A006498 -id:A006498 - cp's OEIS frontend

A242549 T(n,k)=Number of length n+3 0..k arrays with no four elements in a row with pattern aabb (possibly a=b) and new values 0..k introduced in 0..k order.

6, 12, 9, 13, 32, 15, 13, 42, 88, 25, 13, 43, 150, 242, 40, 13, 43, 165, 554, 660, 64, 13, 43, 166, 690, 2072, 1800, 104, 13, 43, 166, 711, 3050, 7808, 4920, 169, 13, 43, 166, 712, 3311, 13988, 29536, 13448, 273, 13, 43, 166, 712, 3339, 16512, 65588, 111878, 36736

Offset: 1

R. H. Hardin, May 17 2014

Table starts
...6.....12......13......13.......13.......13.......13.......13.......13
...9.....32......42......43.......43.......43.......43.......43.......43
..15.....88.....150.....165......166......166......166......166......166
..25....242.....554.....690......711......712......712......712......712
..40....660....2072....3050.....3311.....3339.....3340.....3340.....3340
..64...1800....7808...13988....16512....16968....17004....17005....17005
.104...4920...29536...65588....86671....92360....93103....93148....93149
.169..13448..111878..311431...471698...532175...543773...544920...544975
.273..36736..423969.1489435..2631790..3210136..3362689..3384566..3386262
.441.100352.1607058.7152787.14930915.20066475.21854355.22202698.22241481

			Some solutions for n=4 k=4
..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0
..1....1....0....0....1....1....1....1....0....1....1....0....1....1....1....0
..2....2....1....1....2....1....1....0....1....2....2....1....2....0....0....1
..1....1....2....0....0....1....2....2....2....3....1....2....3....0....1....2
..0....3....0....2....2....2....1....1....1....4....3....3....0....2....1....3
..1....2....0....3....2....0....1....2....1....4....3....3....0....1....0....1
..2....0....0....1....2....0....3....0....1....4....1....0....0....3....2....2

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A006498(n+4)

Empirical for column k:
k=1: a(n) = a(n-1) +a(n-3) +a(n-4)
k=2: a(n) = 2*a(n-1) +4*a(n-3) +4*a(n-4)
k=3: [order 8]
k=4: [order 12]
k=5: [order 16]
k=6: [order 20]
k=7: [order 24]

A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3.

Original entry on oeis.org

1, 2, 4, 8, 16, 24, 36, 54, 81, 135, 225, 375, 625, 1000, 1600, 2560, 4096, 6656, 10816, 17576, 28561, 46137, 74529, 120393, 194481, 314874, 509796, 825384, 1336336, 2161720, 3496900, 5656750, 9150625, 14807375, 23961025, 38773295, 62742241, 101515536

Offset: 1

Jeff Burch

Two consecutive Fibonacci numbers are coprime. This sequence satisfies a 14th-order linear difference equation. Note that it is the fourth sequence in the sequences that begin with the Fibonacci numbers, A006498, and A006500. Subsequent sequences will have orders 22, 32, and 44. - T. D. Noe, Mar 05 2012

Also the number of subsets of the set {1,2,...,n-1} which do not contain two elements whose difference is 4. - David Nacin, Mar 07 2012

			Since F_5 = 5 and F_6 = 8 are consecutive Fibonacci numbers, 8^4 = 4096, 8^3*5 = 2560, 8^2*5^2 = 1600, 8*5^3 = 1000, and 5^4 = 625 are in the sequence.
The number 3^3*8 = 216 is not in the sequence since 3 and 8 are not consecutive.
If n = 6 then this gives the number of subsets of {1,...,5} not containing both 1 and 5. There are 2^3 subsets containing 1 and 5, giving us 2^5 - 2^3 = 24. Thus a(5) = 24. - _David Nacin_, Mar 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, Connections between Combinations Without Specified Separations and Strongly Restricted Permutations, Compositions, and Bit Strings, arXiv:2409.00624 [math.CO], 2024. See p. 16.
M. El-Mikkawy and T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=4.
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-2,2,2,0,2,-2,-2,0,1,-1,-1).

