A022388 Fibonacci sequence beginning 6, 13.

6, 13, 19, 32, 51, 83, 134, 217, 351, 568, 919, 1487, 2406, 3893, 6299, 10192, 16491, 26683, 43174, 69857, 113031, 182888, 295919, 478807, 774726, 1253533, 2028259, 3281792, 5310051, 8591843, 13901894, 22493737, 36395631, 58889368, 95284999, 154174367, 249459366, 403633733, 653093099

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The Pisano periods for this sequence are different from those for the Fibonacci numbers (A001175) for modulus 11 and 22. Furthermore, its Pisano periods are exactly the same as those of the Lucas sequence (A000032), given in A106291. - Klaus Purath, Apr 20 2019

a(n) is the alternating sum of 5 consecutive Lucas numbers (A000032). Also the sum of 4*k consecutive terms of A000285 divided by Fibonacci(2*k) (A000045), k = {1, 2, …}. All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Jul 29 2019

From Wajdi Maaloul, Jun 25 2022: (Start)

For n > 0, a(n) is the number of ways to tile the following figure (a T-shaped horizontal strip of length n beginning with a vertical strip of length 6) with squares and dominoes.

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(End)

Links

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

Crossrefs

Cf. A000032, A000045.

Programs

GAP

List([0..40], n-> 6*Fibonacci(n+2) + Fibonacci(n)); # G. C. Greubel, Jun 30 2019

Magma

[6*Fibonacci(n+2) + Fibonacci(n): n in [0..40]]; // G. C. Greubel, Mar 02 2018

Mathematica

Table[6*Fibonacci[n+2] + Fibonacci[n], {n, 0, 40}] (* or *) LinearRecurrence[{1,1}, {6,13}, 40] (* G. C. Greubel, Mar 02 2018 *)

PARI

vector(40, n, n--; 6*fibonacci(n+2) + fibonacci(n)) \\ G. C. Greubel, Mar 02 2018

Sage

[6*fibonacci(n+2) + fibonacci(n) for n in (0..40)] # G. C. Greubel, Jun 30 2019

Formula

G.f.: (6+7*x)/(1-x-x^2). - Philippe Deléham, Nov 20 2008

a(n) = 6*Fibonacci(n+2) + Fibonacci(n) = 6*Fibonacci(n-1) + 13*Fibonacci(n). - G. C. Greubel, Mar 02 2018

From Klaus Purath, Jul 29 2019: (Start)

L = Lucas (A000032), F = Fibonacci (A000045). All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).

a(n+1) - a(n-4) = L(n)*11.

a(n) = L(n-1) + L(n+4).

a(n) = 3*L(n+1) + L(n+2) = L(n) + 4*L(n+1) = L(n+6) - 4*L(n+2).

a(n) = L(n+1) + 5*F(n+2) = L(n+5) - 5*F(n+1).

a(n) = (7*L(n+1) + 5*F(n+1))/2.

a(n) = (13*L(n+1) + L(n+5) - 5*F(n))/4.

a(n) = 7*F(n) + 6*F(n+1) = 7*F(n+2) - F(n+1).

a(n) = 8*F(n+2) - F(n+3) = 17*F(n+4) - 9*F(n+5).

The following six formulas apply for all sequences of the Fibonacci type.

a(n) = L(2*m)*a(n+2*m) - a(n+4*m).

a(n) = (F(m+2)*a(n+2) - a(m+n+2))/F(m).

a(n) = F(n-m-1)*a(m) + F(n-m)*a(m+1).

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

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

3*a(n+2)*a(n+1)*a(n) = a(n+2)^3 - a(n+1)^3 - a(n)^3. (End)

E.g.f.: exp(-2*x/(1+sqrt(5)))*(-15-sqrt(5)+(45+19*sqrt(5))*exp(sqrt(5)*x))/(5+3*sqrt(5)). - Stefano Spezia, Aug 16 2019

Extensions

Terms a(36) onward added by G. C. Greubel, Mar 02 2018