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%I #30 Sep 08 2022 08:45:15
%S 1,3,15,45,225,675,3375,10125,50625,151875,759375,2278125,11390625,
%T 34171875,170859375,512578125,2562890625,7688671875,38443359375,
%U 115330078125,576650390625,1729951171875,8649755859375,25949267578125
%N A modular recurrence.
%C Interpolated zeros suppressed.
%C The inverse mod 2 binomial transform of 2^n is 1,1,3,3,15,15,... (A100735).
%H G. C. Greubel, <a href="/A100747/b100747.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) = (5 - 2*(n/2 mod 2))b(n-2).
%F a(n) = A101553(2*(n+1))/5.
%F a(2*n) = 15^n, a(2*n+1) = 3 * 15^n. - _Ralf Stephan_, May 16 2007
%F O.g.f.: (1+3*x)/(1-15*x^2). - _R. J. Mathar_, Feb 04 2008
%p a:=n->mul(4+(-1)^j,j=1..n):seq(a(n),n=0..23); # _Zerinvary Lajos_, Dec 13 2008
%t LinearRecurrence[{0, 15}, {1, 3}, 50] (* or *) RecurrenceTable[{a[n] == (5 - 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+3*x)/(1-15*x^2)) \\ _G. C. Greubel_, Apr 16 2018
%o (Magma) I:=[1,3]; [n le 2 select I[n] else 15*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Apr 16 2018
%Y Cf. A101553.
%Y Bisection of A100735.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Dec 06 2004
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