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A226916
Number of (17,11)-reverse multiples with n digits.
7
0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 2, 1, 2, 2, 3, 3, 5, 4, 7, 6, 10, 9, 15, 13, 22, 19, 32, 28, 47, 41, 69, 60, 101, 88, 148, 129, 217, 189, 318, 277, 466, 406, 683, 595, 1001, 872, 1467, 1278, 2150, 1873, 3151, 2745, 4618, 4023, 6768, 5896, 9919, 8641, 14537, 12664, 21305, 18560, 31224, 27201, 45761
OFFSET
0,11
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^4*(1+x)*(1-x+x^3)/(1-x^2-x^6).
a(2n) = A058278(n-1). a(2n+1)=A000930(n-3). - R. J. Mathar, Dec 13 2022
MATHEMATICA
CoefficientList[Series[x^4 (1 - x^2 + x^3 + x^4) / (1 - x^2 - x^6), {x, 0, 70}], x] (* Vincenzo Librandi, Jul 16 2013 *)
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
N. J. A. Sloane, Jun 24 2013
STATUS
approved