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Dying rabbits: a(n) = Fibonacci(n) for n <= 12; for n >= 13, a(n) = a(n-1) + a(n-2) - a(n-13).
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%I #32 Jun 01 2024 11:18:02

%S 0,1,1,2,3,5,8,13,21,34,55,89,144,233,376,608,982,1587,2564,4143,6694,

%T 10816,17476,28237,45624,73717,119108,192449,310949,502416,811778,

%U 1311630,2119265,3424201,5532650,8939375,14443788,23337539,37707610,60926041,98441202,159056294

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

%C 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

%C 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

%H T. D. Noe, <a href="/A107358/b107358.txt">Table of n, a(n) for n = 0..500</a>

%H J. H. E. Cohn, <a href="http://www.fq.math.ca/Scanned/2-2/cohn1.pdf">Letter to the editor</a>, Fib. Quart. 2 (1964), 108.

%H V. E. Hoggatt, Jr. and D. A. Lind, <a href="http://www.fq.math.ca/Scanned/7-5/hoggatt.pdf">The dying rabbit problem</a>, Fib. Quart. 7 (1969), 482-487.

%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,0,0,0,0,0,0,0,0,0,0,-1).

%F 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

%p 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;

%t 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 *)

%o (PARI) Vec(x/(x^13-x^2-x+1)+O(x^99)) \\ _Charles R Greathouse IV_, Jun 10 2011

%Y See A000045 for the Fibonacci numbers. This is a better version of A000044.

%K nonn,easy

%O 0,4

%A _N. J. A. Sloane_, May 25 2005