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

%I A226516 #35 Jan 09 2025 10:18:30
%S A226516 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,
%T A226516 11,17,15,23,20,31,26,41,34,54,45,71,60,94,80,125,106,166,140,220,185,
%U A226516 291,245,385,325,510,431,676,571,896,756,1187,1001,1572,1326,2082,1757,2758,2328,3654,3084,4841,4085,6413
%N A226516 Number of (18,7)-reverse multiples with n digits.
%C A226516 Comment from _Emeric Deutsch_, Aug 21 2016 (Start):
%C A226516 Given an increasing sequence of positive integers S = {a0, a1, a2, ... }, let
%C A226516           F(x) = x^{a0} + x^{a1} + x^{a2} + ... .
%C A226516 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):
%C A226516       (1 + F(x))/(1 - F(x^2))
%C A226516 Playing with this, I have found easily that
%C A226516 1. number of palindromic compositions of n into {3,4,5,...} = A226916(n+4);
%C A226516 2. number of palindromic compositions of n into {1,4,7,10,13,...} = A226916(n+6);
%C A226516 3. number of palindromic compositions of n into {1,4} = A226517(n+10);
%C A226516 4. number of palindromic compositions of n into {1,5} = A226516(n+11).
%C A226516 (End)
%H A226516 Vincenzo Librandi, <a href="/A226516/b226516.txt">Table of n, a(n) for n = 0..1000</a>
%H A226516 V. E. Hogatt, M. Bicknell, <a href="https://fq.math.ca/13-4.html">Palindromic Compositions</a>, Fib. Quart. 13(4) (1975) 350-356
%H A226516 N. J. A. Sloane, <a href="http://arxiv.org/abs/1307.0453">2178 And All That</a>, Fib. Quart., 52 (2014), 99-120.
%H A226516 N. J. A. Sloane, <a href="/A001232/a001232.pdf">2178 And All That</a> [Local copy]
%H A226516 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,0,0,0,0,0,0,1).
%F A226516 G.f.: x^6*(1+x)*(1-x+x^5)/(1-x^2-x^10).
%F A226516 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]
%F A226516 a(2n+1) = A003520(n-5). a(2n) = A098523(n-6). - _R. J. Mathar_, Dec 13 2022
%p A226516 f:=proc(n) option remember;
%p A226516 if
%p A226516 n <= 5 then 0
%p A226516 elif n=6 then 1
%p A226516 elif n <= 10 then 0
%p A226516 elif n <= 12 then 1
%p A226516 else f(n-2)+f(n-10)
%p A226516 fi;
%p A226516 end;
%p A226516 [seq(f(n),n=0..100)]
%t A226516 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 *)
%t A226516 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 *)
%Y A226516 Cf. A214927, A226517, A226916.
%K A226516 nonn,easy,base
%O A226516 0,17
%A A226516 _N. J. A. Sloane_, Jun 16 2013