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.

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A001657 Fibonomial coefficients: column 5 of A010048.

Original entry on oeis.org

1, 8, 104, 1092, 12376, 136136, 1514513, 16776144, 186135312, 2063912136, 22890661872, 253854868176, 2815321003313, 31222272414424, 346260798314872, 3840089017377228, 42587248616222024, 472299787252290712, 5237885063192296801, 58089034826620525728
Offset: 0

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Author

Keywords

Examples

			G.f. = 1 + 8*x + 104*x^2 + 1092*x^3 + 12376*x^4 + 136136*x^5 + 1514513*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).

Crossrefs

Programs

  • Maple
    with(combinat) : a:=n-> 1/30*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3)*fibonacci(n+4): seq(a(n), n=1..19); # Zerinvary Lajos, Oct 07 2007
    A001657:=-1/(z**2+11*z-1)/(z**2-4*z-1)/(z**2+z-1); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_] := Times @@ Fibonacci[Range[n + 1, n + 5]]/30; t = Table[f[n], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
    LinearRecurrence[{8,40,-60,-40,8,1},{1,8,104,1092,12376,136136},20] (* Harvey P. Dale, Nov 30 2019 *)
  • PARI
    a(n)=(n->(n^5-n)/30)(fibonacci(n+3)) \\ Charles R Greathouse IV, Apr 24 2012
    
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 5))  \\ Joerg Arndt, May 08 2016

Formula

a(n) = A010048(5+n, 5) (or fibonomial(5+n, 5)).
G.f.: 1/(1-8*x-40*x^2+60*x^3+40*x^4-8*x^5-x^6) = 1/((1-x-x^2)*(1+4*x-x^2)*(1-11*x-x^2)) (see Comments to A055870).
a(n) = 11*a(n-1) + a(n-2) + ((-1)^n)*fibonomial(n+3, 3), n >= 2; a(0)=1, a(1)=8; fibonomial(n+3, 3)= A001655(n).
a(n) = Fibonacci(n+3)*(Fibonacci(n+3)^4-1)/30. - Gary Detlefs, Apr 24 2012
a(n) = (A049666(n+3) + 2*(-1)^n*A001076(n+3) - 3*A000045(n+3))/150, n >= 0, with A049666(n) = F(5*n)/5, A001076(n) = F(3*n)/2 and A000045(n) = F(n). From the partial fraction decomposition of the o.g.f. and recurrences. - Wolfdieter Lang, Aug 23 2012
a(n) = a(-6-n) * (-1)^n for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(-a(n+1) - 3*a(n+2)) + a(n+1)*(-8*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
G.f.: exp( Sum_{k>=1} F(6*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

Corrected and extended by Wolfdieter Lang, Jun 27 2000

A056573 Sixth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 64, 729, 15625, 262144, 4826809, 85766121, 1544804416, 27680640625, 496981290961, 8916100448256, 160005726539569, 2871098559212689, 51520374361000000, 924491486192068809, 16589354847268067929
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Comments

A divisibility sequence; that is, if n divides m, then a(n) divides a(m).

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).

Crossrefs

Sixth row of array A103323.

Programs

Formula

a(n) = F(n)^6, where F(n) = A000045(n).
G.f.: x*p(6, x)/q(6, x) with p(6, x) := sum_{m=0..5} A056588(5, m)*x^m = (1-x)*(1 - 11*x - 64*x^2 - 11*x^3 + x^4) and q(6, x) := sum_{m=0..7} A055870(7, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..7} A055870(7, m)*a(n-m) = 0, n >= 7; inputs: a(n), n=0..6. a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7).
From Gary Detlefs, Jan 07 2013: (Start)
a(n) = (F(3*n)^2 - (-1)^n*6*F(n)*F(3*n) + 9*F(n)^2)/25.
a(n) = (10*F(n)^3*F(3*n) - F(3*n)^2 + 9*F(n)^2)/25. (End)
a(n+1) = 2*[2*F(n+1)^2-(-1)^n]^3+3*F(n)^2*F(n+1)^2*F(n+2)^2-[F(n)^6+F(n+2)^6] = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^6, for n (this is Theorem 2.2 (vi) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

A001658 Fibonomial coefficients.

