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From a definite integral.
(Formerly M3802 N1553)
11

%I M3802 N1553 #58 Aug 05 2023 13:09:26

%S 1,5,10,30,74,199,515,1355,3540,9276,24276,63565,166405,435665,

%T 1140574,2986074,7817630,20466835,53582855,140281751,367262376,

%U 961505400,2517253800,6590256025,17253514249,45170286749,118257345970

%N From a definite integral.

%C a(n) are the row sums of the elements of the Golden Triangle (A180662) with alternating signs. - _Alexander Adamchuk_, Oct 18 2010

%C Limit_{n->oo} A002570(n)/A002571(n) = 1/sqrt(5). - _Sean A. Irvine_, Apr 09 2014

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992

%H L. R. Shenton, <a href="http://dx.doi.org/10.1017/S0013091500014280">A determinantal expansion for a class of definite integral. Part 5. Recurrence relations</a>, Proc. Edinburgh Math. Soc. (2) 10 (1957), 167-188.

%H L. R. Shenton and K. O. Bowman, <a href="http://www.pphmj.com/abstract/901.htm">Second order continued fractions and Fibonacci numbers</a>, Far East Journal of Applied Mathematics, 20(1), 17-31, 2005.

%F Appears to have g.f. x/((1-3x+x^2)*(1+x)^2). - _Ralf Stephan_, Apr 14 2004

%F a(n) = (-1)^n*Sum_{i=1..n+1} (-1)^(i+1)*Fibonacci(i)*Fibonacci(i+1). - _Alexander Adamchuk_, Jun 16 2006

%F From _Paul D. Hanna_, Feb 20 2009: (Start)

%F Given g.f. A(x), then log(1+A(x)) = Sum_{n>=1} A000204(n)^2 * x^n/n where A000204 is the Lucas numbers.

%F a(n) = (1/n)*(A000204(n)^2 + Sum_{k=1..n-1} A000204(k)^2*a(n-k)) for n>1, with a(1) = 1. (End)

%F G.f.: -1 + 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^A006206(n), where A006206(n) is the number of aperiodic binary necklaces of length n with no subsequence 00. - _Paul D. Hanna_, Jan 07 2012

%F a(n) = 8*a(n-2) - 8*a(n-4) + a(n-6) + 2(-1)^n, n>6. - _Sean A. Irvine_, Apr 09 2014

%F a(n) - a(n-2) = Fibonacci(n+1)^2. - _Peter Bala_, Aug 30 2015

%e From _Paul D. Hanna_, Feb 20 2009: (Start)

%e G.f.: A(x) = x + 5*x^2 + 10*x^3 + 30*x^4 + 74*x^5 + 199*x^6 + ...

%e log(1+A(x)) = x + 3^2*x^2/2 + 4^2*x^3/3 + 7^2*x^4/4 + 11^2*x^5/5 + ... (End)

%e G.f.: A(x) = -1 + 1/((1-x-x^2) * (1-3*x^2+x^4) * (1-4*x^3-x^6) * (1-7*x^4+x^8) * (1-11*x^5-x^10)^2 * (1-18*x^6+x^12)^2 * (1-29*x^7-x^14)^4 * (1-47*x^8+x^16)^5 * (1-76*x^9-x^18)^8 * ...* (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^A006206(n) * ...). - _Paul D. Hanna_, Jan 07 2012

%p A002571:=-(-1-4*z-z**2+z**3)/(z**2-3*z+1)/(1+z)**2; # conjectured (probably correctly) by _Simon Plouffe_ in his 1992 dissertation

%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n,(fibonacci(m+1)+fibonacci(m-1))^2*x^m/m)+x*O(x^n)),n)} \\ _Paul D. Hanna_, Feb 20 2009

%Y Cf. A064831, A077916, A000045, A000204 (Lucas), A006206.

%Y Cf. A001654, A180662 - The Golden Triangle. - _Alexander Adamchuk_, Oct 18 2010

%K nonn

%O 1,2

%A _N. J. A. Sloane_

%E More terms from _Max Alekseyev_ and _Alexander Adamchuk_, Oct 18 2010