cp's OEIS Frontend

A212161 Numbers congruent to 6 or 10 mod 17.

%I A212161 #31 Mar 17 2024 02:16:19
%S A212161 6,10,23,27,40,44,57,61,74,78,91,95,108,112,125,129,142,146,159,163,
%T A212161 176,180,193,197,210,214,227,231,244,248,261,265,278,282,295,299,312,
%U A212161 316,329,333,346,350
%N A212161 Numbers congruent to 6 or 10 mod 17.
%C A212161 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 A212161 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 A212161 Vincenzo Librandi, <a href="/A212161/b212161.txt">Table of n, a(n) for n = 0..1000</a>
%F A212161 Bisection: a(2*n) = 17*n + 6, a(2*n+1) = 17*n + 10, n >= 0.
%F A212161 O.g.f.: (6 + 4*x + 7*x^2)/((1-x)*(1-x^2)).
%F A212161 E.g.f.: ((34*x + 15)*exp(x) + 9*exp(-x))/4. - _David Lovler_, Aug 09 2022
%e A212161 Divisibility of A001844 by 17:
%e A212161 n=0: A001844(6) = 85 = 5*17 == 0 (mod 17).
%e A212161 n=2: A001844(23) = 1105 = 5*13*17 == 0 (mod 17).
%e A212161 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 A212161 Table[1/4*(34*n+9*(-1)^n+15),{n,0,60}] (* _Vincenzo Librandi_, May 24 2012 *)
%o A212161 (Magma) [1/4*(34*n+9*(-1)^n+15): n in [0..60]]; // _Vincenzo Librandi_, May 24 2012
%o A212161 (PARI) a(n) = (34*n + 9*(-1)^n + 15)/4 \\ _David Lovler_, Aug 09 2022
%Y A212161 Cf. A047219 (p=5), A212160 (p=13).
%K A212161 nonn,easy
%O A212161 0,1
%A A212161 _Wolfdieter Lang_, May 09 2012