A197424 -id:A197424 - cp's OEIS frontend

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

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

A060635 a(n) is the number of 2 X 1 domino tilings of the set S in the plane R^2 consisting of the union of the following two rectangles: rectangle1: |x| <= n, |y| <= 1, rectangle2: |x| <= 1, |y| <= n.

Original entry on oeis.org

2, 8, 72, 450, 3200, 21632, 149058, 1019592, 6993800, 47922050, 328499712, 2251473408, 15432082562, 105772401800, 724976569800, 4969058770242, 34058447431808, 233440040239232, 1600021920672450, 10966713178192200, 75166970919070472, 515202081704384258, 3531247605071972352

Offset: 1

Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 16 2001

nonn

The relevant graph has rotational symmetry so the number of tilings is a square or twice a square, in this case by the formula for a(n) it is always twice a square.

			a(1) = 2 because in this case the set S is the unit square and there is one horizontal tiling and one vertical.

Harry J. Smith, Table of n, a(n) for n = 1..200
M. Ciucu, Enumeration of perfect matchings in graphs with reflective symmetry, J. Combin. Theory Ser. A 77 (1997), no. 1, 67-97.
W. Jockusch, Perfect matchings and perfect squares J. Combin. Theory Ser. A 67 (1994), no. 1, 100-115.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045, A001654, A006253, A004003, A006125, A002878, A049658, A197424.

with(combinat): for n from 1 to 40 do printf(`%d,`,2*fibonacci(n)^2*fibonacci(n+1)^2) od:

2*Times @@@ Partition[Fibonacci[Range[25]]^2, 2, 1] (* Paolo Xausa, Jul 03 2025 *)

{ a=1; b=0; c=1; for (n=1, 200, f=a+b; g=b+c; a=b; b=c; c=g; write("b060635.txt", n, " ", 2*f^2*g^2); ) } \\ Harry J. Smith, Jul 08 2009

a(n) = 2 * F(n)^2 * F(n+1)^2 where F(n) is the n-th Fibonacci number - sequence A000045.
G.f.: -2*x*(1-x+x^2) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Jan 30 2011
a(n) = -4*(-1)^n*A002878(n)/25 - 2/25 + 6*A049658(n)/25. - R. J. Mathar, Jan 30 2011
a(n) = 2 * A001654(n)^2 = 2 * A197424(n-2) for n>=2. - Alois P. Heinz, Jul 03 2025

More terms from James Sellers, Apr 16 2001

A363476 a(n) = Fibonacci(n)^2 * Fibonacci(n+1)^3.

Original entry on oeis.org

0, 1, 8, 108, 1125, 12800, 140608, 1565109, 17333064, 192329500, 2132531225, 23651979264, 262296652032, 2908947562937, 32260582549000, 357775937196300, 3967793428038237, 44003514081895936, 488006404120114496, 5412074146674562125, 60020821224245910600

Offset: 0

Feryal Alayont, Jun 03 2023

For n>1, a(n) is the number of edge covers of a caterpillar graph with spine P_(5n-4), one pendant attached at vertex n counting from the left end of the spine, a second one at 2n-1, a third at 3n-2 and a fourth at 4n-3. The caterpillar graph for n=3 is as follows:
        *      *      *      *
        |      |      |      |
  *--*--*--v1--*--v2--*--v3--*--*--*
Each pendant edge must be included in an edge cover and hence allows the left and right sides of a vertex adjacent to a pendant to be independent. Therefore, the caterpillar can be split into independent paths, two P_4's (ends) and three P_5's (middle). Each P_n has F_{n-1} edge covers, resulting in the a(n) expression.
The sequence also counts number of subsets of {1, 2, ..., 5n-7} which do not contain two numbers whose difference is 5, a special instance of a general result given in Math. Mag. Problem 1854 (see Links). The equivalence to the above description can be seen as follows. Every vertex except v1, v2 and v3 is incident with at least one of the pendant edges. Therefore, if we label the middle eight edges in the spine with numbers 4, 1, 6, 2, 7, 3, 8, 5 (starting from the left), the edges have to be chosen so that both 1,6, both 2,7, and both 3,8 cannot be missing. This corresponds to choosing subsets of {1, 2, ..., 8} which do not contain two numbers whose difference is 5.

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

José Heber Nieto, Problem 1854 solution, Mathematics Magazine, Vol. 84, No.4 (2011), pp. 300-301.
Marian Tetiva, Problem 1854, Mathematics Magazine, Vol. 83, No.4 (2010), p. 304.
Index entries for linear recurrences with constant coefficients, signature (8,40,-60,-40,8,1).

Cf. A000045, A066258, A197424.

a[n_] := Fibonacci[n]^2 * Fibonacci[n+1]^3; Array[a, 21, 0] (* Amiram Eldar, Jun 06 2023 *)

G.f.: x*(1+4*x^2+x^3) / ((1-x-x^2)*(1-11*x-x^2)*(1+4*x-x^2)).

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

Extensions

A060635 a(n) is the number of 2 X 1 domino tilings of the set S in the plane R^2 consisting of the union of the following two rectangles: rectangle1: |x| <= n, |y| <= 1, rectangle2: |x| <= 1, |y| <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A363476 a(n) = Fibonacci(n)^2 * Fibonacci(n+1)^3.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Formula