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A010049 Second-order Fibonacci numbers. 19
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.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

T. Amdeberhan, M. B. Can, M. Jensen, Divisors and specializations of Lucas polynomials, arXiv preprint arXiv:1406.0432 [math.CO], 2014.

L. Turban, Lattice animals on a staircase and Fibonacci numbers, 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) = sum(F(k)*F(n-1-k): k=0..n-1) + F(n), 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(sum(F(j-1)*F(i-j-1): j=0..i): i=0..n-1), where F = A000045. - Carlos A. Rico A. Jul 14 2016

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 *)

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(); [a.next() 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

CROSSREFS

Partial sums of A006367.

Cf. A001629, A023610, A094440, A134410, A135830, A029907.

Sequence in context: A298116 A018165 A054179 * A094986 A154949 A107232

Adjacent sequences:  A010046 A010047 A010048 * A010050 A010051 A010052

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane.

EXTENSIONS

More terms from Emeric Deutsch, Dec 10 2003

STATUS

approved

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Last modified August 16 23:58 EDT 2018. Contains 313809 sequences. (Running on oeis4.)