A226517 Number of (19,14)-reverse multiples with n digits.

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 4, 3, 6, 4, 8, 5, 11, 7, 15, 10, 21, 14, 29, 19, 40, 26, 55, 36, 76, 50, 105, 69, 145, 95, 200, 131, 276, 181, 381, 250, 526, 345, 726, 476, 1002, 657, 1383, 907, 1909, 1252, 2635, 1728, 3637, 2385, 5020, 3292, 6929, 4544, 9564, 6272, 13201, 8657, 18221

Offset: 0

Author

N. J. A. Sloane, Jun 16 2013

Keywords

• nonn
• easy
• base

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, 350 - 356):

(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.
• N. J. A. Sloane, 2178 And All That [Local copy]
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,0,0,0,0,1).

Crossrefs

Cf. A214927, A226516, A226916.

Programs

Maple

f:=proc(n) option remember;
if
n <= 5 then 0
elif n=6 then 1
elif n <= 9 then 0
elif n <= 11 then 1
else f(n-2)+f(n-8)
fi;
end;
[seq(f(n),n=0..120)];

Mathematica

CoefficientList[Series[x^6 (1 - x^2 + x^4 + x^5) / (1 - x^2 - x^8), {x, 0, 80}], x] (* Vincenzo Librandi, Jun 18 2013 *)
LinearRecurrence[{0,1,0,0,0,0,0,1},{0,0,0,0,0,0,1,0,0,0,1,1},80] (* Harvey P. Dale, Aug 23 2019 *)

Formula

G.f.: x^6*(1+x)*(1-x+x^4)/(1-x^2-x^8).

a(n) = a(n-2) + a(n-8) for n>11, with initial values a(0)-a(11) = 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1. [Bruno Berselli, Jun 17 2013]

a(2n+1)=A003269(n-4). a(2n)=A103632(n-3). - R. J. Mathar, Dec 13 2022