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A212160 Numbers that are congruent to {2, 10} mod 13.

%I A212160 #26 Mar 05 2024 11:27:19
%S A212160 2,10,15,23,28,36,41,49,54,62,67,75,80,88,93,101,106,114,119,127,132,
%T A212160 140,145,153,158,166,171,179,184,192,197,205,210,218,223,231,236,244,
%U A212160 249,257,262,270,275,283,288,296,301,309,314,322,327,335,340,348
%N A212160 Numbers that are congruent to {2, 10} mod 13.
%C A212160 A001844(N) = N^2 + (N+1)^2 = 4*A000217(N) + 1 is divisible by 13 if and only if N=a(n), n>=0. For the proof it suffices to show that only N=2 and N=10 from {0,1,..,12} satisfy A001844(N)== 0 (mod 13). 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).
%C A212160 Partial sums of the sequence [2,5,8,5,8,5,8,5,8,...] (see the o.g.f., and subtract 2 to see the 5,8 periodicity).
%H A212160 Vincenzo Librandi, <a href="/A212160/b212160.txt">Table of n, a(n) for n = 0..1000</a>
%F A212160 Bisection: a(2*n) = 13*n + 2, a(2*n+1) = 13*n + 10, n>=0.
%F A212160 O.g.f.: (2 + 8*x + 3*x^2)/((1-x)*(1-x^2)).
%F A212160 E.g.f.: ((26*x + 11)*exp(x) - 3*exp(-x))/4. - _David Lovler_, Aug 09 2022
%e A212160 Divisibility of A001844 by 13:
%e A212160 n=0: A001844(2) = 13 == 0 (mod 13).
%e A212160 n=3: A001844(23) = 1105 = 85*13 == 0 (mod 13).
%e A212160 However, 8^2 + 9^2 = 145 == 2 (mod 13) is not divisible by 13 because 8 is not a member of the present sequence.
%p A212160 A212160:=n->(26*n-3*(-1)^n+11)/4; seq(A212160(n), n=0..100); # _Wesley Ivan Hurt_, Feb 26 2014
%t A212160 Table[1/4*(26*n-3*(-1)^n+11),{n,0,60}] (* _Vincenzo Librandi_, May 24 2012 *)
%o A212160 (Magma) [1/4*(26*n-3*(-1)^n+11): n in [0..60]]; // _Vincenzo Librandi_, May 24 2012
%o A212160 (PARI) a(n) = (26*n - 3*(-1)^n + 11)/4 \\ _David Lovler_, Aug 09 2022
%Y A212160 Cf. A047219 (case p=5).
%K A212160 nonn,easy
%O A212160 0,1
%A A212160 _Wolfdieter Lang_, May 09 2012