A027941 -id:A027941 - cp's OEIS frontend

A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154, 4613732, 19544084, 82790070, 350704366, 1485607536, 6293134512, 26658145586, 112925716858, 478361013020, 2026369768940, 8583840088782, 36361730124070, 154030760585064, 652484772464328, 2763969850442378

Offset: 0

Ralf Stephan, Oct 30 2004

Partial sum of the even Fibonacci numbers. - Vladimir Joseph Stephan Orlovsky, Nov 28 2010

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 25.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Project Euler, Problem 2: Even Fibonacci Numbers.
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Partial sums of A014445.
Cf. A000045, A004794, A015448, A049652.
Cf. A087635.
Case k = 3 of partial sums of fibonacci(k*n): A000071, A027941, A058038, A138134, A053606.

[(Fibonacci(3*n+2) - 1)/2: n in [0..30]]; // G. C. Greubel, Jan 17 2018

CoefficientList[Series[2 x/((1 - x) (1 - 4 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *)
LinearRecurrence[{5, -3, -1}, {0, 2, 10}, 30] (* G. C. Greubel, Jan 17 2018 *)
Accumulate[Fibonacci[3Range[0, 19]]] (* Alonso del Arte, Dec 23 2018 *)

a(n) = sum(i=1, n, fibonacci(3*i)); \\ Michel Marcus, Mar 15 2014

a(n) = fibonacci(3*n+2)\2 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = (Fibonacci(3*n + 2) - 1)/2 = (A015448(n+1)-1)/2.
G.f.: 2*x/((1 - x)*(1 - 4*x - x^2)).
a(n) = (F(3n + 2) - 1)/2 = 2 * A049652(n).
a(n) = Sum_{0 <= j <= i <= n} binomial(i, j)*F(i + j). - Benoit Cloitre, May 21 2005
From Gary Detlefs, Dec 08 2010: (Start)
a(n) = 4*a(n - 1) + a(n - 2) + 2, n > 1.
a(n) = 5*a(n - 1) - 3*a(n - 2) - a(n - 3), n > 2.
a(n) = (Fibonacci(3*n + 3) + Fibonacci(3*n) - 2)/4. (End)
a(n) = (-10 + (5 - 3*sqrt(5))*(2 - sqrt(5))^n + (2 + sqrt(5))^n*(5 + 3*sqrt(5)))/20. - Colin Barker, Nov 26 2016
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, Jun 03 2024

a(0) = 0 prepended by Joerg Arndt, Mar 13 2014

A058038 a(n) = Fibonacci(2n)Fibonacci(2*n+2).

Original entry on oeis.org

0, 3, 24, 168, 1155, 7920, 54288, 372099, 2550408, 17480760, 119814915, 821223648, 5628750624, 38580030723, 264431464440, 1812440220360, 12422650078083, 85146110326224, 583600122205488, 4000054745112195, 27416783093579880, 187917426909946968

Offset: 0

N. J. A. Sloane, Jun 09 2002

Partial sums of A033888, i.e., a(n) = Sum_{k=0..n} Fibonacci(4*k). - Vladeta Jovovic, Jun 09 2002
From Paul Weisenhorn, May 17 2009: (Start)
a(n) is the solution of the 2 equations a(n)+1=A^2 and 5*a(n)+1=B^2
which are equivalent to the Pell equation (10*a(n)+3)^2-5*(A*B)^2=4.
(End)
Numbers a(n) such as a(n)+1 and 5*a(n)+1 are perfect squares. - Sture Sjöstedt, Nov 03 2011

			G.f. = 3*x + 24*x^2 + 168*x^3 + 1155*x^4 + 7920*x^5 + 54288*x^6 + ... - _Michael Somos_, Jan 23 2025

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 29.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A033888, A004187, A094874.
Bisection of A059929, A064831 and A080097.
Related to sum of fibonacci(kn) over n; cf. A000071, A099919, A027941, A138134, A053606.

