%I #16 Sep 08 2022 08:44:57

%S 3,5,6,7,11,13,14,15,19,21,22,23,27,29,30,31,35,37,38,39,43,45,46,47,

%T 51,53,54,55,59,61,62,63,67,69,70,71,75,77,78,79,83,85,86,87,91,93,94,

%U 95,99,101,102,103,107,109,110,111,115,117,118,119,123,125

%N Numbers that are congruent to {3, 5, 6, 7} mod 8.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).

%F From _Wesley Ivan Hurt_, May 29 2016: (Start)

%F G.f.: x*(3+2*x+x^2+x^3+x^4) / ((x-1)^2*(1+x+x^2+x^3)).

%F a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.

%F a(n) = (8*n+1-i^(2*n)-(2-i)*i^(-n)-(2+i)*i^n)/4 where i=sqrt(-1).

%F a(2k) = A047550(k), a(2k-1) = A047398(k). (End)

%F E.g.f.: (2 + sin(x) - 2*cos(x) + sinh(x) + 4*x*exp(x))/2. - _Ilya Gutkovskiy_, May 30 2016

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (3*sqrt(2)-2)*Pi/16 - log(2)/8 + sqrt(2)*log(3-2*sqrt(2))/16. - _Amiram Eldar_, Dec 26 2021

%p A047582:=n->(8*n+1-I^(2*n)-(2-I)*I^(-n)-(2+I)*I^n)/4: seq(A047582(n), n=1..100); # _Wesley Ivan Hurt_, May 29 2016

%t Table[(8n+1-I^(2n)-(2-I)*I^(-n)-(2+I)*I^n)/4, {n, 80}] (* _Wesley Ivan Hurt_, May 29 2016 *)

%o (Magma) [n : n in [0..150] | n mod 8 in [3, 5, 6, 7]]; // _Wesley Ivan Hurt_, May 29 2016

%Y Cf. A047398, A047550.

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_