A226516 Number of (18,7)-reverse multiples with n digits.

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

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):

(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 (0,1,0,0,0,0,0,0,0,1).

Crossrefs

Cf. A214927, A226517, A226916.

Programs

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 *)

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