[Fibonacci(2*n)*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

fs4:=n->sum(fibonacci(4*k),k=0..n):seq(fs4(n),n=0..21); # Gary Detlefs, Dec 07 2010

Table[Fibonacci[2 n]*Fibonacci[2 n + 2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
Accumulate[Fibonacci[4*Range[0,30]]] (* or *) LinearRecurrence[{8,-8,1},{0,3,24},30] (* Harvey P. Dale, Jul 25 2013 *)

a(n)=fibonacci(2*n)*fibonacci(2*n+2) \\ Charles R Greathouse IV, Jul 02 2013

a(n) = -3/5 + (1/5*sqrt(5)+3/5)*(2*1/(7+3*sqrt(5)))^n/(7+3*sqrt(5)) + (1/5*sqrt(5)-3/5)*(-2*1/(-7+3*sqrt(5)))^n/(-7+3*sqrt(5)). Recurrence: a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). G.f.: 3*x/(1-7*x+x^2)/(1-x). - Vladeta Jovovic, Jun 09 2002
a(n) = A081068(n) - 1.
a(n) is the next integer from ((3+sqrt(5))*((7+3*sqrt(5))/2)^(n-1)-6)/10. - Paul Weisenhorn, May 17 2009
a(n) = 7*a(n-1) - a(n-2) + 3, n>1. - Gary Detlefs, Dec 07 2010
a(n) = sum_{k=0..n} Fibonacci(4k). - Gary Detlefs, Dec 07 2010
a(n) = (Lucas(4n+2)-3)/5, where Lucas(n)= A000032(n). - Gary Detlefs, Dec 07 2010
a(n) = (1/5)*(Fibonacci(4n+4) - Fibonacci(4n)-3). - Gary Detlefs, Dec 08 2010
a(n) = 3*A092521(n). - R. J. Mathar, Nov 03 2011
a(0)=0, a(1)=3, a(2)=24, a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). - Harvey P. Dale, Jul 25 2013
a(n) = A001906(n)*A001906(n+1). - R. J. Mathar, Jul 09 2019
Sum_{n>=1} 1/a(n) = 2/(3 + sqrt(5)) = A094874 - 1. - Amiram Eldar, Oct 05 2020
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025

A069403 a(n) = 2Fibonacci(2n+1) - 1.

Original entry on oeis.org

1, 3, 9, 25, 67, 177, 465, 1219, 3193, 8361, 21891, 57313, 150049, 392835, 1028457, 2692537, 7049155, 18454929, 48315633, 126491971, 331160281, 866988873, 2269806339, 5942430145, 15557484097, 40730022147, 106632582345, 279167724889, 730870592323, 1913444052081

Offset: 0

R. H. Hardin, Mar 22 2002

Half the number of n X 3 binary arrays with a path of adjacent 1's and a path of adjacent 0's from top row to bottom row.

Indices of A017245 = 9*n + 7 = 7, 16, 25, 34, for submitted A153819 = 16, 34, 88,. A153819(n) = 9*a(n) + 7 = 18*F(2*n+1) -2; F(n) = Fibonacci = A000045, 2's = A007395. Other recurrence: a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3). - Paul Curtz, Jan 02 2009

Colin Barker, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 41, 56.
J. Hietarinta and C.-M. Viallet, Singularity confinement and chaos in discrete systems, Physical Review Letters 81 (1998), pp. 326-328.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000045, A084707.
Cf. 1 X n A000225, 2 X n A016269, vertical path of 1 A069361-A069395, vertical paths of 0+1 A069396-A069416, vertical path of 1 not 0 A069417-A069428, no vertical paths A069429-A069447, no horizontal or vertical paths A069448-A069452.
Equals A052995 - 1.
Bisection of A001595, A062114, A066983.

List([0..30], n-> 2*Fibonacci(2*n+1)-1); # G. C. Greubel, Jul 11 2019

[2*Fibonacci(2*n+1)-1: n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

a[n_]:= a[n] = 3a[n-1] - 3a[n-3] + a[n-4]; a[0] = 1; a[1] = 3; a[2] = 9; a[3] = 25; Table[ a[n], {n, 0, 30}]
Table[2*Fibonacci[2*n+1]-1, {n,0,30}] (* G. C. Greubel, Apr 22 2018 *)
LinearRecurrence[{4,-4,1},{1,3,9},30] (* Harvey P. Dale, Sep 22 2020 *)

a(n) = 2*fibonacci(2*n+1)-1 \\ Charles R Greathouse IV, Jun 11 2015

Vec((1-x+x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

[2*fibonacci(2*n+1)-1 for n in (0..30)] # G. C. Greubel, Jul 11 2019

a(0) = 1, a(1) = 3, a(2) = 9, a(3) = 25; a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).
a(n) = 3*a(n-1) - a(n-2) + 1 for n>1, a(1) = 3, a(0) = 0. - Reinhard Zumkeller, May 02 2006
From R. J. Mathar, Feb 23 2009: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: (1-x+x^2)/((1-x)*(1-3*x+x^2)). (End)
a(n) = 1 + 2*Sum_{k=0..n} Fibonacci(2*k) = 1+2*A027941(n). - Gary Detlefs, Dec 07 2010
a(n) = (2^(-n)*(-5*2^n -(3-sqrt(5))^n*(-5+sqrt(5)) +(3+sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Nov 02 2016

