|
%I #24 Sep 08 2022 08:46:20
%S 1,1,7,27,117,493,2091,8855,37513,158905,673135,2851443,12078909,
%T 51167077,216747219,918155951,3889371025,16475640049,69791931223,
%U 295643364939,1252365390981,5305104928861,22472785106427,95196245354567,403257766524697,1708227311453353,7236167012338111,30652895360805795,129847748455561293,550043889183050965
%N a(n) = 3*a(n-1) + 5*a(n-2) + a(n-3), with a(0) = a(1) = 1 and a(2) = 7, a linear recurrence which is a trisection of A005252.
%H Hermann Stamm-Wilbrandt, <a href="/A294262/a294262_1.svg">6 interlaced bisections</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,5,1).
%F G.f.: (1 - 2*x - x^2)/(1 - 3*x - 5*x^2 - x^3).
%F a(n) = (1/20)*(10*(-1)^n + (2-sqrt(5))^n*(5-sqrt(5)) + (2+sqrt(5))^n*(5+sqrt(5))).
%F a(n) = A005252(3*n).
%F a(n) = 4*a(n-1) + a(n-2) + 2*(-1)^n for n >= 2.
%F a(n) = Sum_{k=0..floor(3*n/4)} binomial(3*n-2*k, 2*k).
%F a(n) = A110679(n) - A001076(n).
%F a(n) = (Fibonacci(3*n + 1) + (-1)^n)/2.
%F a(2*n) = A232970(2*n); a(2*n+1) = A049651(2*n+1). See "6 interlaced bisections" link. - _Hermann Stamm-Wilbrandt_, Apr 18 2019
%t LinearRecurrence[{3,5,1},{1,1,7},30]
%o (bc)
%o a=1
%o b=1
%o c=7
%o print 0," ",a,"\n"
%o print 1," ",b,"\n"
%o print 2," ",c,"\n"
%o for(x=3;x<=1000;++x){
%o d=3*c+5*b+1*a
%o print x," ",d,"\n"
%o a=b
%o b=c
%o c=d
%o } # _Hermann Stamm-Wilbrandt_, Apr 18 2019
%o (PARI) {a(n) = (fibonacci(3*n+1) +(-1)^n)/2}; \\ _G. C. Greubel_, Apr 19 2019
%o (Magma) [(Fibonacci(3*n+1) +(-1)^n)/2 : n in [0..30]]; // _G. C. Greubel_, Apr 19 2019
%o (Sage) [(fibonacci(3*n+1) +(-1)^n)/2 for n in (0..30)] # _G. C. Greubel_, Apr 19 2019
%Y Cf. A001076, A005252, A110679.
%K easy,nonn
%O 0,3
%A _Jean-François Alcover_ and _Paul Curtz_, Oct 26 2017
|
