A056014 -id:A056014 - cp's OEIS frontend

A027994 a(n) = (F(2n+3) - F(n))/2 where F() = Fibonacci numbers A000045.

1, 2, 6, 16, 43, 114, 301, 792, 2080, 5456, 14301, 37468, 98137, 256998, 672946, 1761984, 4613239, 12078110, 31621701, 82787980, 216743836, 567446112, 1485598681, 3889356696, 10182482353, 26658108074, 69791870526, 182717549872, 478360854115, 1252365133866, 3278734743901, 8583839415648, 22472784017272

Offset: 0

Clark Kimberling

Substituting x*(1-x)/(1-2x) into x^2/(1-x^2) yields x^2*(g.f. of sequence).
The number of (s(0), s(1), ..., s(n+1)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n+1, s(0) = 2, s(n+1) = 3. - Herbert Kociemba, Jun 02 2004
Diagonal sums of triangle in A125171. - Philippe Deléham, Jan 14 2014

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Cf. A000667, A059216, A059219, A059502, A027926.

[(Fibonacci(2*n+3)-Fibonacci(n))/2 : n in [0..40]]; // Vincenzo Librandi, Jan 01 2025

Table[(Fibonacci[2n+3]-Fibonacci[n])/2,{n,0,30}] (* or *) LinearRecurrence[{4,-3,-2,1},{1,2,6,16},30] (* Harvey P. Dale, Apr 28 2022 *)

PARI
```
a(n)=(fibonacci(2*n+3)-fibonacci(n))/2
```

G.f.: (1-x)^2/((1-x-x^2)*(1-3*x+x^2)). - Floor van Lamoen and N. J. A. Sloane, Jan 21 2001
a(n) = Sum_{k=0..n} T(n, k)*T(n, n+k), T given by A027926.
a(n) = 2*a(n-1) + Sum_{m < n-1} a(m) + F(n-1) = A059512(n+2) - F(n) where F(n) is the n-th Fibonacci number (A000045). - Floor van Lamoen, Jan 21 2001
a(n) = (2/5)*Sum_{k=1..4} sin(2*Pi*k/5)*sin(3*Pi*k/5)*(1+2*cos(Pi*k/5))^(n+1). - Herbert Kociemba, Jun 02 2004
a(-1-2n) = A056014(2n), a(-2n) = A005207(2n-1).
E.g.f.: exp(3*x/2)*cosh(sqrt(5)*x/2) + exp(x/2)*(2*exp(x) - 1)*sinh(sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Jan 01 2025

A191797 a(n) = binomial(F(n), 2) where F(n) = A000045(n).

Original entry on oeis.org

0, 0, 0, 1, 3, 10, 28, 78, 210, 561, 1485, 3916, 10296, 27028, 70876, 185745, 486591, 1274406, 3337236, 8738290, 22879230, 59901985, 156830905, 410597496, 1074972528, 2814337800, 7368069528, 19289917153, 50501756955, 132215475106, 346144864780, 906219437046

Offset: 0

Emeric Deutsch, Jun 21 2011

			a(7) = binomial(13,2) = 78.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).

Cf. A000045, A000071, A000217, A001622, A056014, A094825 (binomial transform), A122931.

with(combinat): seq(binomial(fibonacci(n), 2), n = 0 .. 30);

Table[Binomial[Fibonacci[n], 2], {n, 0, 39}] (* Alonso del Arte, Apr 04 2013 *)

a(n) = binomial(fibonacci(n), 2); \\ Michel Marcus, Sep 07 2015

concat(vector(3), Vec(x^3 / ((1+x)*(1-x-x^2)*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 26 2017

from sympy import binomial, fibonacci
def a(n): return binomial(fibonacci(n), 2) # Indranil Ghosh, Mar 26 2017

