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A000044 Dying rabbits: a(0) = 1; for 1 <= n <= 12, a(n) = Fibonacci(n); for n >= 13, a(n) = a(n-1) + a(n-2) - a(n-13).
(Formerly M0691 N0255)
5
1, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 232, 375, 606, 979, 1582, 2556, 4130, 6673, 10782, 17421, 28148, 45480, 73484, 118732, 191841, 309967, 500829, 809214, 1307487, 2112571, 3413385, 5515174, 8911138, 14398164, 23263822, 37588502, 60733592, 98130253, 158553878, 256183302, 413927966, 668803781, 1080619176, 1746009572, 2821113574, 4558212008 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

A107358 is a more satisfactory version, but I have left the present sequence unchanged (except for making the definition clearer) since it has been in the OEIS so long.

Number of compositions of n into parts 1, 3, 5, 7, 9, and 11. - Joerg Arndt, Sep 05 2014

If a(0) = 1 then it is not clear why a(2) = 1, it should be equal to a(1) + a(0) = 2. Does the first comment mean that a(0) is erroneous and should read a(0) = 0? In contrast to A107358, the term a(13) = 232 = 144 + 89 - 1 seems correct, since in this month the first and oldest pair of rabbits die. But a(14) should be equal to a(13) + a(12) = 232 + 144 because the first pair (which was also the only one present in month 2) has already died and there is no other pair aged 12 months. In general, the number of pairs which die in month n because they are aged exactly 12 months, equals a(n-14): this is the number of newborn pairs in month n - 12, viz. a(n-12) = a(n-13) [those from preceding month] + a(n-14) [the newborn ones] - #(those which die). - M. F. Hasler, Oct 06 2017

REFERENCES

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

LINKS

Harvey P. Dale, Table of n, a(n) for n = 0..1000

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.: 1/(1 - z - z^3 - z^5 - z^7 - z^9 -z^11).

G.f. A(x) = 1 / (1 - x / (1 - x^2 / (1 + x^10 / (1 + x^2 / (1 - x^2 / (1 + x^6 / (1 + x^2 / (1 - x^2 / (1 + x^2))))))))). - Michael Somos, Jan 04 2013

For n >= 11, a(n) = a(n-1) + a(n-3) + a(n-5) + a(n-7) + a(n-9) + a(n-11). - Eric M. Schmidt, Sep 04 2014

EXAMPLE

G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 5*x^5 + 8*x^6 + 13*x^7 + 21*x^8 + 34*x^9 + ...

MAPLE

with(combinat); f:=proc(n) option remember; if n=0 then RETURN(1); fi; if n <= 12 then RETURN(fibonacci(n)); fi; f(n-1)+f(n-2)-f(n-13); end;

MATHEMATICA

CoefficientList[Series[1/(1 - z - z^3 - z^5 - z^7 - z^9 - z^11), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

LinearRecurrence[{1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {1, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144}, 100] (* Harvey P. Dale, Mar 24 2012 *)

PROG

(MAGMA) [ n eq 1 select 1 else n le 13 select Fibonacci(n-1) else Self(n-1)+Self(n-2)-Self(n-13): n in [1..50] ]; // Klaus Brockhaus, Dec 21 2010

(PARI) Vec(1/(1-z-z^3-z^5-z^7-z^9-z^11)+O(z^50)) \\ Charles R Greathouse IV, Jun 10 2011

CROSSREFS

Cf. A107358. See A000045 for the Fibonacci numbers.

Sequence in context: A268133 A217737 A023442 * A107358 A243063 A248740

Adjacent sequences:  A000041 A000042 A000043 * A000045 A000046 A000047

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane; entry revised May 25 2005

EXTENSIONS

G.f. corrected by Charles R Greathouse IV, Jun 10 2011

STATUS

approved

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Last modified October 17 07:05 EDT 2018. Contains 316276 sequences. (Running on oeis4.)