Simpler definition from Vladeta Jovovic, Mar 19 2003

A200947 Sequence A007924 expressed in decimal.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 16, 17, 18, 20, 32, 33, 64, 65, 66, 68, 128, 129, 256, 257, 258, 260, 512, 513, 514, 516, 517, 520, 1024, 1025, 2048, 2049, 2050, 2052, 2053, 2056, 4096, 4097, 4098, 4100, 8192, 8193, 16384, 16385, 16386, 16388, 32768, 32769, 32770

Offset: 0

Frank M Jackson, Nov 24 2011

nonn

			8=7+1, hence A007924(8)=10001, so a(8)=17.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Wikipedia, "Complete" sequence. [Wikipedia calls a sequence "complete" (sic) if every positive integer is a sum of distinct terms. This name is extremely misleading and should be avoided. - _N. J. A. Sloane_, May 20 2023]

Cf. A003714, A007895, A007924, A027941, A066352, A121559, A345297.

a:= proc(n) option remember; local m, p, r; m:=n; r:=0;
      while m>0 do
        if m=1 then r:=r+1; break fi;
        p:= prevprime(m+1); m:= m-p;
        r:= r+2^numtheory[pi](p)
      od; r
    end:
seq(a(n), n=0..52);  # Alois P. Heinz, Jun 12 2023

cprime[n_Integer] := If[n==0, 1, Prime[n]]; gentable[n_Integer] := (m=n; ptable={}; While[m != 0, (i = 0; While[cprime[i] <= m, i++]; j=0; While[j

a(n) = decimal(A007924(n)).

a(n) mod 2 = A121559(n) for n>=1. - Alois P. Heinz, Jun 12 2023

Edited by N. J. A. Sloane, May 20 2023

A054451 Third column of triangle A054450 (partial row sums of unsigned Chebyshev triangle A049310).

Original entry on oeis.org

1, 1, 4, 5, 12, 17, 33, 50, 88, 138, 232, 370, 609, 979, 1596, 2575, 4180, 6755, 10945, 17700, 28656, 46356, 75024, 121380, 196417, 317797, 514228, 832025, 1346268, 2178293, 3524577, 5702870, 9227464, 14930334, 24157816, 39088150, 63245985, 102334135

Offset: 0

Wolfdieter Lang, Apr 27 2000

Equals triangle A173284 * [1, 2, 3, ...]. - Gary W. Adamson, Mar 03 2010

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-2,-3,1,1).

Cf. A054450, A049310, A000045, A052952.
Cf. A007382.
Cf. A173284.

BB:=1/(1-k^2)^2/(1-k-k^2): seq(coeff(series(BB, k, n+1), k, n), n=0..50); # Zerinvary Lajos, May 16 2007

LinearRecurrence[{1,3,-2,-3,1,1},{1,1,4,5,12,17},40] (* Harvey P. Dale, Oct 06 2024 *)

Vec(-1/((x-1)^2*(x+1)^2*(x^2+x-1)) + O(x^100)) \\ Colin Barker, Jun 14 2015

a(n) = A054450(n+2, 2).
G.f.: Fib(x)/(1-x^2)^2, with Fib(x)=1/(1-x-x^2) = g.f. A000045 (Fibonacci numbers without 0).
a(2*k) = A027941(k)= F(2*k+3)-1; a(2*k+1)= F(2*(k+2))-(k+2)= A054452(k), k >= 0.
a(n-2) = Fibonacci(n+1) - binomial(n-floor(n/2), floor(n/2)), or a(n-2) = Sum_{i=0..floor(n/2)-1} binomial(n-i, i). - Jon Perry, Mar 18 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+2, k). - Paul Barry, Oct 23 2004

More terms from James Sellers, Apr 28 2000

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

A258993 Triangle read by rows: T(n,k) = binomial(n+k,n-k), k = 0..n-1.

Original entry on oeis.org

1, 1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060, 969, 190, 21

Offset: 1

Reinhard Zumkeller, Jun 22 2015

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

			.  n\k |  0  1    2    3     4     5     6     7    8    9  10 11
. -----+-----------------------------------------------------------
.   1  |  1
.   2  |  1  3
.   3  |  1  6    5
.   4  |  1 10   15    7
.   5  |  1 15   35   28     9
.   6  |  1 21   70   84    45    11
.   7  |  1 28  126  210   165    66    13
.   8  |  1 36  210  462   495   286    91    15
.   9  |  1 45  330  924  1287  1001   455   120   17
.  10  |  1 55  495 1716  3003  3003  1820   680  153   19
.  11  |  1 66  715 3003  6435  8008  6188  3060  969  190  21
.  12  |  1 78 1001 5005 12870 19448 18564 11628 4845 1330 231 23  .