Original entry on oeis.org

1, 13, 273, 4641, 85085, 1514513, 27261234, 488605194, 8771626578, 157373300370, 2824135408458, 50675778059634, 909348684070099, 16317540120588343, 292806787575013635, 5254201798026392211, 94282845030238533383, 1691836875411111866723, 30358781826262552258596
Offset: 0

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Author

Keywords

Comments

It appears that a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 6. - John W. Layman, Apr 14 2000
Layman's formula is correct. - Wolfdieter Lang, Jul 13 2000
Layman's formula is a consequence of formula 2.8 (p. 116) of Lind (1971). - Dale Gerdemann, May 08 2016

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).

Crossrefs

Programs

  • Mathematica
    f[n_] := Times @@ Fibonacci[Range[n+1, n+6]]/240; Table[f[n], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
    LinearRecurrence[{13,104,-260,-260,104,13,-1},{1,13,273,4641,85085,1514513,27261234},20] (* Harvey P. Dale, Aug 24 2014 *)
  • PARI
    b(n,k)=prod(j=1,k,fibonacci(n+j)/fibonacci(j));
    vector(20,n,b(n-1,6))  \\ Joerg Arndt, May 08 2016

Formula

From Wolfdieter Lang, Jul 13 2000: (Start)
G.f.: 1/(1-13*x-104*x^2+260*x^3+260*x^4-104*x^5-13*x^6+x^7) = 1/((1+x)*(1-3*x+x^2)*(1+7*x+x^2)*(1-18*x+x^2)) (see Comments to A055870).
a(n) = 5*a(n-1)+F(n-5)*Fibonomial(n+5, 5), n >= 1, a(0) = 1; F(n) = A000045(n) (Fibonacci). a(n) = 18*a(n-1)-a(n-2)+((-1)^n)*Fibonomial(n+4, 4), n >= 2; a(0) = 1, a(1) = 13; Fibonomial(n+4, 4) = A001656(n). (End)
From Gary Detlefs, Dec 03 2012: (Start)
a(n) = F(n+1)*F(n+2)*F(n+3)*F(n+4)*F(n+5)*F(n+6)/240.
a(n) = (F(n+5)^2 - F(n+4)^2)*(F(n+3)^4 - 1)/240, where F(n) = A000045(n). (End)
Conjecture: a(n) = F(7)^(n-6) + Sum_{i=3..n-5} F(i-2)F(6)^{i-1}F(7)^{n-i-5} + Sum_{j=3..i} F(i-2)F(j-2)F(5)^{j-1}F(6)^{i-j}F(7)^{n-i-5} + Sum_{k=3..j} F(i-2)F(j-2)F(k-2)F(4)^{k-1}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5} + Sum_{l=3..k} F(i-2)F(j-2)F(k-2)F(l-2)F(3)^{l-1}F(4)^{k-l}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5} + Sum_{m=3..l} F(i-2)F(j-2)F(k-2)F(l-2)F(m-2)F(m)F(3)^{l-m}F(4)^{k-l}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5}, where F(n)=A000045(n). - Dale Gerdemann, May 08 2016
G.f.: exp( Sum_{k>=1} F(7*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

More terms from Wolfdieter Lang, Jul 13 2000

A056574 Seventh power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 128, 2187, 78125, 2097152, 62748517, 1801088541, 52523350144, 1522435234375, 44231334895529, 1283918464548864, 37281334283719577, 1082404156823183753, 31427428360210000000, 912473096871571914483
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Comments

A divisibility sequence; that is, if n divides m, then a(n) divides a(m).

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).

Crossrefs

Seventh row of array A103323.

