%I #35 Feb 28 2023 02:59:59
%S 31,106,168,243,305,380,442,517,579,654,716,791,853,928,990,1065,1127,
%T 1202,1264,1339,1401,1476,1538,1613,1675,1750,1812,1887,1949,2024,
%U 2086,2161,2223,2298,2360,2435,2497,2572,2634,2709,2771,2846,2908,2983
%N a(n) = 137*(n-1) - a(n-1) with n>1, a(1)=31.
%C Positive numbers k such that k^2 == 2 (mod 137), where the prime 137 == 1 (mod 8).
%C Equivalently, numbers k such that k == 31 or 106 (mod 137).
%C The subsequence of primes begins: 31, 853, 1613, 1949, 2161. - _Jonathan Vos Post_, Mar 09 2012
%H Vincenzo Librandi, <a href="/A206526/b206526.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n) = a(n-2) + 137.
%F G.f.: x*(31+75*x+31*x^2)/((1+x)*(x-1)^2).
%F a(n) = (-137+13*(-1)^n+274*n)/4.
%F a(n) = a(n-1)+a(n-2)-a(n-3).
%F Sum_{n>=1} (-1)^(n+1)/a(n) = cot(31*Pi/137)*Pi/137. - _Amiram Eldar_, Feb 28 2023
%t LinearRecurrence[{1, 1, -1}, {31, 106, 168}, 40] (* or *) CoefficientList[Series[x*(31+75*x+31*x^2)/((1+x)*(x-1)^2), {x, 0, 50}], x] (* or *) a[1] = 31; a[n_] := a[n] = 137*(n-1) - a[n-1]; Table[a[n], {n, 1, 40}]
%o (Magma) [(-137+13*(-1)^n+274*n)/4: n in [1..60]];
%o (Magma) [n: n in [1..3000] | n^2 mod 137 eq 2]; // _Vincenzo Librandi_, Mar 31 2016
%Y Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k+1: A155449, A158803, A159007, A159008, A176010, A206525.
%Y Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k-1: A047341, A155450, A164131, A164135, A167533, A167534, A177044, A177046, A204769.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 09 2012