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

If a diagonal of 1's is added on the right, this becomes A085478.
Essentially the same as A143858.
Cf. A007318, A004736, A094727.
Cf. A027941 (row sums), A117671 (central terms), A143858, A000967, A258708.
T(n,k): A000217 (k=1), A000332 (k=2), A000579 (k=3), A000581 (k=4), A001287 (k=5), A010965 (k=6), A010967 (k=7), A010969 (k=8), A010971 (k=9), A010973 (k=10), A010975 (k=11), A010977 (k=12), A010979 (k=13), A010981 (k=14), A010983 (k=15), A010985 (k=16), A010987 (k=17), A010989 (k=18), A010991 (k=19), A010993 (k=20), A010995 (k=21), A010997 (k=22), A010999 (k=23), A011001 (k=24), A017714 (k=25), A017716 (k=26), A017718 (k=27), A017720 (k=28), A017722 (k=29), A017724 (k=30), A017726 (k=31), A017728 (k=32), A017730 (k=33), A017732 (k=34), A017734 (k=35), A017736 (k=36), A017738 (k=37), A017740 (k=38), A017742 (k=39), A017744 (k=40), A017746 (k=41), A017748 (k=42), A017750 (k=43), A017752 (k=44), A017754 (k=45), A017756 (k=46), A017758 (k=47), A017760 (k=48), A017762 (k=49), A017764 (k=50).
T(n+k,n): A005408 (k=1), A000384 (k=2), A000447 (k=3), A053134 (k=4), A002299 (k=5), A053135 (k=6), A053136 (k=7), A053137 (k=8), A053138 (k=9), A196789 (k=10).
Cf. A165253.

Flat(List([1..12], n-> List([0..n-1], k-> Binomial(n+k,n-k) ))); # G. C. Greubel, Aug 01 2019

a258993 n k = a258993_tabl !! (n-1) !! k
a258993_row n = a258993_tabl !! (n-1)
a258993_tabl = zipWith (zipWith a007318) a094727_tabl a004736_tabl

[Binomial(n+k,n-k): k in [0..n-1], n in [1..12]]; // G. C. Greubel, Aug 01 2019

Table[Binomial[n+k,n-k], {n,1,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,k) = binomial(n+k,n-k);
for(n=1, 12, for(k=0,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019

[[binomial(n+k,n-k) for k in (0..n-1)] for n in (1..12)] # G. C. Greubel, Aug 01 2019

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

A098149 a(0)=-1, a(1)=-1, a(n)=-3*a(n-1)-a(n-2) for n>1.

Original entry on oeis.org

-1, -1, 4, -11, 29, -76, 199, -521, 1364, -3571, 9349, -24476, 64079, -167761, 439204, -1149851, 3010349, -7881196, 20633239, -54018521, 141422324, -370248451, 969323029, -2537720636, 6643838879, -17393796001, 45537549124, -119218851371

Offset: 0

Creighton Dement, Aug 29 2004

Sequence relates bisections of Lucas and Fibonacci numbers.
2*a(n) + A098150(n) = 8*(-1)^(n+1)*A001519(n) - (-1)^(n+1)*A005248(n+1). Apparently, if (z(n)) is any sequence of integers (not all zero) satisfying the formula z(n) = 2(z(n-2) - z(n-1)) + z(n-3) then |z(n+1)/z(n)| -> golden ratio phi + 1 = (3+sqrt(5))/2.
Pisano period lengths: 1, 3, 4, 6, 1, 12, 8, 6, 12, 3, 10, 12, 7, 24, 4, 12, 9, 12, 18, 6, ... . - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Oct 12 2020: (Start)
[X(n) = (-1)^n*(S(n, 3) + S(n-1, 3)), Y(n) = X(n-1)] gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 3*X*Y = +5, for n = -oo..+oo, with Chebyshev S polynomials (A049310), with S(-1, x) = 0,  S(-|n|, x) = - S(|n|-2, x), for |n| >= 2, and S(n,-x) = (-1)^n*S(n, x). The present sequence is a(n) = -X(n-1), for n >= 0. See the formula section.
This binary indefinite quadratic form of discriminant 5, representing 5, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication) See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seong Ju Kim, R. Stees, and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
Tanya Khovanova, Recursive Sequences
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Index entries for linear recurrences with constant coefficients, signature (-3,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A098150, A001519, A005248, A000045, A001906, A049310, A027941.

