o (PARI) a(n) = (34*n + 9*(-1)^n + 15)/4 \\ _David Lovler_, Aug 09 2022

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

%K nonn,easy

%O 0,1

%A _Wolfdieter Lang_, May 09 2012