a(n) = 3*a(n-1) + 1*a(n-2) - 5*a(n-3) - 1*a(n-4) + 1*a(n-5).
G.f.: x^3/(1-3*x-x^2+5*x^3+x^4-x^5) = x^3/((1+x)*(1-x-x^2)*(1-3*x+x^2)).
a(n) + a(n+1) = A056014(n+1). - R. J. Mathar, Jun 24 2011
a(n) = (2*F(n)^2 - F(n+4) + 3*F(n+1))/4, F(n) = A000045(n). - Gary Detlefs, Jan 05 2013
a(n) = Sum_{k=1..n-2} A122931(k). - J. M. Bergot, Apr 05 2013
a(n) = A000217(A000071(n)). - Peter M. Chema, Mar 26 2017
a(n) = (2^(-1-n)*(-(-1)^n*2^(1+n) + sqrt(5)*(1-sqrt(5))^n + (3-sqrt(5))^n - sqrt(5)*(1+sqrt(5))^n + (3+sqrt(5))^n)) / 5. - Colin Barker, Mar 26 2017
a(n) ~ phi^(2*n) / 10, where phi is the golden ratio (A001622). - Amiram Eldar, Aug 26 2025

A056015 a(n) = 6a(n-1) - 5a(n-2) - 4a(n-3) - 3a(n-4) + 2*a(n-5) + a(n-6), with a(0)=...=a(4)=0, a(5)=1.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 6, 31, 152, 730, 3480, 16542, 78544, 372779, 1768958, 8393741, 39827472, 188975588, 896658432, 4254492236, 20186832928, 95783024581, 454473817254, 2156399264651, 10231739547432, 48547824776670, 230350985294584, 1092975362559562

Offset: 0

Asher Auel, Jun 06 2000

With a(0)=0, a(1)=1, a(2)=1, a(3)=2, a(4)=4, a(5)=7, this recurrence produces a(n) = A000073(n+1) (tribonacci numbers).

Index entries for linear recurrences with constant coefficients, signature (6,-5,-4,-3,2,1).

Cf. A000073, A056014.

O.g.f.: -x^5/((x^3+x^2-5*x+1)*(x^3+x^2+x-1)). - R. J. Mathar, Nov 23 2007

New name, from formula, added by Michel Marcus, Mar 19 2024

A059512 For n>=2, the number of (s(0), s(1), ..., s(n-1)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| <= 1 for i = 1,2,....,n-1, s(0) = 2, s(n-1) = 2.

Original entry on oeis.org

0, 1, 1, 3, 7, 18, 46, 119, 309, 805, 2101, 5490, 14356, 37557, 98281, 257231, 673323, 1762594, 4614226, 12079707, 31624285, 82792161, 216750601, 567457058, 1485616392, 3889385353, 10182528721, 26658183099, 69791991919

Offset: 0

Floor van Lamoen, Jan 21 2001

Substituting x(1-x)/(1-2x) into x/(1-x^2) yields g.f. of sequence.

Cf. A000667, A059216, A059219, A059502, A027994.

a(1-2n)=A005207(2n), a(-2n)=A056014(2n+1).

CoefficientList[Series[x(1-x)(1-2x)/((1-x-x^2)(1-3x+x^2)), {x,0,30}], x]  (* Harvey P. Dale, Apr 23 2011 *)

a(n)=(fibonacci(2*n-1)+fibonacci(n-2))/2

a(n) = 2a(n-1) + Sum{mA000045).
G.f.: x(1-x)(1-2x)/((1-x-x^2)(1-3x+x^2)).
a(n+1)=sum{k=0..floor(n/2), C(n,2k)*F(2k+1)}. [From Paul Barry, Oct 14 2009]

A124697 Number of base 4 circular n-digit numbers with adjacent digits differing by 1 or less.

Original entry on oeis.org

1, 4, 10, 22, 54, 134, 340, 872, 2254, 5854, 15250, 39802, 104004, 271964, 711490, 1861862, 4873054, 12755614, 33391060, 87413152, 228841254, 599099054, 1568437210, 4106182322, 10750060804, 28143920884, 73681573690, 192900592822, 505019869254, 1322158472054

Offset: 0

R. H. Hardin, Dec 28 2006

[Empirical] a(base,n)=a(base-1,n)+A002426(n+1) for base>=1.int(n/2)+1

a(n) = T(n, 4) where T(n, k) = Sum_{j=1..k} (1+2*cos(j*Pi/(k+1)))^n. These are the number of smooth cyclic words of length n over the alphabet {1,2,3,4}. See theorem 3.3 in Knopfmacher and others, reference in A124696. - Peter Luschny, Aug 13 2012

Colin Barker, Table of n, a(n) for n = 0..1000
OEIS Wiki, Number of base k circular n-digit numbers with adjacent digits differing by d or less
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Cf. A000032, A005248, A056014.

