A064831 -id:A064831 - cp's OEIS frontend

A002571 From a definite integral.

1, 5, 10, 30, 74, 199, 515, 1355, 3540, 9276, 24276, 63565, 166405, 435665, 1140574, 2986074, 7817630, 20466835, 53582855, 140281751, 367262376, 961505400, 2517253800, 6590256025, 17253514249, 45170286749, 118257345970

Offset: 1

N. J. A. Sloane

nonn

a(n) are the row sums of the elements of the Golden Triangle (A180662) with alternating signs. - Alexander Adamchuk, Oct 18 2010

Limit_{n->oo} A002570(n)/A002571(n) = 1/sqrt(5). - Sean A. Irvine, Apr 09 2014

			From _Paul D. Hanna_, Feb 20 2009: (Start)
G.f.: A(x) = x + 5*x^2 + 10*x^3 + 30*x^4 + 74*x^5 + 199*x^6 + ...
log(1+A(x)) = x + 3^2*x^2/2 + 4^2*x^3/3 + 7^2*x^4/4 + 11^2*x^5/5 + ... (End)
G.f.: A(x) = -1 + 1/((1-x-x^2) * (1-3*x^2+x^4) * (1-4*x^3-x^6) * (1-7*x^4+x^8) * (1-11*x^5-x^10)^2 * (1-18*x^6+x^12)^2 * (1-29*x^7-x^14)^4 * (1-47*x^8+x^16)^5 * (1-76*x^9-x^18)^8 * ...* (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^A006206(n) * ...). - _Paul D. Hanna_, Jan 07 2012

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).

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
L. R. Shenton, A determinantal expansion for a class of definite integral. Part 5. Recurrence relations, Proc. Edinburgh Math. Soc. (2) 10 (1957), 167-188.
L. R. Shenton and K. O. Bowman, Second order continued fractions and Fibonacci numbers, Far East Journal of Applied Mathematics, 20(1), 17-31, 2005.

Cf. A064831, A077916, A000045, A000204 (Lucas), A006206.

Cf. A001654, A180662 - The Golden Triangle. - Alexander Adamchuk, Oct 18 2010

A002571:=-(-1-4*z-z**2+z**3)/(z**2-3*z+1)/(1+z)**2; # conjectured (probably correctly) by Simon Plouffe in his 1992 dissertation

{a(n)=polcoeff(exp(sum(m=1,n,(fibonacci(m+1)+fibonacci(m-1))^2*x^m/m)+x*O(x^n)),n)} \\ Paul D. Hanna, Feb 20 2009

Appears to have g.f. x/((1-3x+x^2)*(1+x)^2). - Ralf Stephan, Apr 14 2004
a(n) = (-1)^n*Sum_{i=1..n+1} (-1)^(i+1)*Fibonacci(i)*Fibonacci(i+1). - Alexander Adamchuk, Jun 16 2006
From Paul D. Hanna, Feb 20 2009: (Start)
Given g.f. A(x), then log(1+A(x)) = Sum_{n>=1} A000204(n)^2 * x^n/n where A000204 is the Lucas numbers.
a(n) = (1/n)*(A000204(n)^2 + Sum_{k=1..n-1} A000204(k)^2*a(n-k)) for n>1, with a(1) = 1. (End)
G.f.: -1 + 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^A006206(n), where A006206(n) is the number of aperiodic binary necklaces of length n with no subsequence 00. - Paul D. Hanna, Jan 07 2012
a(n) = 8*a(n-2) - 8*a(n-4) + a(n-6) + 2(-1)^n, n>6. - Sean A. Irvine, Apr 09 2014
a(n) - a(n-2) = Fibonacci(n+1)^2. - Peter Bala, Aug 30 2015

More terms from Max Alekseyev and Alexander Adamchuk, Oct 18 2010

A077916 Expansion of (1-x)^(-1)/(1 + 2x - 2x^2 - x^3).