a[0] = a[1] = -1; a[n_] := a[n] = -3a[n - 2] - a[n - 1]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Sep 01 2004 *)
LinearRecurrence[{-3,-1},{-1,-1},30] (* Harvey P. Dale, Apr 19 2014 *)
CoefficientList[Series[-(1 + 4 x)/(1 + 3 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 19 2014 *)

G.f.: -(1+4*x)/(1+3*x+x^2). - Philippe Deléham, Nov 19 2006
a(n) = (-1)^n*A002878(n-1). - R. J. Mathar, Jan 30 2011
-a(n+1) = Sum_{k, 0<=k<=n}(-5)^k*Binomial(n+k, n-k) = Sum_{k, 0<=k<=n}(-5)^k*A085478(n, k). - Philippe Deléham, Nov 28 2006
a(n) = (-1)^n*(S(n-1, 3) + S(n-2, 3)) = (-1)^n*S(2*(n-1), sqrt(5)), for n >= 0, with Chebyshev S polynomials (A049310), with S(-1, x) = 0 and S(-2, x) = -1. S(n, 3) = A001906(n+1) = F(2*(n+1)), with F = A000045. - Wolfdieter Lang, Oct 12 2020

Simpler definition from Philippe Deléham, Nov 19 2006

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

A138134 a(n) = Sum_{i=0..n} Fibonacci(5*i).

Original entry on oeis.org

0, 5, 60, 670, 7435, 82460, 914500, 10141965, 112476120, 1247379290, 13833648315, 153417510760, 1701426266680, 18869106444245, 209261597153380, 2320746675131430, 25737475023599115, 285432971934721700, 3165500166305537820

Offset: 0

Gary Detlefs, Dec 07 2010

Partial sums of A102312.
Other sequences in the OEIS related to the sum of Fibonacci(k*n) (although not defined as such) are:
  k = 1: A000071 = Fibonacci(n) - 1 (delete leading 0);
  k = 2: A027941 = Fibonacci(2n+1) - 1;
  k = 3: A099919 = (Fibonacci(3n+2) - 1)/2;
  k = 4: A058038 = Fibonacci(2n)*Fibonacci(2n+2);
  k = 6: A053606 = (Fibonacci(6n+3) - 2)/4.
These sequences appear to be second order linear inhomogeneous sequences of the form: a(0) = 0, a(1) = Fibonacci(k), a(n) = L(k)*a(n-1) + (-1)^(k+1)*a(n-2) + Fibonacci(k), where L(n) = A000032(n), n > 1.
The Koshy reference gives the closed form:
  Sum_{i=0..n} Fibonacci(k*i) = (Fibonacci(n*k+k) - (-1)^k*Fibonacci(n*k) - Fibonacci(k))/(L(k) - (-1)^k - 1).

Thomas Koshy; Fibonacci and Lucas numbers with applications, Wiley,2001, p. 86.

Index entries for linear recurrences with constant coefficients, signature (12,-10,-1).

Cf. A000071, A027941, A099919, A058038, A102312, A053606.

Cf. also A000032, A000045.

with(combinat):fs5:=n-> sum(fibonacci(5*k),k=0..n):
seq(fs5(n),n=0..18)

a(n)=(fibonacci(5*n+5)+fibonacci(5*n)-5)/11 \\ Charles R Greathouse IV, Jun 11 2015

G.f.: 5*x/((x - 1)*(x^2 + 11*x - 1)). - R. J. Mathar, Dec 09 2010
a(n) = 11*a(n) + a(n-1) + 5, n > 1.
a(n) = 12*a(n-1) - 10*a(n-2) - a(n-3), n > 2.
a(n) = 1/11*(Fibonacci(5*n+5) + Fibonacci(5n) - 5).

cp's OEIS Frontend

A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

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A058038 a(n) = Fibonacci(2*n)*Fibonacci(2*n+2).

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A069403 a(n) = 2*Fibonacci(2*n+1) - 1.

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A200947 Sequence A007924 expressed in decimal.

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A054451 Third column of triangle A054450 (partial row sums of unsigned Chebyshev triangle A049310).

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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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A058038 a(n) = Fibonacci(2n)Fibonacci(2*n+2).

A069403 a(n) = 2Fibonacci(2n+1) - 1.

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).