cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 11 results. Next

A056570 Third power of Fibonacci numbers (A000045).

Original entry on oeis.org

0, 1, 1, 8, 27, 125, 512, 2197, 9261, 39304, 166375, 704969, 2985984, 12649337, 53582633, 226981000, 961504803, 4073003173, 17253512704, 73087061741, 309601747125, 1311494070536, 5555577996431, 23533806109393, 99690802348032, 422297015640625
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Comments

Divisibility sequence; that is, if n divides m, then a(n) divides a(m).
In general, cubing the terms of a Horadam sequence with signature (c,d) will result in a fourth-order recurrence with signature (c^3+2*c*d, c^4*d+3*(c*d)^2+2*d^3, -(c*d)^3-2*c*d^4, -d^6). - Gary Detlefs, Nov 12 2021
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/3,2/3)-fences and third-squares (1/3 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/6,1/3)-fences and (1/6,5/6)-fences. - Michael A. Allen, Jan 11 2022

Examples

			a(4) = 27 because the fourth Fibonacci number is 3 and 3^3 = 27.
a(5) = 125 because the fifth Fibonacci number is 5 and 5^3 = 125.

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).

Links

Crossrefs

Cf. A000045, A007598, A056588, A055870, A066259, A066258.
Cf. A346513 (first differences), A005968 (partial sums).
Third row of array A103323.

Programs

Formula

a(n) = A000045(n)^3.
G.f.: x*p(3, x)/q(3, x) with p(3, x) = Sum_{m=0..2} A056588(2, m)*x^m = 1 -2*x -x^2 and q(3, x) = Sum_{m=0..4} A055870(4, m)*x^m = 1 -3*x -6*x^2 +3*x^3 +x^4 = (1+x-x^2)*(1-4*x-x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) -3*a(n-1) -6*a(n-2) +3*a(n-3) +1*a(n-4) = 0, n >= 4, a(0)=0, a(1)=a(2)=1, a(3)=2^3. See 5th row of signed Fibonomial triangle for coefficients: A055870(4, m), m=0..4
a(n) = (Fibonacci(3n) - 3(-1)^n*Fibonacci(n))/5. - Ralf Stephan, May 14 2004
a(n) and a(n+1) are found as rightmost and leftmost terms (respectively) in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., Ma(4) = 27, a(5) = 125. M^4 * [1 0 0 0] = [125 75 45 27]; where 75 = A066259(4) and 45 = A066258(3). The characteristic polynomial of M = x^4 - 3x^3 - 6x^2 + 3x + 1. a(n)/a(n-1) of the sequence and companions tend to 2+sqrt(5) = 4.2360679..., an eigenvalue of M and a root of the polynomial. - Gary W. Adamson, Oct 31 2004
From R. J. Mathar, Oct 16 2006: (Start)
Sum_{j=0..n} binomial(n,j)*a(j) = (2^n*A001906(n) + 3*A000045(n))/5.
Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = ((-2)^n*A000045(n) - 3*A001906(n))/5. (End)
G.f.: x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)). - Colin Barker, Feb 28 2012
a(n) = F(n-2)*F(n+1)^2 + F(n-1)*(-1)^n. - J. M. Bergot, Mar 17 2016
a(n) = ((-3*(1/2*(-1-sqrt(5)))^n-(2-sqrt(5))^n+3*(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)) / (5*sqrt(5)). - Colin Barker, Jun 04 2016
a(n) = F(n-1)*F(n)*F(n+1) + F(n)*(-1)^(n-1). - Tony Foster III, Apr 11 2018
5*a(n) = L(2*n-1)*F(n+2) - L(2*n+1)*F(n-2) - 7*(-1)^n*F(n), where L(n) = A000032(n). - Peter Bala, Nov 12 2019
F(n+1)*F(n)*F(n-1) = 2*Sum_{j=1..n-1} P(j)*a(n-j) for n>0, where Pell number P(n) = A000129(n). - Michael A. Allen, Jan 11 2022

A001655 Fibonomial coefficients: a(n) = F(n+1) * F(n+2) * F(n+3)/2, where F() = Fibonacci numbers A000045.

