A022100 Fibonacci sequence beginning 1, 10.

1, 10, 11, 21, 32, 53, 85, 138, 223, 361, 584, 945, 1529, 2474, 4003, 6477, 10480, 16957, 27437, 44394, 71831, 116225, 188056, 304281, 492337, 796618, 1288955, 2085573, 3374528, 5460101, 8834629, 14294730, 23129359, 37424089, 60553448, 97977537, 158530985

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(10; n-1-k, k), n >= 1, with a(-1)=9. These are the SW-NE diagonals in P(10; n, k), the (10,1) Pascal triangle A093645. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

In general, for b Fibonacci sequence beginning with 1, h, we have:

b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011

Links

• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (1, 1).

Crossrefs

Cf. A000045.

a(n) = A109754(9, n+1) = A101220(9, 0, n+1).

Programs

Magma

a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

Mathematica

LinearRecurrence[{1,1},{1,10},40] (* Harvey P. Dale, May 17 2017 *)

Formula

a(n) = a(n-1) + a(n-2) for n >= 2, a(0)=1, a(1)=10, a(-1):=9.

G.f.: (1 + 9*x)/(1 - x - x^2).

a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*(9*binomial(1, k) - 8*binomial(0, k)). - Paul Barry, May 05 2005

a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (9/2)*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009

From Bruno Berselli, Feb 20 2017: (Start)

a(n) = 9*A000045(n) + A000045(n+1).

a(n) = 11*A000045(n) - A000045(n-2). (End)