Original entry on oeis.org

1, -1, 5, -10, 30, -74, 199, -515, 1355, -3540, 9276, -24276, 63565, -166405, 435665, -1140574, 2986074, -7817630, 20466835, -53582855, 140281751, -367262376, 961505400, -2517253800, 6590256025, -17253514249, 45170286749, -118257345970, 309601751190, -810547907570

Offset: 0

N. J. A. Sloane, Nov 17 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,4,-1,-1).

Cf. A002571.
Bisections are A103433 and A103434.
Cf. A064831, A000045. A203803, A207969, A207970.

a[0] = 1; a[1] = -1; a[2] = 5; a[3] = -10; a[n_] := a[n] = -a[n-1] + 4 a[n-2] - a[n-3] - a[n-4]; Table[a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
CoefficientList[Series[(1 - x)^(-1)/(1 + 2*x - 2*x^2 - x^3), {x, 0, 50}], x] (* G. C. Greubel, Dec 25 2017 *)
Table[If[OddQ[n], (Fibonacci[2n+2]+n+1)/5, -(Fibonacci[2n+2]-n-1)/5], {n,1,20}] (* Rigoberto Florez, May 09 2019 *)

Vec((1-x)^(-1)/(1+2*x-2*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

Vec(1/((1-x)^2*(1+3*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 28 2015

a(n-1) = Sum_{i=1..n} (-1)^(i+1)*Fibonacci(i)*Fibonacci(i+1), n >= 1. - Alexander Adamchuk, Jun 16 2006
From R. J. Mathar, Mar 14 2011: (Start)
G.f.: 1/((1-x)^2*(1+3*x+x^2)).
a(n) = ((-1)^n*A001906(n+2)+n+2)/5. (End)
O.g.f.: exp( Sum_{n >= 1} Lucas(n)^2*(-x)^n/n ) = 1 - x + 5*x^2 - 10*x^3 + .... Cf. A203803. See also A207969 and A207970. - Peter Bala, Apr 03 2014
From Vladimir Reshetnikov, Oct 28 2015: (Start)
Recurrence (5-term): a(0) = 1, a(1) = -1, a(2) = 5, a(3) = -10, a(n) = -a(n-1) + 4*a(n-2) - a(n-3) - a(n-4).
Recurrence (4-term): a(0) = 1, a(1) = -1, a(2) = 5, n*a(n) = (1-2*n)*a(n-1) + (3*n+3)*a(n-2) + (n+1)*a(n-3).
(End)
a(n) = (F(2n+2)+n+1)/5 if n is odd and a(n)= -(F(2n+2)-n-1)/5 if n is even, where F(n) = Fibonacci numbers (A000045). - Rigoberto Florez, May 09 2019

A080144 a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 if n odd, a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 - F(5) if n even, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

0, 8, 21, 63, 165, 440, 1152, 3024, 7917, 20735, 54285, 142128, 372096, 974168, 2550405, 6677055, 17480757, 45765224, 119814912, 313679520, 821223645, 2149991423, 5628750621, 14736260448, 38580030720, 101003831720

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 31 2003

The present sequence is a member of the m-family of sums b(m,n) := Sum_{k=1..n} F(k+m)*F(k) for m >= 0, n >= 0 (see the B. Cloitre formula given below (m=5)). These sums are (F(m)*A027941(n) + L(m)*A001654(n))/2, with F = A000045 and the L = A000032. Proof by induction on m using the recurrence. - Wolfdieter Lang, Jul 27 2012
The o.g.f. of b(m,n) is A(m,x) = x*(F(m+1) - F(m-1)*x)/((1-x^2)*(1-3*x+x^2)), m >= 0, with F(-1)=1. - Wolfdieter Lang, Jul 30 2012
b(m,n) = ((-1)^(n+1)*L(m) - 5*F(m) + 2*L(m + 2*n + 1))/10. - Ehren Metcalfe, Aug 21 2017

G. C. Greubel, Table of n, a(n) for n = 0..1000
S. Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Index entries for linear recurrences with constant coefficients, signature (3, 0, -3, 1).

Cf. A000032, A000045, A059840, A064831, A080143.

F:=Fibonacci;; List([0..30], n-> (2*F(n+3)^2 -5-3*(-1)^n)/2); # G. C. Greubel, Jul 23 2019

F:=Fibonacci; [(2*F(n+3)^2 -5-3*(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x*(8+5*x-3*x^2)/((1-x^2)*(1-2x-2x^2+x^3)), {x, 0, 30}], x]
With[{F=Fibonacci}, Table[(2*F[n + 3]^2 -5-3*(-1)^n)/2, {n,0,30}]] (* G. C. Greubel, Jul 23 2019 *)

my(x='x+O('x^30)); concat([0],Vec(x*(8-3*x)/((1-x^2)*(1-3*x+x^2)) )) \\ G. C. Greubel, Mar 04 2017

vector(30, n, n--; f=fibonacci; (2*f(n+3)^2 -5-3*(-1)^n)/2) \\ G. C. Greubel, Jul 23 2019

f=fibonacci; [(2*f(n+3)^2 -5-3*(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Jul 23 2019

