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%I #51 Dec 07 2019 12:18:24
%S 1,15,224,3345,49951,745920,11138849,166336815,2483913376,37092363825,
%T 553901543999,8271430796160,123517560398401,1844491975179855,
%U 27543862067299424,411313439034311505
%N A Chebyshev S-sequence with Diophantine property.
%C a(n) gives the general (positive integer) solution of the Pell equation b^2 - 221*a^2 = +4 with companion sequence b(n)=A078365(n+1), n>=0.
%C This is the m=17 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..16 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362 and A007655. The m=1..3 (signed) sequences are A049347, A056594, A010892.
%C For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 15's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - _John M. Campbell_, Jul 08 2011
%C For n>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,14}. - _Milan Janjic_, Jan 23 2015
%H Harvey P. Dale, <a href="/A078364/b078364.txt">Table of n, a(n) for n = 0..850</a>
%H A. F. Horadam, <a href="http://www.fq.math.ca/Scanned/5-5/horadam.pdf">Special properties of the sequence W_n(a,b; p,q)</a>, Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=15, q=-1.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H W. Lang, <a href="http://www.fq.math.ca/Scanned/38-5/lang.pdf">On polynomials related to powers of the generating function of Catalan's numbers</a>, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=17.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (15,-1).
%F a(n) = 15*a(n-1) - a(n-2), n>= 1; a(-1)=0, a(0)=1.
%F a(n) = S(2*n+1, sqrt(17))/sqrt(17) = S(n, 15); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
%F a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (15+sqrt(221))/2 and am = (15-sqrt(221))/2.
%F G.f.: 1/(1 - 15*x + x^2). - _Philippe Deléham_, Nov 17 2008
%F a(n) = Sum_{k=0..n} A101950(n,k)*14^k. - _Philippe Deléham_, Feb 10 2012
%F Product {n >= 0} (1 + 1/a(n)) = 1/13*(13 + sqrt(221)). - _Peter Bala_, Dec 23 2012
%F Product {n >= 1} (1 - 1/a(n)) = 1/30*(13 + sqrt(221)). - _Peter Bala_, Dec 23 2012
%F For n>=1, a(n) = U(n-1,15/2), where U(k,x) is Chebyshev polynomial of the second kind. - _Milan Janjic_, Jan 23 2015
%t LinearRecurrence[{15,-1},{1,15},30] (* _Harvey P. Dale_, Oct 16 2011 *)
%o (Sage) [lucas_number1(n,15,1) for n in range(1,20)] # _Zerinvary Lajos_, Jun 25 2008
%Y a(n) = sqrt((A078365(n+1)^2 - 4)/221), n>=0, (Pell equation d=221, +4).
%Y Cf. A077428, A078355 (Pell +4 equations).
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Nov 29 2002
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