Original entry on oeis.org

1, 3, 15, 60, 260, 1092, 4641, 19635, 83215, 352440, 1493064, 6324552, 26791505, 113490195, 480752895, 2036500788, 8626757644, 36543528780, 154800876945, 655747029795, 2777789007071, 11766903040368, 49845401197200, 211148507782800, 894439432403425
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

In a triangle having sides of F(n+1), 2*F(n+2) and F(n+3), the product of the area and circumradius will be a(n). For example: a triangle having sides of 5, 16 and 13 will have an area of 4*sqrt(51), a circumradius of 65*sqrt(51)/51, and the product is 4*65 = 260. - Gary Detlefs, Dec 14 2010
Explanation of this comment: if a triangle with sides (a, b, c) has a circumradius R and an area A, then A*R = abc/4; here, with a = F(n+1), b=2*F(n+2) and c=F(n+3), this gives a(n)= A*R. - Bernard Schott, Jan 26 2023

Examples

			G.f. = 1 + 3*x + 15*x^2 + 60*x^3 + 260*x^4 + 1092*x^5 + 4641*x^6 + ...

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

Crossrefs

Cf. A066258 (first differences), A215037 (partial sums), A363753 (alternating sums).
Cf. A000045, A055870, A065563, A079586, A114525, A256178.

Programs

  • Magma
    [Fibonacci(n+3)*Fibonacci(n+2)*Fibonacci(n+1)/2: n in [0..30]]; // Vincenzo Librandi, May 09 2016
  • Maple
    A001655:=1/(z**2-z-1)/(z**2+4*z-1); # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1])/2, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{3, 6, -3, -1}, {1, 3, 15, 60}, 25] (* Jean-François Alcover, Sep 23 2017 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j)); vector(20, n, b(n-1, 3))  \\ Joerg Arndt, May 08 2016

Formula

G.f.: 1/(1-3*x-6*x^2+3*x^3+x^4) = 1/((1+x-x^2)*(1-4*x-x^2)) (see Comments to A055870).
a(n) = A010048(n+3, 3) = fibonomial(n+3, 3).
a(n) = (1/2) * A065563(n).
a(n) = 4*a(n-1) + a(n-2) + ((-1)^n)*F(n+1), n >= 2; a(0)=1, a(1)=3.
a(n) = (F(n+3)^3 - F(n+2)^3 - F(n+1)^3)/6. - Gary Detlefs, Dec 24 2010
a(n-1) = Sum_{k=0..n} F(k+1)*F(k)^2, n >= 1. - Wolfdieter Lang, Aug 01 2012
From Wolfdieter Lang, Aug 09 2012: (Start)
a(n-1)*(-1)^n = Sum_{k=0..n} (-1)^k*F(k+1)^2*F(k), n >= 1. See the link under A215037, eq. (25).
a(n) = (F(3*(n+2)) + 2*(-1)^n*F(n+2))/10, n >= 0. See the same link, eq. (32). (End)
a(n) = -a(-4-n)*(-1)^n for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(-a(n+1) - a(n+2)) + a(n+1)*(-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
O.g.f.: exp( Sum_{n >= 1} L(n)*L(2*n)*x^n/n ), where L(n) = A000032(n) is a Lucas number. Cf. A114525, A256178. - Peter Bala, Mar 18 2015
Sum_{n>=0} (-1)^n/a(n) = 2 * A079586 - 6. - Amiram Eldar, Oct 04 2020
The formula by Gary Detlefs above is valid for all sequences of the Fibonacci type f(n) = f(n-1) + f(n-2): 3*f(n+2)*f(n+1)*f(n) = f(n+2)^3 - f(n+1)^3 - f(n)^3. - Klaus Purath, Mar 25 2021
a(n) = sqrt(Sum_{j=1..n+1} F(j)^3*F(j+1)^3). See Treeby link. - Michel Marcus, Apr 10 2022
a(n) = Sum_{k=1..n+1} A000129(k)*A056570(n+2-k). - Michael A. Allen, Jan 25 2023
G.f.: exp( Sum_{k>=1} F(4*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A066259 a(n) = Fibonacci(n)*Fibonacci(n+1)^2.

Original entry on oeis.org

1, 4, 18, 75, 320, 1352, 5733, 24276, 102850, 435655, 1845504, 7817616, 33116057, 140281700, 594243090, 2517253683, 10663258432, 45170286424, 191344405725, 810547906740, 3433536036866, 14544692047439, 61612304237568, 260993908980000, 1105587940186225, 4683345669678532
Offset: 1

Views

Author

Len Smiley, Dec 09 2001

Keywords

Links

Crossrefs

Cf. A000045, A065563, A066258, A000045, A056570.