G.f.: x*(8-3*x)/((1-x^2)*(1-3*x+x^2)) (see the comment section). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{i=0..n} A000045(i+5)*A000045(i). - Benoit Cloitre, Jun 14 2004
a(n) = (5*A027941(n) + 11*A001654(n))/2, n >= 0. See A080143 and A080097. See the comment section for the general formula for such sums. - Wolfdieter Lang, Jul 27 2012
a(n) = (2*Lucas(2*n + 6) + 11*(-1)^(n+1) - 25)/10. - Ehren Metcalfe, Aug 21 2017
a(n) = (2*Fibonacci(n+3)^2 - 5 - 3*(-1)^n)/2. - G. C. Greubel, Jul 23 2019

A005313 Maximal sum of inverse squares of the singular values of triangular anti-Hadamard matrices of order n.

Original entry on oeis.org

1, 3, 6, 13, 29, 70, 175, 449, 1164, 3035, 7931, 20748, 54301, 142143, 372114, 974185, 2550425, 6677074, 17480779, 45765245, 119814936, 313679543, 821223671, 2149991448, 5628750649, 14736260475, 38580030750, 101003831749, 264431464469, 692290561630, 1812440220391, 4745030099513

Offset: 1

N. J. A. Sloane

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

Noga Alon and Van H. Vu, Anti-Hadamard Matrices, Coin Weighing, Threshold Gates, and Indecomposable Hypergraphs, Journal of Combinatorial Theory, Series A, Volume 79, Issue 1, July 1997, Pages 133-160.
R. L. Graham and N. J. A. Sloane, Anti-Hadamard matrices, Linear Alg. Applic., 62 (1984), 113-137, Table 1.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (4,-3,-3,4,-1).

Cf. A005312, A064831.

a(n) = A064831(n-1) + n.

G.f.: [x(1-x-3x^2+x^3)]/[(1-3x+x^2)(1+x)(1-x)^2]. - Conjectured by Simon Plouffe in his 1992 dissertation. [This is in fact the correct g.f. - N. J. A. Sloane, Jan 28 2022]

Definition corrected by Stefano Spezia, Jan 30 2022

A097083 Positive values of k such that there is exactly one permutation p of (1,2,3,...,k) such that i+p(i) is a Fibonacci number for 1<=i<=k.

Original entry on oeis.org

1, 2, 3, 5, 9, 15, 24, 39, 64, 104, 168, 272, 441, 714, 1155, 1869, 3025, 4895, 7920, 12815, 20736, 33552, 54288, 87840, 142129, 229970, 372099, 602069, 974169, 1576239, 2550408, 4126647, 6677056, 10803704, 17480760, 28284464, 45765225

Offset: 1

John W. Layman, Jul 23 2004

nonn

Numbers k such that A097082(k) = 1. If f is a Fibonacci number and k < f <= 2k, then a permutation for f-k-1 may be extended to a permutation for k, with p(i) = f-i for f-k < i <= k. This explains the sparseness of this sequence. - David Wasserman, Dec 19 2007

If the formula is correct, the bisections give A059840 and A064831. - David Wasserman, Dec 19 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,0,-1).

Cf. A073364, A000045.

a=b=c=d=0;Table[e=a+b+d+1;a=b;b=c;c=d;d=e,{n,100}] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2011 *)
CoefficientList[Series[x/((x - 1)*(x^2 + 1)*(x^2 + x - 1)), {x,0,50}], x] (* G. C. Greubel, Mar 05 2017 *)
LinearRecurrence[{2,-1,1,0,-1},{1,2,3,5,9},50] (* Harvey P. Dale, Nov 09 2024 *)

x='x+O('x^50); Vec(x/((x - 1)*(x^2 + 1)*(x^2 + x - 1))) \\ G. C. Greubel, Mar 05 2017