Cf. A000045, A006498, A006500.

A031923 := proc(n)
    local n0,i,r,s,m ;
    n0 := n-1 ;
    i := floor(n0/4) ;
    r := combinat[fibonacci](i+2) ;
    s := combinat[fibonacci](i+3) ;
    m := modp(n0,4) ;
    r^(4-m)*s^m ;
end proc:
seq(A031923(n),n=1..50) ; # R. J. Mathar, Jan 23 2022

f = Fibonacci[Range[12]]; m = Most[f]; r = Rest[f]; Union[m^4, m^3 r, m^2 r^2, m r^3] (* T. D. Noe, Mar 05 2012 *)
LinearRecurrence[{1, 1, 0, -2, 2, 2, 0, 2, -2, -2, 0, 1, -1, -1}, {1, 2, 4, 8, 16, 24, 36, 54, 81, 135, 225, 375, 625, 1000}, 40] (* T. D. Noe, Mar 05 2012 *)
Table[Fibonacci[Floor[n/4] + 3]^Mod[n, 4]*Fibonacci[Floor[n/4] + 2]^(4 - Mod[n, 4]), {n, 0, 40}] (* David Nacin, Mar 07 2012 *)
cfn[{a_,b_}]:={a^4,a^3 b,a^2 b^2,a b^3}; Flatten[cfn/@Partition[ Fibonacci[ Range[20]],2,1]]//Union (* Harvey P. Dale, Feb 03 2019 *)

for(m=2,10,r=fibonacci(m);s=fibonacci(m+1);print(r^4," ",r^3*s," ",r^2*s^2," ",r*s^3)) \\ Michael B. Porter, Mar 04 2012

def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:0, 5:4, 6:15, 7:37, 8:87, 9:200}):
    if n in adict:
        return adict[n]
    adict[n]=3*a(n-1)-2*a(n-2)+2*a(n-3)-4*a(n-4)+2*a(n-5)-2*a(n-6)-4*a(n-7)-a(n-8)+a(n-9)+2*a(n-10)
    return adict[n] # David Nacin, Mar 07 2012

a(n) = F(floor((n-1)/4) + 3)^(n-1 mod 4)*F(floor((n-1)/4) + 2)^(4 - (n-1 mod 4)) where F(n) is the n-th Fibonacci number. - David Nacin, Mar 07 2012
a(n) = a(n-1) + a(n-2) - 2*a(n-4) + 2*a(n-5) + 2*a(n-6) + 2*a(n-8) - 2*a(n-9) - 2*a(n-10) + a(n-12) - a(n-13) - a(n-14). - David Nacin, Mar 07 2012
G.f.: x*(2 + 2*x + 2*x^2 + 4*x^3 + 4*x^4 - 2*x^6 - 1*x^7 - 4*x^8 - 3*x^9 - x^10 - x^11 - 2*x^12 - x^13)/((1 - x)*(1 + x)*(1 + x^2)*(1 - x - x^2)*(1 + 3*x^4 + x^8)). - David Nacin, Mar 08 2012
a(4*k-3) = F(k+1)^4, a(4*k-2) = F(k+1)^3*F(k+2), a(4*k-1) = F(k+1)^2*F(k+2)^2, a(4*k) = F(k+1)*F(k+2)^3, k >= 1, where F = A000045. - Jianing Song, Feb 06 2019
a(4n+1)= A056571(n+2). a(4n+3)=A197424(n). - R. J. Mathar, Jan 23 2022

a(19) changed from 10416 to 10816 by David Nacin, Mar 04 2012

A074677 a(n) = Sum_{i = 0..floor(n/2)} (-1)^(i + floor(n/2)) F(2*i + e), where F = A000045 (Fibonacci numbers) and e = (1-(-1)^n)/2.