Programs

  • Mathematica
    First[#]Last[#]^2&/@Partition[Fibonacci[Range[30]],2,1]  (* Harvey P. Dale, Mar 04 2011 *)
  • PARI
    a(n) = { fibonacci(n) * fibonacci(n+1)^2 } \\ Harry J. Smith, Feb 07 2010

Formula

O.g.f.: (x+x^2)/(1-3x-6x^2+3x^3+x^4) = x(1+x)/((1+x-x^2)(1-4x-x^2)).
a(n) = second term from left in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., a(4) = 75 since M^4 * [1 0 0 0] = [125 75 45 27] = [A056570(5) a(4) A066258(3) A056570(4)]. - Gary W. Adamson, Oct 31 2004
a(n) = (1/5)*(F(3n+2) - (-1)^n*F(n-1)). - Ralf Stephan, Jul 26 2005
a(n) = (F(n+2)^3 - 2*F(n)^3 - F(n-1)^3)/6. - Greg Dresden, Aug 12 2022

A197424 Number of subsets of {1, 2, ..., 4*n + 2} which do not contain two numbers whose difference is 4.

Original entry on oeis.org

4, 36, 225, 1600, 10816, 74529, 509796, 3496900, 23961025, 164249856, 1125736704, 7716041281, 52886200900, 362488284900, 2484529385121, 17029223715904, 116720020119616, 800010960336225, 5483356589096100, 37583485459535236, 257601040852192129
Offset: 0

Views

Author

John W. Layman, Oct 14 2011

Keywords

Comments

This sequence is an instance of a general result given in Math. Mag. Problem 1854 (see Links).
From Feryal Alayont, May 20 2023: (Start)
a(n) is the number of edge covers of a caterpillar graph with spine P_(4n+5), one pendant attached at vertex n+2 counting from the left end of the spine, a second one at 2n+3 and a third at 3n+4. The caterpillar graph for n=1 is as follows:
* * *
| | |
*--*--*--v1--*--v2--*--*--*
Each pendant edge must be included in an edge cover leaving only the middle six edges flexible. Every vertex except v1 and v2 is incident with at least one of the pendant edges. Therefore, if we label the middle six edges in the spine with numbers 3, 1, 5, 2, 6, 4 (starting from the left), the edges have to be chosen so that both 1,5 and 2,6 cannot be missing. This corresponds to choosing subsets of {1, 2, ..., 6} which do not contain two numbers whose difference is 4. (End)

References

  • F. Alayont and E. Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs; submitted.

Links

Crossrefs

Cf. A000045, A001654, A060635, A066258.

Programs

  • Mathematica
    Table[(1/25) (LucasL[2 (2 n + 5)] - 2 (-1)^n LucasL[2 n + 5] - 1), {n, 0, 20}] (* Michael De Vlieger, Mar 27 2016 *)
  • PARI
    Vec((4+16*x-15*x^2-5*x^3+x^4) / ((1-x)*(1-7*x+x^2)*(1+3*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 26 2016
  • PARI
    a(n) = (fibonacci(n+2)*fibonacci(n+3))^2; \\ Altug Alkan, Mar 26 2016

Formula

a(n) = F(n+2)^2*F(n+3)^2 = A001654(n+2)^2, where F(n) denotes the n-th Fibonacci number A000045(n).
G.f.: ( -4-16*x+15*x^2+5*x^3-x^4 ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Oct 15 2011
Empirical: a(n) = A189145(2n+3). - R. J. Mathar, Oct 15 2011
For L=Lucas, a(n) = (1/25)*(L(2*(2*n+5)) - 2*(-1)^n*L(2*n+5) - 1), an instance of (F(n+p)*F(n+q))^2 = (1/25)*(L(2*(2*n+p+q)) - 2*(-1)^(n+q)*L(p-q)*L(2*n+p+q) + L(2*(p-q)) + 4*(-1)^(p-q)) which follows from squaring a specialization of identity 17b in the Vajda reference at A000045, F(n+p)*F(n+q) = (1/5)*(L(2*n+p+q) - (-1)*(n+q)*L(p-q)), then applying Vajda 17c, L(n)^2 = L(2*n) + 2*(-1)^n, to the expansion. - Ehren Metcalfe, Mar 26 2016
a(n) = A060635(n+2)/2. - Alois P. Heinz, Jul 03 2025

A099014 a(n) = Fibonacci(n)*(Fibonacci(n-1)^2 + Fibonacci(n+1)^2).

