A107358 - OEIS

A107358

Dying rabbits: a(n) = Fibonacci(n) for n <= 12; for n >= 13, a(n) = a(n-1) + a(n-2) - a(n-13).

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 376, 608, 982, 1587, 2564, 4143, 6694, 10816, 17476, 28237, 45624, 73717, 119108, 192449, 310949, 502416, 811778, 1311630, 2119265, 3424201, 5532650, 8939375, 14443788, 23337539, 37707610, 60926041, 98441202, 159056294

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OFFSET

0,4

COMMENTS

In the limit, the growth rate is 1.61575... per generation as opposed to 1.61803... for Fibonacci numbers. - T. D. Noe, Jan 22 2009

If the rabbits die after 12 months, then those that were there in month 1 should die in month 13, whence a(13) = 144 + 89 - 1 = 232 and not 233. In month 14, no rabbits die because the only pair which was there in month 2 already dies. Then in month 15, the one pair born in month 3 will die. In general, the number of rabbits which die in month n (because they are aged 12 months) is equal to the number of newborn rabbits in month n - 12, which is the number of rabbits present in month n - 14. (Recall that a(n - 12) = a(n - 13) + a(n - 14) - #(dying rabbits) = #(rabbits from previous month) + #(newborn rabbits) - #(dying rabbits).) So the recurrence should read a(n) = a(n - 1) + a(n - 2) - a(n - 14). - M. F. Hasler, Oct 06 2017

LINKS

T. D. Noe, Table of n, a(n) for n = 0..500

J. H. E. Cohn, Letter to the editor, Fib. Quart. 2 (1964), 108.

V. E. Hoggatt, Jr. and D. A. Lind, The dying rabbit problem, Fib. Quart. 7 (1969), 482-487.

Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,0,0,0,0,0,0,0,0,-1).

FORMULA

G.f.: x/((x-1)*(1+x)*(x^11+x^9+x^7+x^5+x^3+x-1)). - R. J. Mathar, Jul 27 2009

MAPLE

with(combinat); f:=proc(n) option remember; if n <= 12 then RETURN(fibonacci(n)); fi; f(n-1)+f(n-2)-f(n-13); end;

MATHEMATICA

LinearRecurrence[{1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1}, Fibonacci[Range[0, 12]], 50] (* Harvey P. Dale, Feb 28 2013 *)

PROG