A002571 From a definite integral.
1, 5, 10, 30, 74, 199, 515, 1355, 3540, 9276, 24276, 63565, 166405, 435665, 1140574, 2986074, 7817630, 20466835, 53582855, 140281751, 367262376, 961505400, 2517253800, 6590256025, 17253514249, 45170286749, 118257345970
Offset: 1
Keywords
Examples
From _Paul D. Hanna_, Feb 20 2009: (Start) G.f.: A(x) = x + 5*x^2 + 10*x^3 + 30*x^4 + 74*x^5 + 199*x^6 + ... 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) 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
References
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- L. R. Shenton, A determinantal expansion for a class of definite integral. Part 5. Recurrence relations, Proc. Edinburgh Math. Soc. (2) 10 (1957), 167-188.
- L. R. Shenton and K. O. Bowman, Second order continued fractions and Fibonacci numbers, Far East Journal of Applied Mathematics, 20(1), 17-31, 2005.
Crossrefs
Programs
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Maple
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
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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
Formula
Appears to have g.f. x/((1-3x+x^2)*(1+x)^2). - Ralf Stephan, Apr 14 2004
a(n) = (-1)^n*Sum_{i=1..n+1} (-1)^(i+1)*Fibonacci(i)*Fibonacci(i+1). - Alexander Adamchuk, Jun 16 2006
From Paul D. Hanna, Feb 20 2009: (Start)
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.
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
a(n) = 8*a(n-2) - 8*a(n-4) + a(n-6) + 2(-1)^n, n>6. - Sean A. Irvine, Apr 09 2014
a(n) - a(n-2) = Fibonacci(n+1)^2. - Peter Bala, Aug 30 2015
Extensions
More terms from Max Alekseyev and Alexander Adamchuk, Oct 18 2010
Comments