Original entry on oeis.org

0, 1, 5, 20, 87, 365, 1552, 6565, 27825, 117844, 499235, 2114729, 8958240, 37947545, 160748653, 680941780, 2884516383, 12219006325, 51760543280, 219261176861, 928805254905, 3934482189716, 16666734024715, 70601418270865
Offset: 0

Views

Author

Paul Barry, Sep 22 2004

Keywords

Comments

Form the matrix A=[1,1,1,1;3,2,1,0;3,1,0,0;1,0,0,0]=(binomial(3-j,i)). Then a(n)=(2,3)-element of A^n.

Links

Crossrefs

Cf. A000045, A056570, A066258, A066259.

Programs

  • Magma
    [Fibonacci(n)*(Fibonacci(n-1)^2+Fibonacci(n+1)^2): n in [0..30]]; // Vincenzo Librandi, Jun 05 2011
  • Mathematica
    CoefficientList[Series[x*(1 + 2*x - x^2)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4), {x, 0, 50}], x] (* G. C. Greubel, Dec 31 2017 *)
Join[{0},#[[2]](#[[1]]^2+#[[3]]^2)&/@Partition[Fibonacci[ Range[ 0,30]],3,1]] (* or *) LinearRecurrence[{3,6,-3,-1},{0,1,5,20},30] (* Harvey P. Dale, Oct 17 2021 *)
  • PARI
    a(n)=fibonacci(n)*(fibonacci(n-1)^2+fibonacci(n+1)^2) \\ Charles R Greathouse IV, Jun 05 2011

Formula

G.f.: x*(1+2*x-x^2)/(1-3*x-6*x^2+3*x^3+x^4) = x*(1+2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)).
a(n) = Sum_{k=0..n} (-1)^(k+1)*Fib(k)*(0^(n-k) + 6*A001076(n-k)).
a(n) = ((-1)^n*Fib(n) + 3*Fib(3*n))/5. - Ehren Metcalfe, May 21 2016

A220362 a(n) = Fibonacci(n-1) * Fibonacci(n) * Fibonacci(n+2).

Original entry on oeis.org

0, 3, 10, 48, 195, 840, 3536, 15015, 63546, 269280, 1140535, 4831632, 20466720, 86699067, 367262090, 1555748880, 6590255259, 27916773720, 118257343984, 500946159615, 2122041966330, 8989114051008, 38078498128175, 161303106631968, 683290924545600
Offset: 1

Views

Author

Michel Marcus, Dec 12 2012

Keywords

Comments

An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons.
Type B have sides A056570(n+2), A056570(n+2), A220363(n+2), A056570(n+2), A056570(n+2), and opposite diagonals a(n+2), A066258(n+2), A066258(n+2), A066258(n+2), a(n+2), for n=1,2,...

References

  • R. K. Guy, Unsolved Problems in Number Theory, D20.

Links

Programs

  • Mathematica
    Table[Fibonacci[n - 1] * Fibonacci[n] * Fibonacci[n + 2], {n, 30}] (* T. D. Noe, Dec 13 2012 *)
#[[1]]*#[[2]]*#[[4]]&/@Partition[Fibonacci[Range[0,30]],4,1] (* Harvey P. Dale, Jan 16 2016 *)
  • PARI
    x='x+O('x^99); concat(0, Vec(x*(3+x)/((x^2-x-1)*(x^2+4*x-1)))) \\ Altug Alkan, Mar 26 2016

Formula

a(n) = 3*a(n-1) + 6*a(n-2) -3*a(n-3) - a(n-4); g.f.: (3x+x^2)/(1-3x-6x^2+3x^2+x^4) = x(3+x)/( (x^2-x-1)(x^2+4x-1) ). [Ron Knott, Jun 27 2013]
a(n) = 2*(-1)^n*Lucas(n-3)/25 + 3*(-1)^n*Lucas(n+1)/25 + Fibonacci(3*n+1)/5 = (1/5)*(-(-1)^n*Fibonacci(n-1) + 3*(-1)^n*Fibonacci(n) + Fibonacci(3*n+1)). - Ehren Metcalfe, Mar 26 2016

A099015 a(n) = Fib(n+1)*(2*Fib(n)^2 + Fib(n)*Fib(n-1) + Fib(n-1)^2).

