A226916 Number of (17,11)-reverse multiples with n digits.

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

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.

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

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

Cf. A214927, A226516, A226517.

Sequence in context: A161052 A161256 A161281 * A003113 A225502 A152227

Adjacent sequences: A226913 A226914 A226915 * A226917 A226918 A226919

Keyword

nonn,base,easy

Author

N. J. A. Sloane, Jun 24 2013

Status

approved