It appears that {a(n)} satisfies a(1)=1, a(2)=2 and, for n>2, a(n) = F(n+2) - a(n-2) - 1, where {F(k)} is the sequence of Fibonacci numbers, i.e, that the sequence is the partial sums of A006498.
If the partial sum assumption is correct: a(n) = floor(phi^(n+3)/5), where phi=(1+sqrt(5))/2 = A001622, and a(n) = a(n-1) + a(n-2) + ( (n*(n+1)/2) mod 2). - Gary Detlefs, Mar 12 2011
From R. J. Mathar, Mar 13 2011: (Start)
If the partial sum assumption is correct: a(n)= +2*a(n-1) -a(n-2) +a(n-3) -a(n-5).
G.f.: x/( (x-1)*(x^2+1)*(x^2+x-1) ).
a(n) = A000032(n+3)/5 -(-1)^n*A112030(n)/10 - 1/2. (End)
Conjecture: a(n) = floor(F(n+3)/sqrt(5)), where F(n) = A000045(n) are Fibonacci numbers. - Vladimir Reshetnikov, Nov 05 2015

a(9) from Ray Chandler, Jul 29 2004
More terms from David Wasserman, Dec 19 2007
Terms > 90000 assuming the partial sums formula by Vladimir Joseph Stephan Orlovsky, Feb 26 2011

A180664 Golden Triangle sums: a(n) = a(n-1) + A001654(n+1) with a(0)=0.

Original entry on oeis.org

0, 2, 8, 23, 63, 167, 440, 1154, 3024, 7919, 20735, 54287, 142128, 372098, 974168, 2550407, 6677055, 17480759, 45765224, 119814914, 313679520, 821223647, 2149991423, 5628750623, 14736260448, 38580030722, 101003831720

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn13 sums of the Golden Triangle A180662. See A180662 for information about these knight and other chess sums.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A000032, A000045, A005248, A064831, A115730, A180662, A180664, A180665, A180666.

[(1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)): n in [0..40]]; // G. C. Greubel, Jan 21 2022

nmax:=26: with(combinat): for n from 0 to nmax+1 do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax do a(n) := a(n-1)+A001654(n+1) od: seq(a(n),n=0..nmax);

Table[Sum[Fibonacci[i+2]*Fibonacci[i+3], {i,0,n-1}], {n,0,40}] (* Rigoberto Florez, Jul 07 2020 *)
LinearRecurrence[{3,0,-3,1},{0,2,8,23},30] (* Harvey P. Dale, Mar 30 2023 *)

[(1/10)*((-1)^n - 15 + 2*lucas_number2(2*n+4,1,-1)) for n in (0..40)] # G. C. Greubel, Jan 21 2022

a(n+1) = Sum_{k=0..n} A180662(2*n-k+2, k+2).
a(n) = (-15 + (-1)^n + (6-2*A)*A^(-n-1) + (6-2*B)*B^(-n-1))/10 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.
G.f.: (2*x+2*x^2-x^3)/(1-3*x-x^4+3*x^3).
a(n) = Sum_{i=0..n-1} F(i+2)*F(i+3), where F(i) = A000045(i). - Rigoberto Florez, Jul 07 2020
a(n) = (1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)). - G. C. Greubel, Jan 21 2022

A080143 a(n) = F(3)F(n)F(n+1) + F(4)F(n+1)^2 - F(4) if n even, F(3)F(n)F(n+1) + F(4)F(n+1)^2 if n odd, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

0, 5, 13, 39, 102, 272, 712, 1869, 4893, 12815, 33550, 87840, 229968, 602069, 1576237, 4126647, 10803702, 28284464, 74049688, 193864605, 507544125, 1328767775, 3478759198, 9107509824, 23843770272, 62423800997, 163427632717

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 30 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
S. Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A064831, A059840, A080097, A080144.

F:=Fibonacci;; List([0..30], n-> (2*F(n+2)*F(n+3) -3 -(-1)^n)/2); # G. C. Greubel, Jul 23 2019

F:=Fibonacci; [(2*F(n+2)*F(n+3) -3 -(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x*(5+3*x-2*x^2)/((1-x^2)*(1-2*x-2*x^2+x^3)), {x, 0, 30}], x]
With[{F=Fibonacci}, Table[(2*F[n+2]*F[n+3] -3 -(-1)^n)/2, {n,0,30}]] (* G. C. Greubel, Jul 23 2019 *)

my(x='x+O('x^30)); concat([0], Vec(x*(5+3*x-2*x^2)/((1-x^2)*(1- 2*x-2*x^2+x^3)))) \\ G. C. Greubel, Mar 05 2017

vector(30, n, n--; f=fibonacci; (2*f(n+2)*f(n+3) -3 -(-1)^n)/2) \\ G. C. Greubel, Jul 23 2019

f=fibonacci; [(2*f(n+2)*f(n+3) -3 -(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Jul 23 2019

