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

%I A226517 #34 Jan 09 2025 10:19:06
%S A226517 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,
%T A226517 29,19,40,26,55,36,76,50,105,69,145,95,200,131,276,181,381,250,526,
%U A226517 345,726,476,1002,657,1383,907,1909,1252,2635,1728,3637,2385,5020,3292,6929,4544,9564,6272,13201,8657,18221
%N A226517 Number of (19,14)-reverse multiples with n digits.
%C A226517 Comment from _Emeric Deutsch_, Aug 21 2016 (Start):
%C A226517 Given an increasing sequence of positive integers S = {a0, a1, a2, ... }, let
%C A226517           F(x) = x^{a0} + x^{a1} + x^{a2} + ... .
%C A226517 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):
%C A226517       (1 + F(x))/(1 - F(x^2))
%C A226517 Playing with this, I have found easily that
%C A226517 1. number of palindromic compositions of n into {3,4,5,...} = A226916(n+4);
%C A226517 2. number of palindromic compositions of n into {1,4,7,10,13,...} = A226916(n+6);
%C A226517 3. number of palindromic compositions of n into {1,4} = A226517(n+10);
%C A226517 4. number of palindromic compositions of n into {1,5} = A226516(n+11).
%C A226517 (End)
%H A226517 Vincenzo Librandi, <a href="/A226517/b226517.txt">Table of n, a(n) for n = 0..1000</a>
%H A226517 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 A226517 N. J. A. Sloane, <a href="http://arxiv.org/abs/1307.0453">2178 And All That</a>, Fib. Quart., 52 (2014), 99-120.
%H A226517 N. J. A. Sloane, <a href="/A001232/a001232.pdf">2178 And All That</a> [Local copy]
%H A226517 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,0,0,0,0,1).
%F A226517 G.f.: x^6*(1+x)*(1-x+x^4)/(1-x^2-x^8).
%F A226517 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]
%F A226517 a(2n+1)=A003269(n-4). a(2n)=A103632(n-3). - _R. J. Mathar_, Dec 13 2022
%p A226517 f:=proc(n) option remember;
%p A226517 if
%p A226517 n <= 5 then 0
%p A226517 elif n=6 then 1
%p A226517 elif n <= 9 then 0
%p A226517 elif n <= 11 then 1
%p A226517 else f(n-2)+f(n-8)
%p A226517 fi;
%p A226517 end;
%p A226517 [seq(f(n),n=0..120)];
%t A226517 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 *)
%t A226517 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 *)
%Y A226517 Cf. A214927, A226516, A226916.
%K A226517 nonn,easy,base
%O A226517 0,15
%A A226517 _N. J. A. Sloane_, Jun 16 2013