Original entry on oeis.org

1, 2, 8, 33, 140, 592, 2509, 10626, 45016, 190685, 807764, 3421728, 14494697, 61400482, 260096680, 1101787113, 4667245276, 19770767984, 83750317589, 354772037730, 1502838469496, 6366125914117, 26967342128548
Offset: 0

Views

Author

Paul Barry, Sep 22 2004

Keywords

Comments

Form the matrix A=[1,1,1,1;3,2,1,0;3,1,0,0;1,0,0,0]=(binomial(3-j,i)). Then a(n)=(2,2)-element of A^n.

Links

Crossrefs

Cf. A000045, A056570, A066258, A066259, A099014, A033887.

Programs

  • Magma
    [Fibonacci(n+1)*(2*Fibonacci(n)^2 + Fibonacci(n)*Fibonacci(n-1) + Fibonacci(n-1)^2): n in [0..30]]; // Vincenzo Librandi, Jun 05 2011
  • Mathematica
    LinearRecurrence[{3,6,-3,-1},{1,2,8,33},30] (* Harvey P. Dale, Nov 28 2015 *)
CoefficientList[Series[(1-x-4*x^2)/((1+x-x^2)*(1-4*x-x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 31 2017 *)
  • PARI
    a(n)=my(e=fibonacci(n-1),f=fibonacci(n));(e+f)*(2*f^2+f*e+e^2) \\ Charles R Greathouse IV, Jun 05 2011
  • PARI
    first(n) = Vec((1 - x - 4*x^2)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4) + O(x^n)) \\ Iain Fox, Dec 31 2017

Formula

G.f.: (1-x-4*x^2)/((1+x-x^2)*(1-4*x-x^2)).
G.f.: (1-x-4*x^2)/(1-3*x-6*x^2+3*x^3+x^4).
a(n) = (3*Fib(3*n+1) + (-1)^n*Fib(n-3))/5.
a(n) = (2+sqrt(5))^n*(3/10 + 3*sqrt(5)/50) + (2-sqrt(5))^n*(3/10 - 3*sqrt(5)/50) + (-1)^n*((1/2 - sqrt(5)/2)^n*(1/5 + 2*sqrt(5)/25) + (1/5 - 2*sqrt(5)/25)*(1/2 + sqrt(5)/2)^n).

A220363 a(n) = Fibonacci(n)^3 + (-1)^n*Fibonacci(n+2).

Original entry on oeis.org

1, -1, 4, 3, 35, 112, 533, 2163, 9316, 39215, 166519, 704736, 2986361, 12648727, 53583620, 226979403, 961507387, 4072998992, 17253519469, 73087050795, 309601764836, 1311494041879, 5555578042799, 23533806034368, 99690802469425, 422297015444207
Offset: 0

Views

Author

Michel Marcus, Dec 12 2012

Keywords

Comments

An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons.
Type B have sides A056570(n+2), A056570(n+2), a(n+2), A056570(n+2), A056570(n+2), and opposite diagonals A220362(n+2), A066258(n+2), A066258(n+2), A066258(n+2), A220362(n+2), for n=1,2,...

References

  • R. K. Guy, Unsolved Problems in Number Theory, D20.

Links

Programs

  • Mathematica
    Table[Fibonacci[n]^3 + (-1)^n * Fibonacci[n + 2], {n, 0, 30}] (* T. D. Noe, Dec 13 2012 *)
LinearRecurrence[{3,6,-3,-1},{1,-1,4,3},30] (* Harvey P. Dale, Mar 19 2022 *)
  • PARI
    Vec((x^2-4*x+1)/((x^2-x-1)*(x^2+4*x-1)) + O(x^100)) \\ Colin Barker, Sep 23 2014
  • PARI
    a(n) = fibonacci(n)^3 + (-1)^n*fibonacci(n+2) \\ Charles R Greathouse IV, Feb 14 2017

Formula

a(n) = 3*a(n-1)+6*a(n-2)-3*a(n-3)-a(n-4). G.f.: (x^2-4*x+1) / ((x^2-x-1)*(x^2+4*x-1)). - Colin Barker, Sep 23 2014

A177727 a(0)=1; a(n) = a(n-1) * Fibonacci(3+n) * Fibonacci(1+n) / (Fibonacci(n))^2, n > 1.