Original entry on oeis.org

0, 1, 1, 1, 2, 4, 6, 9, 15, 25, 40, 64, 104, 169, 273, 441, 714, 1156, 1870, 3025, 4895, 7921, 12816, 20736, 33552, 54289, 87841, 142129, 229970, 372100, 602070, 974169, 1576239, 2550409, 4126648, 6677056, 10803704, 17480761, 28284465, 45765225, 74049690

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 30 2002

Essentially the same as A006498 (g.f. 1/(1-x-x^3-x^4)).

a(n) is the convolution of F(n) with the sequence (1,0,-1,0,1,0,-1,0,...), A056594.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Victoria Zhuravleva, Diophantine approximations with Fibonacci numbers, Journal de théorie des nombres de Bordeaux, 25 no. 2 (2013), p. 499-520. See Lemma 5.1.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A000045, A056594.

a074677 n = a074677_list !! (n-1)
a074677_list = 0 : 1 : 1 : 1 : zipWith (+) a074677_list
   (zipWith (+) (tail a074677_list) (drop 3 a074677_list))
-- Reinhard Zumkeller, Dec 28 2011

[&+[(-1)^(i+Floor(n/2))*Fibonacci(2*i+(1-(-1)^n) div 2): i in [0..Floor(n/2)]]: n in [0..50]]; // Bruno Berselli, Mar 15 2016

CoefficientList[Series[x/(1 - x - x^3 - x^4), {x, 0, 40}], x]

concat(0, Vec(x/((1+x^2)*(1-x-x^2)) + O(x^50))) \\ Colin Barker, Mar 15 2016

[sum((-1)^(i+floor(n/2))*fibonacci(2*i+(1-(-1)^n)/2) for i in (0..floor(n/2))) for n in [0..50]]; # Bruno Berselli, Mar 15 2016

a(n) = a(n-1) + a(n-3) + a(n-4) for n>3, a(0)=0, a(1)=1, a(2)=1, a(3)=1.
G.f.: x/(1 - x - x^3 - x^4).
a(n) = Fibonacci(ceiling(n/2))*Fibonacci(floor(n/2+1)). - Alois P. Heinz, Jan 15 2024

A089928 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.

Original entry on oeis.org

1, 2, 4, 10, 25, 60, 144, 348, 841, 2030, 4900, 11830, 28561, 68952, 166464, 401880, 970225, 2342330, 5654884, 13652098, 32959081, 79570260, 192099600, 463769460, 1119638521, 2703046502, 6525731524, 15754509550, 38034750625, 91824010800

Offset: 0

Paul Barry, Nov 15 2003

a(n) is the number of tilings of an n-board (a board of size n X 1) using white squares, black squares, and white (1,1)-fences. A (1,1)-fence is a tile composed of two squares separated by a gap of width 1. - Michael A. Allen, Mar 12 2021

a(n) is the number of tilings of an n-board using white squares, black squares, white trominoes, black trominoes, and white tetrominoes. - Michael A. Allen, Mar 12 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, J. Int. Seq. 24 (2021) Article 21.3.8.
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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (2,0,2,1).

Cf. A000129, A001333, A006498, A057077, A079291, A114620.

[(Evaluate(DicksonFirst(n+2,-1), 2) + 2*(-1)^Binomial(n,2))/8: n in [0..40]]; // G. C. Greubel, Aug 18 2022

CoefficientList[Series[1/(1-2x-2x^3-x^4),{x,0,30}],x] (* Michael A. Allen, Mar 12 2021 *)
LinearRecurrence[{2,0,2,1}, {1,2,4,10}, 41] (* G. C. Greubel, Aug 18 2022 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,2d+2b+a}; NestList[nxt,{1,2,4,10},30][[;;,1]] (* Harvey P. Dale, Jul 18 2024 *)

[(lucas_number2(n+2,2,-1) +2*(-1)^binomial(n,2))/8 for n in (0..40)] # G. C. Greubel, Aug 18 2022

a(n) = ( (1+sqrt(2))^(n+2) + (1-sqrt(2))^(n+2) + 2*(-1)^floor(n/2) )/8.
a(n) = (-i)^n*Sum_{k=0..floor(n/2)} U(n-2*k, i) with i^2 = -1.
a(n) + a(n+2) = A000129(n+3). - Alex Ratushnyak, Aug 06 2012
G.f.: 1/ ( (1+2*x)*(1-2*x-x^2) ). - R. J. Mathar, Apr 26 2013
4*a(n) = A057077(n) + A001333(n+2). - R. J. Mathar, Apr 26 2013
a(2*n) = (A000129(n+1))^2 = A079291(n+1). - Michael A. Allen, Mar 12 2021
a(2*n+1) = A000129(n+1)*A000129(n+2) = A114620(n+1). - Michael A. Allen, Mar 12 2021

Formula corrected by Max Alekseyev, Aug 22 2013

A115008 a(n) = a(n-1) + a(n-3) + a(n-4).

