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Numbers that are congruent to {1, 3, 4, 7} mod 8.
1

%I #18 Sep 08 2022 08:44:57

%S 1,3,4,7,9,11,12,15,17,19,20,23,25,27,28,31,33,35,36,39,41,43,44,47,

%T 49,51,52,55,57,59,60,63,65,67,68,71,73,75,76,79,81,83,84,87,89,91,92,

%U 95,97,99,100,103,105,107,108,111,113,115,116,119,121,123,124

%N Numbers that are congruent to {1, 3, 4, 7} mod 8.

%H G. C. Greubel, <a href="/A047544/b047544.txt">Table of n, a(n) for n = 1..1000</a>

%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*(1+2*x+x^2+3*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-5+i^(2*n)+i^(1-n)-i^(1+n))/4 where i=sqrt(-1).

%F a(2k) = A004767(k-1) for n>0, a(2k-1) = A047461(k). (End)

%F E.g.f.: (2 + sin(x) + (4*x - 3)*sinh(x) + (4*x - 2)*cosh(x))/2. - _Ilya Gutkovskiy_, May 29 2016

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(2)+3)*Pi/16 - log(2)/4 + sqrt(2)*log(sqrt(2)+1)/8. - _Amiram Eldar_, Dec 24 2021

%p A047544:=n->(8*n-5+I^(2*n)+I^(1-n)-I^(1+n))/4: seq(A047544(n), n=1..100); # _Wesley Ivan Hurt_, May 29 2016

%t Table[(8n-5+I^(2n)+I^(1-n)-I^(1+n))/4, {n, 80}] (* _Wesley Ivan Hurt_, May 29 2016 *)

%t LinearRecurrence[{1, 0, 0, 1, -1}, {1, 3, 4, 7, 9}, 50] (* _G. C. Greubel_, May 29 2016 *)

%o (Magma) [n : n in [0..150] | n mod 8 in [1, 3, 4, 7]]; // _Wesley Ivan Hurt_, May 29 2016

%Y Cf. A004767, A047461.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_