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A226516
Number of (18,7)-reverse multiples with n digits.
4
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 3, 4, 4, 6, 5, 8, 6, 10, 8, 13, 11, 17, 15, 23, 20, 31, 26, 41, 34, 54, 45, 71, 60, 94, 80, 125, 106, 166, 140, 220, 185, 291, 245, 385, 325, 510, 431, 676, 571, 896, 756, 1187, 1001, 1572, 1326, 2082, 1757, 2758, 2328, 3654, 3084, 4841, 4085, 6413
OFFSET
0,17
COMMENTS
Comment from Emeric Deutsch, Aug 21 2016 (Start):
Given an increasing sequence of positive integers S = {a0, a1, a2, ... }, let
F(x) = x^{a0} + x^{a1} + x^{a2} + ... .
Then the g. f. for the number of palindromic compositions of n with parts in S is (see Hoggatt and Bicknell, Fibonacci Quarterly, 13(4), 1975):
(1 + F(x))/(1 - F(x^2))
Playing with this, I have found easily that
1. number of palindromic compositions of n into {3,4,5,...} = A226916(n+4);
2. number of palindromic compositions of n into {1,4,7,10,13,...} = A226916(n+6);
3. number of palindromic compositions of n into {1,4} = A226517(n+10);
4. number of palindromic compositions of n into {1,5} = A226516(n+11).
(End)
LINKS
V. E. Hogatt, M. Bicknell, Palindromic Compositions, Fib. Quart. 13(4) (1975) 350-356
N. J. A. Sloane, 2178 And All That, Fib. Quart., 52 (2014), 99-120.
FORMULA
G.f.: x^6*(1+x)*(1-x+x^5)/(1-x^2-x^10).
a(n) = a(n-2) + a(n-10) for n>12, with initial values a(0)-a(12) equal to 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1. [Bruno Berselli, Jun 17 2013]
a(2n+1) = A003520(n-5). a(2n) = A098523(n-6). - R. J. Mathar, Dec 13 2022
MAPLE
f:=proc(n) option remember;
if
n <= 5 then 0
elif n=6 then 1
elif n <= 10 then 0
elif n <= 12 then 1
else f(n-2)+f(n-10)
fi;
end;
[seq(f(n), n=0..100)]
MATHEMATICA
CoefficientList[Series[x^6 (1 - x^2 + x^5 + x^6) / (1 - x^2 - x^10), {x, 0, 80}], x] (* Vincenzo Librandi, Jun 18 2013 *)
LinearRecurrence[{0, 1, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1}, 80] (* Harvey P. Dale, Jun 17 2015 *)
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
N. J. A. Sloane, Jun 16 2013
STATUS
approved