G.f.: x*(5-2*x)/((1-x^2)*(1-3*x+x^2)), see a comment on A080144 for A(4,x). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{i=0..n} ( A000045(i+4)*A000045(i) ). - Benoit Cloitre, Jun 14 2004
a(n) = (3*A027941(n) + 7*A001654(n))/2, n >= 0. Proof: from the preceding sum given by B. Cloitre via recurrence on the first factor under the sum. See also A080097 and A059840(n+2). - Wolfdieter Lang, Jul 27 2012
a(n) = (2*Lucas(2*n + 5) + 7*(-1)^(n+1) - 15)/10. - Ehren Metcalfe, Aug 21 2017
a(n) = (2*Fibonacci(n+2)*Fibonacci(n+3) - 3 - (-1)^n)/2. - G. C. Greubel, Jul 23 2019

A115730 a(n) = a(n-3) + A001654(n-1) with a(0)=0, a(1)=0 and a(2)=1.

Original entry on oeis.org

0, 0, 1, 2, 6, 16, 42, 110, 289, 756, 1980, 5184, 13572, 35532, 93025, 243542, 637602, 1669264, 4370190, 11441306, 29953729, 78419880, 205305912, 537497856, 1407187656, 3684065112, 9645007681, 25250957930, 66107866110

Offset: 0

Roger L. Bagula, Mar 13 2006

The a(n+1) represent the Ca2 and Ze4 sums of the Golden Triangle A180662. Furthermore the a(3*n) represent the Ze1 (terms doubled) and Ca3 sums of the Golden triangle. See A180662 for more information about these and other triangle sums.

			G.f. = x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 42*x^6 + 110*x^7 + 289*x^8 + ... - _Michael Somos_, Sep 05 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,0,-2,-2,1).

Cf. A000045, A001654, A064831, A079962, A180662, A180664, A180665, A180666, A182890.

function A115730(n)
  if n lt 3 then return Floor(n/2);
  else return A115730(n-3) + Fibonacci(n-1)*Fibonacci(n);
  end if; return A115730;
end function;
[A115730(n): n in [0..40]]; // G. C. Greubel, Jan 20 2022

nmax:=31: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: a(1):=0: a(2):=1: for n from 3 to nmax do a(n):=a(n-3) + A001654(n-1) od: seq(a(n),n=0..nmax);

LinearRecurrence[{2,2,0,-2,-2,1}, {0,0,1,2,6,16}, 40] (* modified by G. C. Greubel, Jan 20 2022 *)
a[ n_] := Floor[(2*Fibonacci[2*n+1] + Fibonacci[2*n+2] + 2)/20]; (* Michael Somos, Sep 05 2023 *)

{a(n) = (2*fibonacci(2*n+1) + fibonacci(2*n+2) + 2)\20}; /* Michael Somos, Sep 05 2023 */

U=chebyshev_U
def A115730(n): return (1/60)*((-1)^n*(6 - 5*U(n, 1/2) + 10*U(n-1, 1/2)) - (10 - 9*U(n, 3/2) + 6*U(n-1, 3/2)))
[A115730(n) for n in (0..40)] # G. C. Greubel, Jan 20 2022

a(n) = -floor(g(Fibonacci(n+1))) where g(x) = (1-x^2)^2/(-4*x^2).
G.f.: x^2/( (1-x)*(1+x)*(1+x+x^2)*(1-3*x+x^2) ). - R. J. Mathar, Jun 20 2015
a(n) - a(n-2) = A182890(n-1). - R. J. Mathar, Jun 20 2015
a(n) = (1/60)*((-1)^n*(6 - 5*ChebyshevU(n, 1/2) + 10*ChebyshevU(n-1, 1/2)) - (10 - 9*ChebyshevU(n, 3/2) + 6*ChebyshevU(n-1, 3/2))). - G. C. Greubel, Jan 20 2022
a(n) = floor((2*Fibonacci(2*n+1) + Fibonacci(2*n+2) + 2)/20). - Michael Somos, Sep 05 2023

Corrected and information added by Johannes W. Meijer, Sep 22 2010

Edited by Editors-in-Chief. - N. J. A. Sloane, Jun 20 2015

A119996 Numerator of Sum_{k=1..n} 1/(Fibonacci(k)*Fibonacci(k+2)).

Original entry on oeis.org

1, 5, 14, 39, 103, 272, 713, 1869, 4894, 12815, 33551, 87840, 229969, 602069, 1576238, 4126647, 10803703, 28284464, 74049689, 193864605, 507544126, 1328767775, 3478759199, 9107509824, 23843770273, 62423800997, 163427632718

Offset: 1

Alexander Adamchuk, Aug 03 2006

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A000045, A059248, A064831, A001654, A045468, A064739, A091729.

