A178898 -id:A178898 - cp's OEIS frontend

A081016 a(n) = (Lucas(4n+3) + 1)/5, or Fibonacci(2n+1)Fibonacci(2n+2), or A081015(n)/5.

1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830

Offset: 0

R. K. Guy, Mar 01 2003

a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.
Numbers k such that 1 - 2*k + 5*k^2 is a square. - Artur Jasinski, Oct 26 2008
Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - Carmine Suriano, Jun 23 2010

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.
Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.
Cf. A145995, A178898.

List([0..30], n-> (Lucas(1,-1,4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019

[(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+3)+1)/5) od: # James Sellers, Mar 03 2003

LinearRecurrence[{8,-8,1}, {1,6,40}, 30] (* Bruno Berselli, Aug 31 2017 *)

a(n)=([0,1,0; 0,0,1; 1,-8,8]^n*[1;6;40])[1,1] \\ Charles R Greathouse IV, Sep 28 2015

first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017

[(lucas_number2(4*n+3,1,-1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).
a(n) = F(1) + F(5) + F(9) +...+ F(4*n+1) = F(2*n)*F(2*n+3) + 1, where F(j) = Fibonacci(j).
a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - R. J. Mathar, Nov 07 2015

A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

Original entry on oeis.org

0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855, 116139356908771352, 796030994547383610

Offset: 0

R. K. Guy, Mar 01 2003

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010
Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014
Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015
The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

			G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Colin Barker, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Dae S. Hong, When is the Generating Function of the Fibonacci Numbers an Integer?, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Cf. A001622, A081007, A081016, A178898.

List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # G. C. Greubel, Jul 14 2019

[(Lucas(4*n+1)-1)/5: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # James Sellers, Mar 03 2003

(LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* G. C. Greubel, Aug 24 2015, modified Jul 14 2019 *)

concat(0, Vec(x*(2-x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ Colin Barker, Dec 20 2014

[(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Mar 30 2012
E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 24 2015
From - Michael Somos, Aug 27 2015: (Start)
a(n) = -A081016(-1-n) for all n in Z.
0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - Stefano Spezia, Mar 11 2021
Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021

More terms from James Sellers, Mar 03 2003

A179015 Number of ways in which n^2 can be expressed as the sum of exactly five positive squares.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 5, 2, 6, 6, 9, 9, 15, 8, 25, 20, 21, 25, 39, 26, 46, 44, 57, 49, 71, 52, 102, 81, 81, 99, 145, 92, 156, 126, 164, 160, 204, 151, 247, 217, 236, 245, 326, 211, 357, 319, 381, 360, 416, 344, 518, 446, 476, 450, 670, 468, 675, 607, 661, 668, 825, 625

Offset: 1

Carmine Suriano, Jun 24 2010

nonn

As n goes to infinity the ratio of a(n)/a(n) of sequence A178898 (using all different squares) tends to 5/4.

Alois P. Heinz, Table of n, a(n) for n = 1..740

Cf. A000290, A178898.
Cf. A000132. - R. J. Mathar, Jun 26 2010
Cf. A065458, A065459.

a(8) = 5 since 64 can be expressed in five different ways as the sum of 5 squares (order is ignored): 8^2 = 7^2+3^2+2^2+1^2+1^2 = 6^2+5^2+1^2+1^2+1^2 = 6^2+4^2+2^2+2^2+2^2 = 6^2+3^2+3^2+3^2+1^2 = 5^2+5^2+3^2+1^2+1^2.

Asymptotic behavior for large values of n is a(n) = n^2/2-47n/2+243.
a(n) = A025429(n^2). - R. J. Mathar, Jun 26 2010
a(n) = A065459(n) - A065458(n). - Alois P. Heinz, Oct 25 2018

cp's OEIS Frontend

A081016 a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.

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A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

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Programs

GAP

Magma

Maple

Mathematica

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A179015 Number of ways in which n^2 can be expressed as the sum of exactly five positive squares.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A081016 a(n) = (Lucas(4n+3) + 1)/5, or Fibonacci(2n+1)Fibonacci(2n+2), or A081015(n)/5.