Programs

Formula

a(n) = F(n)^7, where F(n) = A000045(n).
G.f.: x*p(7, x)/q(7, x) with p(7, x) := sum_{m=0..6} A056588(6, m)*x^m = 1 - 20*x - 166*x^2 + 318*x^3 + 166*x^4 - 20*x^5 - x^6 and q(7, x) := sum_{m=0..8} A055870(8, m)*x^m = (1 + x - x^2)*(1 - 4*x - x^2)*(1 + 11*x - x^2)*(1 - 29*x - x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..8} A055870(8, m)*a(n-m) = 0, n >= 8; inputs: a(n), n=0..7. a(n) = 21*a(n-1) + 273*a(n-2) - 1092*a(n-3) - 1820*a(n-4) + 1092*a(n-5) + 273*a(n-6) - 21*a(n-7) - a(n-8).
a(n+1) = F(n)^7+F(n+1)^7+7*F(n)*F(n+1)*F(n+2)*[2*F(n+1)^2-(-1)^n]^2 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^7, for n>=0 (This is Theorem 2.3 (iv) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

A003268 Central Fibonomial coefficients.

Original entry on oeis.org

1, 2, 6, 15, 60, 260, 1820, 12376, 136136, 1514513, 27261234, 488605194, 14169550626, 411591708660, 19344810307020, 908637119420910, 69056421075989160, 5249543573067466872, 645693859487298425256, 79413089729752455762384, 15803204856220738696714416
Offset: 0

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Author

Keywords

References

  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 74.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Central column of A010048, |A055870|.
Cf. A062073.

Programs

  • Mathematica
    Table[Product[Fibonacci[k],{k,Floor[n/2]+1,n}] / Product[Fibonacci[k],{k,1,Ceiling[n/2]}],{n,2,20}] (* Vaclav Kotesovec, Apr 10 2015 *)

Formula

a(n) = (Product_{k=floor(n/2)+1..n} Fibonacci(k)) / (Product_{k=1..ceiling(n/2)} Fibonacci(k)).
a(n) ~ ((1+sqrt(5))/2)^(n^2/4 + n + 1 - (1-(-1)^n)/8) / A062073, where A062073 = 1.2267420107203532444176302... is the Fibonacci factorial constant. - Vaclav Kotesovec, May 01 2015

Extensions

More terms from Vaclav Kotesovec, May 01 2015

A056565 Fibonomial coefficients.

Original entry on oeis.org

1, 21, 714, 19635, 582505, 16776144, 488605194, 14169550626, 411591708660, 11948265189630, 346934172869802, 10072785423545712, 292460526776698763, 8491396839675395415, 246543315138161480670, 7158243695757340957617, 207835653079349665473587
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Crossrefs

Programs

  • Magma
    [ &*[Fibonacci(n+k): k in [0..6]]/3120: n in [1..16] ]; // Bruno Berselli, Apr 11 2011
    
  • Mathematica
    (Times@@@Partition[Fibonacci[Range[30]],7,1])/3120  (* Harvey P. Dale, Apr 10 2011 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 7))  \\ Joerg Arndt, May 08 2016

Formula

a(n) = A010048(n+7, 7) =: Fibonomial(n+7, 7).
G.f.: 1/p(8, n) with p(8, n) = 1 - 21*x - 273*x^2 + 1092*x^3 + 1820*x^4 - 1092*x^5 - 273*x^6 + 21*x^7 + x^8 = (1 + x - x^2) * (1 - 4*x - x^2) * (1 + 11*x - x^2) * (1 - 29*x - x^2) (n=8 row polynomial of signed Fibonomial triangle A055870; see this entry for Knuth and Riordan references).
a(n) = 29*a(n-1) + a(n-2) + ((-1)^n) * A001657(n), n >= 2, a(0)=1, a(1)=21.
G.f.: exp( Sum_{k>=1} F(8*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

Offset corrected by Seiichi Manyama, May 07 2025

A215039 a(n) = Fibonacci(2*n)^3, n>=0.

