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A007598 Squared Fibonacci numbers: F(n)^2 where F = A000045.
(Formerly M3364)
94
0, 1, 1, 4, 9, 25, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

a(n)*(-1)^(n+1) = (2*(1-T(n,-3/2))/5), n>=0, with Chebyshev's polynomials T(n,x) of the first kind, is the r=-1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found. - Wolfdieter Lang, Oct 18 2004

From Giorgio Balzarotti, Mar 11 2009: (Start)

Determinant of power series with alternate signs of gamma matrix with determinant 1!.

a(n) = Determinant(A - A^2 + A^3 - A^4 + A^5 - ... - (-1)^n*A^n) where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant.

A = [[1,1,1,1,1,1,...], [1,2,1,2,1,2,...], [1,2,3,1,2,3,...], [1,2,3,4,1,2,...], [1,2,3,4,5,1,...], [1,2,3,4,5,6,...], ...]; note: Determinant A(1..n,1..n) = (n-1)!.

a(n) is even with respect to signs of power of A.

See A158039...A158050 for sequence with matrix 2!, 3!, ... (End)

Equals the INVERT transform of (1, 3, 2, 2, 2, ...). Example: a(7) = 169 = (1, 1, 4, 9, 25, 64) dot (2, 2, 2, 2, 3, 1) = (2 + 2 + 8 + 18 + 75 + 64). - Gary W. Adamson, Apr 27 2009

This is a divisibility sequence.

a(n+1)*(-1)^n, n>=0, is the sequence of the alternating row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010

a(n+1) is the number of tilings of a 2 X 2n rectangle with n tetrominoes of any shape, cf. A230031. - Alois P. Heinz, Nov 29 2013

This is the case P1 = 1, P2 = -6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014

Differences between successive golden rectangle numbers A001654. - Jonathan Sondow, Nov 05 2015

REFERENCES

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.

R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.

Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop, Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp 4623-4627.

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

R. P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.

LINKS

Indranil Ghosh, Table of n, a(n) for n = 0..2389 (terms 0..200 from T. D. Noe)

Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.

Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.

Paul S. Bruckman, Problem B-1023: And a cubic as a sum of two squares, Fibonacci Quarterly, Vol. 45, Number 2; May 2007; p. 186.

Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math. 199 (1999), no. 1-3, 217--220. MR1675924 (99k:05016).

D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux,

Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4

T. Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.

T. Mansour, Squaring the terms of an l-th order linear recurrence, arXiv:math/0303138 [math.CO], 2003.

P. Stanica, Generating functions, weighted and non-weighted sums of powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume

Index entries for two-way infinite sequences

Index to divisibility sequences

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

FORMULA

G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).

a(n) = 2*a(n-1)+2*a(n-2)-a(n-3), n>2. a(0)=0, a(1)=1, a(2)=1.

a(-n) = a(n) for all n in Z.

L.g.f.: 1/5*log((1+3*x+x^2)/(1-6*x+x^2)) = sum(n>=0, a(n)/n*x^n); special case of l.g.f. given in A079291. - Joerg Arndt, Apr 13 2011

a(0) = 0, a(1) = 1; a(n) = a(n-1) + Sum(a(n-i)) + k, 0 <= i < n where k = 1 when n is odd, or k = -1 when n is even. E.g. a(2) = 1 = 1 + (1 + 1 + 0) - 1, a(3) = 4 = 1 + (1 + 1 + 0) + 1, a(4) = 9 = 4 + (4 + 1 + 1 + 0) - 1, a(5) = 25 = 9 + (9 + 4 + 1 + 1 + 0) + 1. - Sadrul Habib Chowdhury (adil040(AT)yahoo.com), Mar 02 2004

a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - Ralf Stephan, May 14 2004

a(n) = F(n-1)*F(n+1) - (-1)^n = A059929(n-1) - A033999(n).

a(n) = right term of M^n * [1 0 0] where M = the 3X3 matrix [1 2 1 / 1 1 0 / 1 0 0]. M^n * [1 0 0] = [a(n+1) A001654(n) a(n)]. E.g., a(4) = 9 since M^4 * [1 0 0] = [25 15 9] = [a(5) A001654(4) a(4)]. - Gary W. Adamson, Dec 19 2004