Original entry on oeis.org

1, 0, 2, 4, 5, 7, 13, 22, 34, 54, 89, 145, 233, 376, 610, 988, 1597, 2583, 4181, 6766, 10946, 17710, 28657, 46369, 75025, 121392, 196418, 317812, 514229, 832039, 1346269, 2178310, 3524578, 5702886, 9227465, 14930353, 24157817, 39088168

Offset: 0

Creighton Dement, Feb 23 2006

a(n+2) - a(n+1) - a(n) gives match to A000034, apart from signs.

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

Cf. A000034, A000045, A001519, A001906, A006498, A116697, A116698, A116699.

A115008:= func< n | Fibonacci(n+1) - (n mod 2) + 2*0^((n+1) mod 4) >;
[A115008(n): n in [0..50]]; // G. C. Greubel, Aug 24 2025

Table[Fibonacci[n+1] -I^(n-1)*Mod[n,2], {n,0,50}] (* G. C. Greubel, Aug 24 2025 *)

def A115008(n): return fibonacci(n+1) -i**(n-1)*(n%2)
print([A115008(n) for n in range(51)]) # G. C. Greubel, Aug 24 2025

a(2*n) = A000045(2*n+1) = A001519(n).
G.f.: (1-x+2*x^2+x^3)/((1+x^2)*(1-x-x^2)).
a(2*n+1) = (-1)^(n+1) + A001906(n+1) (compare with a similar property for A116697) - Creighton Dement, Mar 31 2006
From G. C. Greubel, Aug 24 2025: (Start)
a(n) = A000045(n+1) - i^(n-1)*(n mod 2).
E.g.f.: exp(x/2)*(cosh(p*x) + (1/(2*p))*sinh(p*x)) - sin(x), where 2*p = sqrt(5). (End)

A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

Original entry on oeis.org

1, 1, -2, 2, -2, 5, -9, 13, -20, 34, -56, 89, -143, 233, -378, 610, -986, 1597, -2585, 4181, -6764, 10946, -17712, 28657, -46367, 75025, -121394, 196418, -317810, 514229, -832041, 1346269, -2178308, 3524578, -5702888

Offset: 0

Creighton Dement, Feb 23 2006

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

Cf. A000045, A001519, A006498, A056594, A115008, A116698, A116699, A128535.

Cf. A186679 (first differences).

a116697 n = a116697_list !! n
a116697_list = [1,1,-2,2]
               ++ (zipWith (-) a116697_list
                               $ zipWith (+) (tail a116697_list)
                                             (drop 3 a116697_list))
a128535_list = 0 : (map negate $ map a116697 [0,2..])
a001519_list = 1 : map a116697 [1,3..]
a186679_list = zipWith (-) (tail a116697_list) a116697_list
a128533_list = map a186679 [0,2..]
a081714_list = 0 : (map negate $ map a186679 [1,3..])
a075193_list = 1 : -3 : (zipWith (+) a186679_list $ drop 2 a186679_list)
-- Reinhard Zumkeller, Feb 25 2011

A116697:= func< n | (-1)^Floor((n+1)/2)*(1+(-1)^n)/2 -(-1)^n*Fibonacci(n) >;
[A116697(n): n in [0..50]]; // G. C. Greubel, Jun 08 2025

LinearRecurrence[{-1,0,-1,1},{1,1,-2,2},40] (* Harvey P. Dale, Nov 04 2011 *)

def A116697(n): return (-1)^(n//2)*((n+1)%2) - (-1)^n*fibonacci(n)
print([A116697(n) for n in range(51)]) # G. C. Greubel, Jun 08 2025

