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A modular recurrence.
2

%I #29 Sep 08 2022 08:45:16

%S 1,5,15,75,225,1125,3375,16875,50625,253125,759375,3796875,11390625,

%T 56953125,170859375,854296875,2562890625,12814453125,38443359375,

%U 192216796875,576650390625,2883251953125,8649755859375,43248779296875

%N A modular recurrence.

%C Interpolated zeros suppressed.

%H G. C. Greubel, <a href="/A101553/b101553.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,15).

%F a(n) = b(2*n) where b(0)=1, b(1)=0, b(n) = (3 + 2*(n/2 mod 2))*b(n-2).

%F a(n) = A100747(2(n+1))/3.

%F a(2n) = 15^n, a(2n+1) = 5*15^n. - _Ralf Stephan_, May 16 2007

%F O.g.f.: (1+5*x)/(1-15*x^2). - _Philippe Deléham_, Dec 02 2011

%p a:=n->mul(4-(-1)^j,j=1..n):seq(a(n),n=0..23); # _Zerinvary Lajos_, Dec 13 2008

%t CoefficientList[Series[(1+5x)/(1-15x^2),{x,0,30}],x] (* or *) LinearRecurrence[ {0,15},{1,5},30] (* _Harvey P. Dale_, Oct 14 2013 *)

%t RecurrenceTable[{a[n] == (3 + 2*Mod[n/2, 2])*a[n - 2], a[0] == 1, a[1] == 0}, a, {n, 0, 50}][[1 ;; ;; 2]] (* _G. C. Greubel_, Apr 16 2018 *)

%o (PARI) x='x+O('x^30); Vec((1+5*x)/(1-15*x^2)) \\ _G. C. Greubel_, Apr 16 2018

%o (Magma) I:=[1,5]; [n le 2 select I[n] else 15*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Apr 16 2018

%K easy,nonn

%O 0,2

%A _Paul Barry_, Dec 06 2004