Original entry on oeis.org

0, 1, 27, 512, 9261, 166375, 2985984, 53582633, 961504803, 17253512704, 309601747125, 5555577996431, 99690802348032, 1788878864685457, 32100128763082731, 576013438873664000, 10336141770970357629, 185474538438612378103
Offset: 0

Views

Author

Wolfdieter Lang, Aug 10 2012

Keywords

Comments

Bisection (even part) of A056570. From this follows the o.g.f., and its partial fraction decomposition leads to the explicit formula given below. The recurrences for F(2*n) and F(6*n)/8 are used in this computation. They follow from the fact that F(2*n) = S(n-1,3), and F(6*n)/8 = S(n-1,18), with Chebyshev's S(n,x) = U(n,x/2) polynomial of the second kind (see A001906 and A049660, respectively).

Crossrefs

Cf. A000045, A056570, A163198 (partial sums).

Programs

  • GAP
    List([0..20], n-> Fibonacci(2*n)^3 ); # G. C. Greubel, Dec 22 2019
  • Magma
    [Fibonacci(2*n)^3: n in [0..20]]; // G. C. Greubel, Dec 22 2019
    
  • Maple
    with(combinat); seq( fibonacci(2*n)^3, n=0..20); # G. C. Greubel, Dec 22 2019
  • Mathematica
    Fibonacci[2*(Range[21]-1)]^3 (* G. C. Greubel, Dec 22 2019 *)
  • PARI
    vector(21, n, fibonacci(2*(n-1)) ); \\ G. C. Greubel, Dec 22 2019
    
  • Sage
    [fibonacci(2*n)^3 for n in (0..20)] # G. C. Greubel, Dec 22 2019
    

Formula

a(n) = F(2*n)^3, n>=0, with F=A000045.
O.g.f.: x*(1+6*x+x^2)/((1-3*x+x^2)*(1-18*x+x^2)) (from the bisection (even part) of A056570).
a(n) = (F(6*n) - 3*F(2*n))/5, n>=0.
a(n+2) - 18*a(n+1) + a(n) - 9*F(2*(n+1)) = 0, n>=0. From the F_n^3 recurrence (see a comment and references on A055870, use row n=4) together with the recurrence appearing in the solution of exercise 6.58, p. 315, on p. 556 of the second edition of the Graham-Knuth-Patashnik book (reference given on A007318), both with n -> 2*n. See also Koshy's book (reference given on A065563) p. 87, 1. and p. 89, 32. (with a - sign) and 33. - Wolfdieter Lang, Aug 11 2012

A056566 Fibonomial coefficients.

Original entry on oeis.org

1, 34, 1870, 83215, 3994320, 186135312, 8771626578, 411591708660, 19344810307020, 908637119420910, 42689423937884208, 2005443612183077232, 94214069697350815795, 4426039514623184676790, 207929935924379904006970, 9768275694729434277258589, 458901121999204061365680096
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Crossrefs

Cf. A010048, A000045, A001654-8, A056565, A001906 (signed), A004187, A049660 (signed), A049668.

Programs

  • Mathematica
    a[n_] := (1/65520) Times @@ Fibonacci[n + Range[8]]; Array[a, 20, 0] (* Giovanni Resta, May 08 2016 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 8)) \\ Joerg Arndt, May 08 2016

Formula

a(n) = A010048(n+8, 8) = Fibonomial(n+8, 8).
G.f.: 1/p(9, n) with p(9, n)= 1 - 34*x - 714*x^2 + 4641*x^3 + 12376*x^4 - 12376*x^5 - 4641*x^6 + 714*x^7 + 34*x^8 - x^9 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2)* (1 + 18*x + x^2)*(1 - 47*x + x^2) (n=9 row polynomial of signed Fibonomial triangle A055870; see this entry for Knuth and Riordan references).
Recursion: a(n) = 47*a(n-1) - a(n-2) + ((-1)^n)*A001658(n), n >= 2, a(0)=1, a(1)=34.
G.f.: exp( Sum_{k>=1} F(9*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A056585 Eighth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 256, 6561, 390625, 16777216, 815730721, 37822859361, 1785793904896, 83733937890625, 3936588805702081, 184884258895036416, 8686550888106661441, 408066367122340274881, 19170731299728100000000
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Comments

A divisibility sequence; that is, if n divides m, then a(n) divides a(m).