G.f.: (1 + 2*x - x^2 + x^3)/((1 + x^2)*(1 + x - x^2)).
a(2*n+1) = A000045(2*n+1) = A001519(n).
a(2*n) = - A128535(n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A056594(n) - (-1)^n*A000045(n). - Bruno Berselli, Feb 26 2011
E.g.f.: cos(x) + (2/sqrt(5))*exp(-x/2)*sinh(sqrt(5)*x/2). - G. C. Greubel, Jun 08 2025

A263424 T(n,k)=Number of nXk arrays of permutations of 0..n*k-1 with each element moved a city block distance of 0 or 2.

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 4, 36, 36, 4, 6, 196, 1272, 196, 6, 9, 961, 26244, 26244, 961, 9, 15, 5329, 532620, 1993744, 532620, 5329, 15, 25, 29584, 11751184, 141086884, 141086884, 11751184, 29584, 25, 40, 160000, 256052940, 10808097444, 34502512032

Offset: 1

R. H. Hardin, Oct 17 2015

Table starts
..1......1..........2............4.............6.............9...............15
..1......4.........36..........196...........961..........5329............29584
..2.....36.......1272........26244........532620......11751184........256052940
..4....196......26244......1993744.....141086884...10808097444.....831331827076
..6....961.....532620....141086884...34502512032.9414955824400.2581177525941432
..9...5329...11751184..10808097444.9414955824400
.15..29584..256052940.831331827076
.25.160000.5521233025
.40.868624
.64

			Some solutions for n=3 k=4
..2..9..0..1....8..9..7..6....0..4..7..6....8..4..0..3....5..6.10.11
..4..7.11.10....1..5.11..2....9..2..1.10....1..5.11..7....1..0..4..7
..5..6..8..3....0..4.10..3....5.11..8..3...10..9..2..6....8..9..2..3

R. H. Hardin, Table of n, a(n) for n = 1..60

Column 1 is A006498.

Empirical for column k:
k=1: a(n) = a(n-1) +a(n-3) +a(n-4)
k=2: [order 15]
k=3: [order 64]

A208742 Number of subsets of the set {1,2,...,n} which do not contain two elements whose difference is 5.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 48, 72, 108, 162, 243, 405, 675, 1125, 1875, 3125, 5000, 8000, 12800, 20480, 32768, 53248, 86528, 140608, 228488, 371293, 599781, 968877, 1565109, 2528253, 4084101, 6612354, 10705716, 17333064, 28063056, 45435424, 73498480, 118894600

Offset: 0

David Nacin, Mar 01 2012

			If n=6 then we must count all subsets not containing both 1 and 6.  There are 2^4 subsets containing 1 and 6, giving us 2^6 - 2^4 = 48.  Thus a(6) = 48.

M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=5.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.
Index entries for linear recurrences with constant coefficients, signature (1, 1, 0, 0, -3, 3, 3, 0, 0, 6, -6, -6, 0, 0, 3, -3, -3, 0, 0, -1, 1, 1).

Cf. A006498, A006500, A208741, A208743.

Table[Fibonacci[Floor[n/5] + 3]^Mod[n, 5] * Fibonacci[Floor[n/5] + 2]^(5 - Mod[n, 5]), {n, 1, 40}]
LinearRecurrence[{1, 1, 0, 0, -3, 3, 3, 0, 0, 6, -6, -6, 0, 0, 3, -3, -3, 0, 0, -1, 1, 1}, {2, 4, 8, 16, 32, 48, 72, 108, 162, 243, 405, 675, 1125, 1875, 3125, 5000, 8000, 12800, 20480, 32768, 53248, 86528, 140608, 228488, 371293, 599781, 968877}, 80]

a(n)=fibonacci(n\5+3)^(n%5)*fibonacci(n\5+2)^(5-n%5) \\ Charles R Greathouse IV, Mar 05 2012

