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%I #70 Jul 14 2023 05:21:08
%S 0,1,34,1155,39236,1332869,45278310,1538129671,52251130504,
%T 1775000307465,60297759323306,2048348816684939,69583562007964620,
%U 2363792759454112141,80299370259431848174,2727814796061228725775,92665403695822344828176,3147895910861898495432209
%N a(n) = Pell(4n) / Pell(4).
%C A000129(k*n)/A000129(k) = ((sqrt(2)-1)^k(-1)^k-(sqrt(2)+1)^k)((sqrt(2)-1)^(k*n)(-1)^(k*n)-(sqrt(2)+1)^(k*n))/((sqrt(2)-1)^(2k)+(sqrt(2)+1)^(2k)-2(-1)^k).
%C All squares of the form (3m-1)^3 + (3m)^3 + (3m+1)^3 (cf. A116108) are given for m = 24 b, where b is a square of this sequence. From Ribenboim & McDaniel, it follows there are no squares > 1 in this sequence. - _M. F. Hasler_, Jun 05 2007
%C A divisibility sequence, cf. R. K. Guy's post to the SeqFan list. - _M. F. Hasler_, Feb 05 2013
%C a(n) gives all nonnegative solutions of the Pell equation b(n)^2 - 32*(3*a(n))^2 = +1, together with b(n) = A056771(n). - _Wolfdieter Lang_, Mar 09 2019
%H M. F. Hasler, <a href="/A091761/b091761.txt">Table of n, a(n) for n = 0..99</a>
%H R. K. Guy, <a href="http://list.seqfan.eu/oldermail/seqfan/2013-February/010759.html">A new sequence</a>, post to the SeqFan list, Feb 05 2013.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Paulo Ribenboim and Wayne L. McDaniel, <a href="http://dx.doi.org/10.1006/jnth.1996.0068">The Square Terms in Lucas Sequences</a>, Journal of Number Theory 58, 104-123 (1996).
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (34,-1).
%F G.f.: x/(1-34*x+x^2).
%F a(n) = A000129(4n)/A000129(4).
%F a(n) = ((1+sqrt(2))^(4n) - (1-sqrt(2))^(4n))*sqrt(2)/48.
%F From _M. F. Hasler_, Jun 05 2007: (Start)
%F a(n) = n (mod 2^m) for any m >= 0.
%F a(n) = sinh(4n*log(sqrt(2)+1)/(12 sqrt(2)).
%F a(n) = A[1,1], first element of the 2 X 2 matrix A = (34,1;-1,0)^(n-1). (End)
%F a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=1. - _Philippe Deléham_, Nov 03 2008
%F A029547(n) = a(n+1). - _M. F. Hasler_, Feb 05 2013
%F a(n) = sqrt((A056771(n)^2 - 1)/(32*9)), n >= 0. See the Pell remark above. - _Wolfdieter Lang_, Mar 11 2019
%F E.g.f.: exp(17*x)*sinh(12*sqrt(2)*x)/(12*sqrt(2)). - _Stefano Spezia_, Apr 16 2023
%F a(n) = A002965(8*n)/12. - _Gerry Martens_, Jul 14 2023
%p with (combinat):seq(fibonacci(4*n,2)/12, n=0..17); # _Zerinvary Lajos_, Apr 21 2008
%t LinearRecurrence[{34,-1}, {0,1}, 20] (* _G. C. Greubel_, Mar 11 2019 *)
%o (PARI) A091761(n, x=[ -1,17],A=[17,72*4;1,17]) = vector(n,i,(x*=A)[1]) \\ _M. F. Hasler_, May 26 2007
%o (PARI) A091761(n)=([34,1;-1,0]^(n-1))[1,1] \\ _M. F. Hasler_, Jun 05 2007
%o (Sage) [lucas_number1(n,34,1) for n in range(0, 16)]# _Zerinvary Lajos_, Nov 07 2009
%o (Magma) I:=[0,1]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // _G. C. Greubel_, Mar 11 2019
%Y A029547 is an essentially identical sequence, cf. formula.
%Y Cf. A000129, A001109, A002965, A029547, A041085, A056771, A116108.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Feb 06 2004
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