A187541 - OEIS

%I #23 Jul 06 2016 00:49:21

%S 0,4,1,12,16,20,3,28,32,36,5,44,48,52,7,60,64,68,9,76,80,84,11,92,96,

%T 100,13,108,112,116,15,124,128,132,17,140,144,148,19,156,160,164,21,

%U 172,176,180,23,188,192,196,25,204,208,212,27,220,224,228,29,236,240,244,31,252

%N a(4n+2) = 2n+1, otherwise a(n) = 4n.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,2,0,0,0,-1).

%F a(n) = 2*a(n-4) - a(n-8) for n>7.

%F G.f.: x*(4+x+12*x^2+16*x^3+12*x^4+x^5+4*x^6)/(1-x^4)^2; a(n) = (n/8)*(32 -7*(1+(-1)^n)*(1-i^n)) where i=sqrt(-1). - _Bruno Berselli_, Mar 15 2011

%F From _Paul Curtz_, Mar 22 2011: (Start)

%F A060819(n)*a(n) = 0,4,1,36,16,100, = 0,4, followed by A061038(n+2).

%F a(n) = a(n-4) + period 4: repeat [16, 16, 2, 16]. Note that a(n) = 4*n/(period 4: repeat [1, 1, 8, 1]), Hence 16's = A010855. (End)

%F a(n) = 16*n/(11+7*(I^(2*n)-I^(-n)-I^n)). - _Wesley Ivan Hurt_, Jul 05 2016

%p A187541:=n->16*n/(11+7*(I^(2*n)-I^(-n)-I^n)): seq(A187541(n), n=0..100); # _Wesley Ivan Hurt_, Jul 05 2016

%t Table[16n/(11+7*(I^(2*n)-I^(-n)-I^n)), {n, 0, 80}] (* _Wesley Ivan Hurt_, Jul 05 2016 *)

%Y Cf. A008586, A010855, A060819, A061038.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Mar 11 2011

%E Edited by _N. J. A. Sloane_, Mar 15 2011