This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A266238 #42 Dec 27 2017 03:22:10 %S A266238 1,1,9,23,151,361,2409,5783,38551,92521,616809,1480343,9868951, %T A266238 23685481,157903209,378967703,2526451351,6063483241,40423221609, %U A266238 97015731863,646771545751,1552251709801,10348344732009,24836027356823,165573515712151,397376437709161 %N A266238 a(n+1) = 2^(2*n - 1) + (-1)^n * a(n), a(1) = 1. %H A266238 Iain Fox, <a href="/A266238/b266238.txt">Table of n, a(n) for n = 1..1662</a> %H A266238 Iain Fox, <a href="/A266238/a266238.txt">Proof for generating function and recurrence relation</a> %H A266238 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,15,0,16) %F A266238 From _Colin Barker_, Dec 21 2017: (Start) %F A266238 G.f.: x*(1 + x - 6*x^2 + 8*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + x^2)). [Proved by _Iain Fox_, Dec 21 2017] %F A266238 a(n) = ((7+7*i)*(-i)^n + (7-7*i)*i^n + (-1)^(1+n)*4^n + 4^(1+n)) / 34 where i=sqrt(-1). %F A266238 a(n) = 15*a(n-2) + 16*a(n-4) for n > 4. [Proved by _Iain Fox_, Dec 21 2017] (End) %e A266238 a(4) = 2^(2*3 - 1) + (-1)^3 * 9 = 23. %p A266238 f:= gfun:-rectoproc({a(n) = 15*a(n-2) + 16*a(n-4),a(1)=1,a(2)=1,a(3)=9,a(4)=23},a(n),remember): %p A266238 map(f, [$1..50]); # _Robert Israel_, Dec 25 2017 %t A266238 RecurrenceTable[{a[1]==1,a[n+1]==2^(2n-1)+(-1)^n a[n]},a,{n,30}] (* _Harvey P. Dale_, Dec 20 2017 *) %t A266238 f[n_]:= ((7 +7I)(-I)^n + (7 -7I)*I^n +(-1)^(1 +n) 2^(2n) +2^(2 +2n))/34; Array[f, 26] (* or *) %t A266238 CoefficientList[ Series[ -(8x^3 -6x^2 +x +1)/(16x^4 +15x^2 -1), {x, 0, 25}], x] (* or *) %t A266238 LinearRecurrence[{0, 15, 0, 16}, {1, 1, 9, 23}, 26] (* _Robert G. Wilson v_, Dec 24 2017 *) %o A266238 (PARI) a=vector(10^3); a[1]=1; for(n=2, #a, a[n] = 2^(2*n-3)-(-1)^n*a[n-1]); a \\ _Altug Alkan_, Dec 20 2017 %o A266238 (PARI) first(n) = Vec(x*(1 + x - 6*x^2 + 8*x^3)/((1 - 4*x)*(1 + 4*x)*(1 + x^2)) + O(x^(n+1))) \\ _Iain Fox_, Dec 21 2017 %K A266238 nonn,easy %O A266238 1,3 %A A266238 _Ben Paul Thurston_, Dec 25 2015 %E A266238 Corrected by _Harvey P. Dale_, Dec 20 2017