Sum_{j=0..2*n} binomial(2*n,j)*a(j) = 5^(n-1)*A005248(n+1) for n>=1. [P. Stanica; Sum_{j=0..2*n+1} binomial(2*n+1,j)*a(j) = 5^n*A001519(n+1) [P. Stanica]. - R. J. Mathar, Oct 16 2006

a(n) = ( A005248(n)-2*(-1)^n)/5. - R. J. Mathar, Sep 12 2010

a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k, - Gary Detlefs, Dec 13 2010

a(n) = 3*a(n-1)-a(n-2)+2*(-1)^(n+1), n>1. - Gary Detlefs, Dec 20 2010

a(n) = Fibonacci(2*n-2)+a(n-2). - Gary Detlefs, Dec 20 2010

a(n) = (Fibonacci(3*n)-3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n>0. - Gary Detlefs, Dec 20 2010

a(n) = (Fibonacci(n)*Fibonacci(n+4)-3*Fibonacci(n)*Fibonacci(n+1))/2. - Gary Detlefs, Jan 17 2011

a(n) = (((3+sqrt(5))/2)^n+((3-sqrt(5))/2)^n-2*(-1)^n)/5; without leading zero we would have a(n) = ((3+sqrt(5))*((3+sqrt(5))/2)^n+(3-sqrt(5))*((3-sqrt(5))/2)^n+4*(-1)^n)/10. - Tim Monahan, Jul 17 2011

E.g.f.: (exp((phi+1)*x) + exp((2-phi)*x) - 2*exp(-x))/5, with the golden section phi:=(1+sqrt(5))/2. From the Binet-de Moivre formula for F(n). - Wolfdieter Lang, Jan 13 2012

Starting with "1" = triangle A059260 * the Fibonacci sequence as a vector. - Gary W. Adamson, Mar 06 2012

a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - Thomas Ordowski, Jan 06 2013

a(n)+a(n-1) = A001519(n), n>0. - R. J. Mathar, Mar 19 2014

From Peter Bala, Mar 31 2014: (Start)

a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = 3/2 and beta = -1 and T(n,x) denotes the Chebyshev polynomial of the first kind.

a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/2; 1, 1/2].

a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.

See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)

a(n) = (F(n+2)*F(n+3)-L(n)*L(n+1))/3 for F(k)=A000045(k) and L(n)=A000032(k). - J. M. Bergot, Jun 02 2014

0 = a(n)*(+a(n) - 2*a(n+1) - 2*a(n+2)) + a(n+1)*(+a(n+1) - 2*a(n+2)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Jun 03 2014

(F(n)*b(n+2))^2 + (F(n+1)*b(n-1))^2 = F(2*n+1)^3 = A001519(n+1)^3, with b(n) = a(n) + 2*(-1)^n and F(n)=A000045(n) (see Bruckman link). - Michel Marcus, Jan 24 2015

a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A001254. - Peter Bala, Aug 18 2015

a(n) = F(n)*F(n+1) - F(n-1)*F(n). - Jonathan Sondow, Nov 05 2015

For n>2 a(n)=F(n-2)*(3*F(n-1) + F(n-3)) + F(2*n-5). Also, for n>2 a(n)=2*F(n-3)*F(n) + F(2*n-3) -(2)*(-1)^n. - J. M. Bergot, Nov 05 2015

a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - J. M. Bergot, Nov 08 2015

a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - J. M. Bergot, Mar 17 2016

a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2, with F(k)=A000045(k). - J. M. Bergot, May 25 2017

EXAMPLE

G.f. = x + x^2 + 4*x^3 + 9*x^4 + 25*x^5 + 64*x^6 + 169*x^7 + 441*x^8 + ...

MAPLE

with (combinat):seq(fibonacci(n)^2, n=0..27); # Zerinvary Lajos, Sep 21 2007

MATHEMATICA

f[n_]:=Fibonacci[n]^2; Array[f, 4!, 0] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2009 *)

LinearRecurrence[{2, 2, -1}, {0, 1, 1}, 41] (* Harvey P. Dale, May 18 2011 *)

PROG

(PARI) {a(n) = fibonacci(n)^2};

(PARI) concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^50))) \\ Altug Alkan, Nov 06 2015

(Sage) [(fibonacci(n))^2 for n in xrange(0, 28)]# Zerinvary Lajos, May 15 2009

(MAGMA) [Fibonacci(n)^2: n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

(Haskell)

a007598 = (^ 2) . a000045  -- Reinhard Zumkeller, Sep 01 2013

CROSSREFS

Cf. A001254, A079291.

Bisection of A006498 and A074677. First differences of A001654.

Equals A080097(n-2) + 1. Cf. A061646, A065885.

Cf. A056570.

Cf. A001654.

Second row of array A103323.

Half of A175395.

Cf. A047946.

Cf. A059260.

Sequence in context: A181357 A244558 A175627 * A121648 A133022 A184326

Adjacent sequences:  A007595 A007596 A007597 * A007599 A007600 A007601

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane, Robert G. Wilson v

STATUS

approved

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Last modified August 18 17:37 EDT 2017. Contains 290732 sequences.