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A022100
Fibonacci sequence beginning 1, 10.
8
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
0,2
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
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)
MATHEMATICA
LinearRecurrence[{1, 1}, {1, 10}, 40] (* Harvey P. Dale, May 17 2017 *)
PROG
(Magma) a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
CROSSREFS
Cf. A000045.
a(n) = A109754(9, n+1) = A101220(9, 0, n+1).
Sequence in context: A298849 A277588 A339138 * A041475 A041204 A041202
KEYWORD
nonn,easy
STATUS
approved