LinearRecurrence[{4,-3,-2,1},{1,4,10,22,54},30] (* Harvey P. Dale, Oct 14 2016 *)

Vec(-(3*x^4-4*x^3-3*x^2+1)/((x^2-3*x+1)*(x^2+x-1)) + O(x^40)) \\ Colin Barker, Jul 19 2015

G.f.: A(x) = (3*x^4-4*x^3-3*x^2+1) / ((x^2-3*x+1)*(1-x-x^2)). - Colin Barker, Jul 19 2015
From Peter Bala, Nov 08 2022: (Start)
a(n) = Lucas(n) + Lucas(2*n) = A000032(n) + A005248(n) for n >= 1.
A(x) = 1 + x*B'(x)/B(x), where B(x) = 1/((1 - x - x^2)*(1 - 3*x + x^2)) = 1 + 4*x + 13*x^2 + 38*x^3 + ... has integral coefficients. See A056014.
It follows that the Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r. (End)

A097040 a(n) = 2sum(C(n,2k+1)F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045.

Original entry on oeis.org

0, 0, 2, 8, 26, 76, 212, 576, 1542, 4092, 10802, 28424, 74648, 195808, 513242, 1344672, 3521994, 9223284, 24151052, 63235040, 165562430, 433465780, 1134856802, 2971140048, 7778620656, 20364814656, 53315973362, 139583348216

Offset: 1

Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004

Create a triangle with first column T(n,1)=A000045(n) for n=0,1,2... The remaining terms T(r,c)=T(r,c-1)+T(r-1,c-1). The sum of all terms for the first n+1 rows of this triangle=a(n+2). The sum of the terms in row(n+1)= 0, 2, 6, 18, 50, 136, 364...with partial sums of these sums duplicating this sequence 0, 2, 8, 26, 76, 212, 576... - J. M. Bergot, Dec 19 2012

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

f[n_] := f[n] = f[n - 1] + f[n - 2]; f[0] = 0; f[1] = 1; Table[2 Sum[Binomial[n, 2k + 1]f[2k], {k, 0, Floor[(n - 1)/2]}], {n, 1, 30}]
Table[Fibonacci[2n-1]-Fibonacci[n+1],{n,30}] (* Harvey P. Dale, Oct 05 2011 *)
LinearRecurrence[{4, -3, -2, 1}, {0, 0, 2, 8}, 29] (* Robert G. Wilson v, Dec 26 2012 *)

a(n) = F(2n-1)-F(n+1) = 2*A056014(n).

G.f. -2*x^3 / ( (x^2-3*x+1)*(x^2+x-1) ). - R. J. Mathar, Jan 08 2013

cp's OEIS Frontend

A027994 a(n) = (F(2n+3) - F(n))/2 where F() = Fibonacci numbers A000045.

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A191797 a(n) = binomial(F(n), 2) where F(n) = A000045(n).

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A056015 a(n) = 6*a(n-1) - 5*a(n-2) - 4*a(n-3) - 3*a(n-4) + 2*a(n-5) + a(n-6), with a(0)=...=a(4)=0, a(5)=1.

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Extensions

A059512 For n>=2, the number of (s(0), s(1), ..., s(n-1)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| <= 1 for i = 1,2,....,n-1, s(0) = 2, s(n-1) = 2.

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Mathematica

PARI

Formula

A124697 Number of base 4 circular n-digit numbers with adjacent digits differing by 1 or less.

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Programs

Mathematica

PARI

Formula

A097040 a(n) = 2*sum(C(n,2k+1)*F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045.

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Mathematica

Formula

A056015 a(n) = 6a(n-1) - 5a(n-2) - 4a(n-3) - 3a(n-4) + 2*a(n-5) + a(n-6), with a(0)=...=a(4)=0, a(5)=1.

A097040 a(n) = 2sum(C(n,2k+1)F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045.