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 A010049 Second-order Fibonacci numbers. 20
 0, 1, 1, 3, 5, 10, 18, 33, 59, 105, 185, 324, 564, 977, 1685, 2895, 4957, 8462, 14406, 24465, 41455, 70101, 118321, 199368, 335400, 563425, 945193, 1583643, 2650229, 4430290, 7398330, 12342849, 20573219, 34262337, 57013865, 94800780, 157517532, 261545777 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Number of parts in all compositions of n+1 with no 1's. E.g. a(5)=10 because in the compositions of 6 with no part equal to 1, namely 6,4+2,3+3,2+4,2+2+2, the total number of parts is 10. - Emeric Deutsch, Dec 10 2003 REFERENCES D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 83. Cornelius Gerrit Lekkerkerker, Voorstelling van natuurlyke getallen door een som van getallen van Fibonacci, Simon Stevin, Vol. 29 (1952), pp. 190-195. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Carlos A. Rico A., Ana Paula Chaves, Double-Recurrence Fibonacci Numbers and Generalizations, arXiv:1903.07490 [math.NT], 2019. T. Amdeberhan, M. B. Can, M. Jensen, Divisors and specializations of Lucas polynomials, arXiv preprint arXiv:1406.0432 [math.CO], 2014. Mengmeng Liu, Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8. Jia Huang, Compositions with restricted parts, arXiv:1812.11010 [math.CO], 2018. L. Turban, Lattice animals on a staircase and Fibonacci numbers, arXiv:cond-mat/0011038 [cond-mat.stat-mech], 2000; J.Phys. A 33 (2000) 2587-2595. Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1) FORMULA First differences of A001629. a(n) = ((2*n+3)*F(n) - n*F(n-1))/5, F(n)=A000045(n) (Fibonacci numbers) (Turban reference eq.(2.12)). - Wolfdieter Lang, May 03 2000 G.f.: x*(1-x)/(1-x-x^2)^2 (Turban reference eq.(2.10)). - Wolfdieter Lang, May 03 2000 Recurrence: a(0)=0, a(1)=1, a(2)=1, a(n+2) = a(n+1) + a(n) + F(n). - Benoit Cloitre, Sep 02 2002 Set A(n) = a(n+1) + a(n-1), B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + Lucas(n) and B(n+2) = B(n+1) + B(n) + Fibonacci(n). The polynomials F_2(n,-x) = Sum_{k=0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials F(n,-x) defined in A094440. For a similar conjecture for polynomials involving the second-order Lucas numbers see A134410. - Peter Bala, Oct 24 2007 a(n) = -A001629(n+2) + 2*A001629(n+1) + A000045(n+1). - R. J. Mathar, Nov 16 2007 Starting (1, 1, 3, 5, 10, ...), = row sums of triangle A135830. - Gary W. Adamson, Nov 30 2007 a(n) = F(n) + Sum_{k=0..n-1} F(k)*F(n-1-k), where F = A000045. - Reinhard Zumkeller, Nov 01 2013 a(n) = Sum_{k=0..n-1} (k+1)*binomial(n-k-1, k). - Peter Luschny, Nov 20 2013 a(n) = Sum_{i=0..n-1} Sum_{j=0..i} F(j-1)*F(i-j-1), where F = A000045. - Carlos A. Rico A., Jul 14 2016 a(n) = Sum_{k = F(n+1)..F(n+2)-1} A007895(k), where F(n) is the n-th Fibonacci number (Lekkerkerker, 1952). - Amiram Eldar, Jan 11 2020 MAPLE with(combinat): A010049 := proc(n) options remember; if n <= 1 then n else A010049(n-1)+A010049(n-2)+fibonacci(n-2); fi; end; MATHEMATICA CoefficientList[Series[(z - z^2)/(z^2 + z - 1)^2, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *) CoefficientList[Series[x (1 - x) / (1 - x - x^2)^2, {x, 0, 60}], x] (* Vincenzo Librandi, Jun 11 2013 *) LinearRecurrence[{2, 1, -2, -1}, {0, 1, 1, 3}, 38] (* Amiram Eldar, Jan 11 2020 *) PROG (Haskell) a010049 n = a010049_list !! n a010049_list = uncurry c \$ splitAt 1 a000045_list where    c us (v:vs) = (sum \$ zipWith (*) us (1 : reverse us)) : c (v:us) vs -- Reinhard Zumkeller, Nov 01 2013 (Sage) def A010049():     a, b, c, d = 0, 1, 1, 3     while True:         yield a         a, b, c, d = b, c, d, 2*(d-b)+c-a a = A010049(); [next(a) for i in range(38)]  # Peter Luschny, Nov 20 2013 (PARI) a(n)=([0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 0, 1; -1, -2, 1, 2]^n*[0; 1; 1; 3])[1, 1] \\ Charles R Greathouse IV, Jul 20 2016 (MAGMA) [((2*n+3)*Fibonacci(n)-n*Fibonacci(n-1))/5: n in [0..40]]; // Vincenzo Librandi, Dec 31 2018 (GAP) a:=List([0..40], n->Sum([0..n-1], k->(k+1)*Binomial(n-k-1, k)));; Print(a); # Muniru A Asiru, Dec 31 2018 CROSSREFS Partial sums of A006367. Cf. A000045, A001629, A007895, A023610, A029907, A094440, A134410, A135830. Sequence in context: A298116 A018165 A054179 * A094986 A154949 A318248 Adjacent sequences:  A010046 A010047 A010048 * A010050 A010051 A010052 KEYWORD nonn,easy AUTHOR EXTENSIONS More terms from Emeric Deutsch, Dec 10 2003 STATUS approved

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Last modified October 23 07:22 EDT 2020. Contains 337964 sequences. (Running on oeis4.)