F:=Fibonacci;; List([1..30], n-> F(n+1)*F(n+2)-1); # G. C. Greubel, Jul 23 2019

[Fibonacci(n+1)* Fibonacci(n+2)-1: n in [1..30]]; // Vincenzo Librandi, Aug 14 2012

with(combinat): seq(fibonacci(n+1)*fibonacci(n+2)-1, n=1..30); # Zerinvary Lajos, Jan 31 2008

Numerator[Table[Sum[1/(Fibonacci[k]*Fibonacci[k+2]),{k,n}],{n,30}]]
LinearRecurrence[{3,0,-3,1},{1,5,14,39},30] (* Harvey P. Dale, Aug 22 2011 *)

vector(30, n, f=fibonacci; f(n+1)*f(n+2)-1) \\ G. C. Greubel, Jul 23 2019

f=fibonacci; [f(n+1)*f(n+2)-1 for n in (1..30)] # G. C. Greubel, Jul 23 2019

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4); a(0)=1, a(1)=5, a(2)=14, a(3)=39. - Harvey P. Dale, Aug 22 2011
G.f.: ((x-2)*x-1)/(x^4 - 3*x^3 + 3*x - 1). - Harvey P. Dale, Aug 22 2011
a(n) = Fibonacci(n+1)*Fibonacci(n+2) - 1. - Gary Detlefs, Mar 31 2012
a(n) = Sum_{k=1..n} Fibonacci(k+1)^2. Can be proved by induction from Gary Detlefs formula. - Joel Courtheyn, Mar 15 2021

A180665 Golden Triangle sums: a(n)=a(n-2)+A001654(n) with a(0)=0 and a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 17, 47, 121, 320, 835, 2190, 5730, 15006, 39282, 102847, 269252, 704917, 1845491, 4831565, 12649195, 33116030, 86698885, 226980636, 594243012, 1555748412, 4073002212, 10663258237, 27916772486, 73087059235

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n) are the Kn21, Kn22, Kn23, Fi2, and Ze2 sums of the Golden Triangle A180662. Furthermore the a(2*n) are the Kn3, Fi1 (terms doubled) and Ze3 (terms tripled) sums. See A180662 for information about these and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (2, 3, -3, -2, 1).

Cf. A064831, A180664, A180665, A115730, A180666.

nmax:=27: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n)*fibonacci(n+1) od: a(0):=0: a(1):=1: for n from 2 to nmax do a(n) := a(n-2) + A001654(n) od: seq(a(n),n=0..nmax);

a(n) = a(n-2)+A001654(n) with a(0)=0 and a(1)=1.
GF(x) = (-x)/((x-1)*(x+1)^2*(x^2-3*x+1)).
a(n) = ((-1)^(-n-1)*(15+10*n)-25+(16-4*A)*A^(-n-1)+(16-4*B)*B^(-n-1))/100 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.

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A002571 From a definite integral.

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A077916 Expansion of (1-x)^(-1)/(1 + 2*x - 2*x^2 - x^3).

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A080144 a(n) = F(4)*F(n)*F(n+1) + F(5)*F(n+1)^2 if n odd, a(n) = F(4)*F(n)*F(n+1) + F(5)*F(n+1)^2 - F(5) if n even, where F(n) is the n-th Fibonacci number (A000045).

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A005313 Maximal sum of inverse squares of the singular values of triangular anti-Hadamard matrices of order n.

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A097083 Positive values of k such that there is exactly one permutation p of (1,2,3,...,k) such that i+p(i) is a Fibonacci number for 1<=i<=k.

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A180664 Golden Triangle sums: a(n) = a(n-1) + A001654(n+1) with a(0)=0.

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A080143 a(n) = F(3)*F(n)*F(n+1) + F(4)*F(n+1)^2 - F(4) if n even, F(3)*F(n)*F(n+1) + F(4)*F(n+1)^2 if n odd, where F(n) is the n-th Fibonacci number (A000045).

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A077916 Expansion of (1-x)^(-1)/(1 + 2x - 2x^2 - x^3).

A080144 a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 if n odd, a(n) = F(4)F(n)F(n+1) + F(5)F(n+1)^2 - F(5) if n even, where F(n) is the n-th Fibonacci number (A000045).

A080143 a(n) = F(3)F(n)F(n+1) + F(4)F(n+1)^2 - F(4) if n even, F(3)F(n)F(n+1) + F(4)F(n+1)^2 if n odd, where F(n) is the n-th Fibonacci number (A000045).