a(n) = F(floor(n/5) + 3)^(n mod 5)*F(floor(n/5) + 2)^(5 - (n mod 5)) where F(n) is the n-th Fibonacci number.
a(n) = a(n-1) + a(n-2) - 3*a(n-5) + 3*a(n-6) + 3*a(n-7) + 6*a(n-10) - 6*a(n-11) - 6*a(n-12) + 3*a(n-15) - 3*a(n-16) - 3*a(n-17) - a(n-20) + a(n-21) + a(n-22).
G.f.: 1-x*(x^21 +2*x^20 +x^19 +x^18 +x^17 -2*x^16 -6*x^15 -4*x^14 -3*x^13 -3*x^12 -9*x^11 -12*x^10 -3*x^9 -6*x^8 -6*x^7 -2*x^6 +6*x^5 +8*x^4 +4*x^3 +2*x^2 +2*x +2) / ((x^2 +x -1) * (x^10 -4*x^5 -1) * (x^10 +x^5 -1)). - Colin Barker, Jun 02 2013

a(0)=1 prepended by Alois P. Heinz, Sep 17 2024

A209398 Number of subsets of {1,...,n} containing two elements whose difference is 2.

Original entry on oeis.org

0, 0, 0, 2, 7, 17, 39, 88, 192, 408, 855, 1775, 3655, 7478, 15228, 30898, 62511, 126177, 254223, 511472, 1027840, 2063600, 4140015, 8300767, 16635087, 33324462, 66736764, 133615658, 267461287, 535294673, 1071191415, 2143357000, 4288290240, 8579130888

Offset: 0

David Nacin, Mar 07 2012

Also, the number of bitstrings of length n containing either 101 or 111.

			For n=3 the subsets containing 1 and 3 are {1,3} and {1,2,3} so a(3)=2.

David Nacin, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (3,-2,1,-1,-2).

Cf. A006498, A209399, A209400.

[2^n - Fibonacci(Floor(n/2) + 2)*Fibonacci(Floor((n + 1)/2) + 2): n in [0..30]]; // G. C. Greubel, Jan 03 2018

Table[2^n -Fibonacci[Floor[n/2] + 2]*Fibonacci[Floor[(n + 1)/2] + 2], {n, 0,30}]
LinearRecurrence[{3, -2, 1, -1, -2}, {0, 0, 0, 2, 7}, 40]
CoefficientList[ Series[x^3 (x +2)/(2x^5 +x^4 -x^3 +2x^2 -3x +1), {x, 0, 33}], x] (* Robert G. Wilson v, Jan 03 2018 *)
a[n_] := Floor[ N[(2^-n ((50 - 14 Sqrt[5]) (1 - Sqrt[5])^n + ((-1 + 2I) (-2I)^n - (1 + 2I) (2I)^n + 5 4^n) (15 + 11 Sqrt[5]) - 2 (1 + Sqrt[5])^n (85 + 37 Sqrt[5])))/(150 + 110 Sqrt[5])]]; Array[a, 33] (* Robert G. Wilson v, Jan 03 2018 *)

x='x+O('x^30); concat([0,0,0], Vec(x^3*(2+x)/((1-2*x)*(1+x^2)*(1-x-x^2)))) \\ G. C. Greubel, Jan 03 2018

#Through Recurrence
def a(n, adict={0:0, 1:0, 2:0, 3:2, 4:7}):
 if n in adict:
  return adict[n]
 adict[n]=3*a(n-1)-2*a(n-2)+a(n-3)-a(n-4)-2*a(n-5)
 return adict[n]

#Returns the actual list of valid subsets
def contains101(n):
 patterns=list()
 for start in range (1,n-1):
  s=set()
  for i in range(3):
   if (1,0,1)[i]:
    s.add(start+i)
  patterns.append(s)
 s=list()
 for i in range(2,n+1):
  for temptuple in comb(range(1,n+1),i):
   tempset=set(temptuple)
   for sub in patterns:
    if sub <= tempset:
     s.append(tempset)
     break
 return s
#Counts all such subsets using the preceding function
def countcontains101(n):
 return len(contains101(n))