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).

Crossrefs

Programs

Formula

a(n) = F(n)^8, F(n)=A000045(n).
G.f.: x*p(8, x)/q(8, x) with p(8, x) := sum_{m=0..7} A056588(7, m)*x^m = (1+x)*(1 - 34*x - 458*x^2 + 2242*x^3 - 458*x^4 - 34*x^5 + x^6) and q(8, x) := sum_{m=0..9} A055870(9, m)*x^m = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2)*(1 + 18*x + x^2)*(1 - 47*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..9} A055870(9, m)*a(n-m) = 0, n >= 9; inputs: a(n), n=0..8. a(n) = 34*a(n-1) + 714*a(n-2) - 4641*a(n-3) - 12376*a(n-4) + 12376*a(n-5) + 4641*a(n-6) - 714*a(n-7) - 34*a(n-8) + a(n-9).
a(n+1) = 8*F(n)^2*F(n+1)^2*[F(n)^4+F(n+1)^4+4*F(n)^2*F(n+1)^2+3*F(n)*F(n+1)*F(n+2)]-[F(n)^8+F(n+2)^8]+2*[2*F(n+1)^2-(-1)^n]^4 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^8, for n>=0 (This is Theorem 2.2 (vii) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

A217471 Partial sum of fifth power of the even-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 244, 33012, 4117113, 507401488, 62424765712, 7678070811369, 944346243245076, 116147016764564500, 14285140634333292625, 1756956185432949082176, 216091326285380812359744, 26577476188001703626949937
Offset: 0

Views

Author

Wolfdieter Lang, Oct 11 2012

Keywords

Comments

For the o.g.f. for general powers of Fibonacci numbers F=A000045 see A056588 (row polynomials as numerators) and A055870 (row polynomials as denominator). The even part of the bisection leads to the o.g.f. for powers of F(2*n), and the partial sums of these powers are then given by dividing this o.g.f. by (1-x). For the o.g.f.s for F(n)^5 and F(2*n)^5 see A056572 and A215044, respectively.
The tables of the coefficient of the polynomials which appear in Ozeki's formula and in Melham's conjecture are found in A217472 and A217475, respectively (see References).

Examples

			a(2) = 244 = 2*(8-3)/5 - 610/20 + (832040-6765)/55^2 - 7/22.
a(2) = 244 = (1/11)*5^5 - (15/44)*5^3 + (25/44)*5 - 7/22.
a(2) = 244 = (5-1)^2*(4*5^3 + 8*5^2 - 3*5 - 14)/44
           = (4*5^3 + 8*5^2 - 3*5 - 14)*(4/11).
		

Crossrefs

Cf. A163198 (third powers).

Programs

  • Mathematica
    Table[Sum[Fibonacci[2*k]^5, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Apr 12 2017 *)
    Accumulate[Fibonacci[Range[0,30,2]]^5] (* Harvey P. Dale, Jun 30 2025 *)
  • PARI
    a(n) = sum(k=1, n, fibonacci(2*k)^5); \\ Michel Marcus, Feb 29 2016

Formula

a(n) = Sum_{k=0..n} F(2*k)^5, n>=0.
O.g.f.: x*(1+99*x+416*x^2+99*x^3+x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)*(1-x)).
a(n) = 2*(F(2*(n+1)) - F(2*n))/5 - F(3*(2*n+1))/20 +
(F(10*(n+1)) - F(10*n))/F(10)^2 - 7/22 (from the partial fraction decomposition of the o.g.f.).
a(n) = (1/11)*F(2*n+1)^5 - (15/44)*F(2*n+1)^3 + (25/44)*F(2*n+1) - 7/22 (from Ozeki reference, Theorem 2, p. 109 --- with a misprint -- and from Prodinger reference, p. 207).
a(n) =(F(2*n+1)-1)^2*(4*F(2*n+1)^3 + 8*F(2*n+1)^2 - 3*F(2*n+1) - 14)/44 (an example for Melham's conjecture, see the reference, eq. (2.7) for m=2).
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