Original entry on oeis.org

1, 3, 30, 180, 1300, 8736, 60333, 412335, 2829310, 19384200, 132882696, 910735488, 6242420665, 42785803515, 293259265950, 2010026277756, 13776931957468, 94428478367520, 647222466507045, 4436128656563175, 30405678471399166, 208403619747957648, 1428419662108160400
Offset: 0

Views

Author

Roger L. Bagula, May 12 2010

Keywords

Comments

Similar recurrences a(n) = a(n-1)*F(a0+n-1)*F(b0+n-1)/(F(n)*F(c0+n-1)) are:
{a0,b0,c0} = {3,2,1} in A066258.
{a0,b0,c0} = {3,1,1} in A001654.
{a0,b0,c0} = {4,1,1} in A001655 and next for 5,6 as well.

References

  • Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986, p. 93.

Links

Crossrefs

Cf. A066258, A001654, A001655, A001656, A001657.

Programs

  • Magma
    I:=[1, 3, 30, 180, 1300]; [n le 5 select I[n] else 5*Self(n-1)+15*Self(n-2)-15*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Nov 18 2011
  • Maple
    with (combinat):
A177727 := proc(n)
   if n = 0 then
           1;
   else
           procname(n-1)*fibonacci(3+n)*fibonacci(1+n)/fibonacci(n)^2 ;
   end if;
end proc:
seq(A177727(n),n=0..10) ; # R. J. Mathar, Nov 17 2011
  • Mathematica
    a0 = 4; b0 = 2; c0 = 1;
a[0] = 1;
a[n_] := a[n] = (Fibonacci[(a0 + n - 1)]*Fibonacci[( b0 + n - 1)]/(Fibonacci[n]*Fibonacci[(c0 + n - 1)]))*a[n - 1];
Table[a[n], {n, 0, 30}]
LinearRecurrence[{5,15,-15,-5,1},{1,3,30,180,1300},30] (* Vincenzo Librandi, Nov 18 2011 *)

Formula

G.f.: ( -1+2*x ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Nov 17 2011
a(n) = A001656(n) - 2*A001656(n-1). - R. J. Mathar, Nov 17 2011

A215038 Partial sums of A066259: a(n) = Sum_{k=0..n} F(k+1)^2*F(k), n>=0, with the Fibonacci numbers F=A000045.

Original entry on oeis.org

0, 1, 5, 23, 98, 418, 1770, 7503, 31779, 134629, 570284, 2415788, 10233404, 43349461, 183631161, 777874251, 3295127934, 13958386366, 59128672790, 250473078515, 1061020985255, 4494557022121, 19039249069560, 80651553307128
Offset: 0

Views

Author

Wolfdieter Lang, Aug 09 2012

Keywords

Comments

For a derivation of the explicit form of this sum see the link under A215308 on the partial summation formula, eq. (7).

Examples

			a(2) = 0 + 1^2*1 + 2^2*1 = 1 + 4 = 5.

Links

Crossrefs

Cf. A000045, A001655, A008346, A066258, A066259, A215308, A215037.

Formula

a(n) = Sum_{k=0..n} A066259(k) = Sum_{k=0..n} F(k+1)^2*F(k), n >= 0, with A066259(0)=0.
a(n) = (F(n+2)*F(n+1)^2 - (-1)^n*F(n) - 1)/2 = (A066258(n+1) - (-1)^n*A008346(n))/2, n >= 0.
O.g.f.: x*(1+x)/((1+x-x^2)*(1-4*x-x^2)*(1-x)) (from A066259).
E.g.f.: (2*exp(-x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2)) + exp(2*x)*(15*cosh(sqrt(15)*x) + 7*sqrt(5)*sinh(sqrt(5)*x)) - 25*exp(x))/50. - Stefano Spezia, Oct 28 2024
Showing 1-10 of 11 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.