a(n) = 3*a(n-1) - 2*a(n-2) + a(n-3) - a(n-4) - 2*a(n-5), a(0)=0, a(1)=0, a(2)=0, a(3)=2, a(4)=7.
a(n) = 2^n - F(2+floor(n/2))*F(floor(2+(n+1)/2)), where F(n) are the Fibonacci numbers.
a(n) = 2^n - A006498(n+2).
G.f.: (2*x^3 + 1*x^4)/(1 - 3*x + 2*x^2 - x^3 + x^4 + 2*x^5) = x^3*(2 + x) / ((1 - 2*x)*(1 + x^2)*(1 - x - x^2)).
E.g.f.: (2*cos(x) + 5*cosh(2*x) + sin(x) + 5*sinh(2*x) - exp(x/2)*(7*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2)))/5. - Stefano Spezia, Mar 12 2024

A209434 Table T(n,m), read by antidiagonals, is the number of subsets of {1,...,n} which do not contain two elements whose difference is m+1.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 5, 4, 2, 1, 8, 6, 4, 2, 1, 13, 9, 8, 4, 2, 1, 21, 15, 12, 8, 4, 2, 1, 34, 25, 18, 16, 8, 4, 2, 1, 55, 40, 27, 24, 16, 8, 4, 2, 1, 89, 64, 45, 36, 32, 16, 8, 4, 2, 1, 144, 104, 75, 54, 48, 32, 16, 8, 4, 2, 1, 233, 169, 125, 81, 72, 64, 32

Offset: 0

David Nacin, Mar 09 2012

1st column is the Fibonacci sequence.

			Table begins:
1,   1,   1,   1,   1,   1,   1,   1,   1,   1,    1,    ...
2,   2,   2,   2,   2,   2,   2,   2,   2,   2,    2,    ...
3,   4,   4,   4,   4,   4,   4,   4,   4,   4,    4,    ...
5,   6,   8,   8,   8,   8,   8,   8,   8,   8,    8,    ...
8,   9,   12,  16,  16,  16,  16,  16,  16,  16,   16,   ...
13,  15,  18,  24,  32,  32,  32,  32,  32,  32,   32,   ...
21,  25,  27,  36,  48,  64,  64,  64,  64,  64,   64,   ...
34,  40,  45,  54,  72,  96,  128, 128, 128, 128,  128,  ...
55,  64,  75,  81,  108, 144, 192, 256, 256, 256,  256,  ...
89,  104, 125, 135, 162, 216, 288, 384, 512, 512,  512,  ...
144, 169, 200, 225, 243, 324, 432, 576, 768, 1024, 1024, ...
............................................................

M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1.

G. C. Greubel, Table of n, a(n) for the first 100 antidiagonals, flattened
Katharine A. Ahrens, Combinatorial Applications of the k-Fibonacci Numbers: A Cryptographically Motivated Analysis, Ph. D. thesis, North Carolina State University (2020).
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.

Cf. A209435, A209436, A209437. Columns: A006498, A006500, A031923, A208742, A208743, A009641

a[n_, m_] := Product[Fibonacci[Floor[(n + i)/(m + 1) + 2]], {i, 0, m}]; Flatten[Table[a[j - i, i], {j, 0, 30}, {i, 0, j}]]

T(n,m) = Product_{i=0 to m} F(floor[(n + i)/(m + 1) + 2]) where F(n) is the n-th Fibonacci number.

cp's OEIS Frontend

A242549 T(n,k)=Number of length n+3 0..k arrays with no four elements in a row with pattern aabb (possibly a=b) and new values 0..k introduced in 0..k order.

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A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3.

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A074677 a(n) = Sum_{i = 0..floor(n/2)} (-1)^(i + floor(n/2)) F(2*i + e), where F = A000045 (Fibonacci numbers) and e = (1-(-1)^n)/2.

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A089928 a(n) = 2*a(n-1) + 2*a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.

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A115008 a(n) = a(n-1) + a(n-3) + a(n-4).

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A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

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A263424 T(n,k)=Number of nXk arrays of permutations of 0..n*k-1 with each element moved a city block distance of 0 or 2.

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A089928 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.