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%I #28 Dec 07 2019 12:18:25
%S 1,111,11211,1122211,112232211,11223332211,1122334332211,
%T 112233444332211,11223344544332211,1122334455544332211,
%U 112233445565544332211,11223344556665544332211,1122334455667665544332211,112233445566777665544332211,11223344556677877665544332211
%N Expansion of 1/((1-x)(1-10x)(1-100x)).
%C Partial sums of A109241.
%H Andrew Howroyd, <a href="/A109242/b109242.txt">Table of n, a(n) for n = 0..100</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).
%F a(n) = 10^(2*n+3)/891 - 10^(n+1)/81 + 1/891.
%F a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
%F a(n) = 110*a(n-1) - 1000*a(n-2) + 1, n >= 2. - _Vincenzo Librandi_, Mar 18 2011
%e The numbers of 1's, 2's, 3's etc. appearing occur according to
%e 1:1,3,4,4,4,4,4,4,...
%e 2:0,0,1,3,4,4,4,4,...
%e 3:0,0,0,0,1,3,4,4,...
%e 4:0,0,0,0,0,0,1,3,... etc. up to term 17, where 9->10 etc. changes the pattern.
%o (Sage) [gaussian_binomial(n,2,10) for n in range(2,14)] # _Zerinvary Lajos_, May 27 2009
%o (PARI) a(n) = {10^(2*n+3)/891 - 10^(n+1)/81 + 1/891} \\ _Andrew Howroyd_, Nov 08 2019
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jun 23 2005
%E Terms a(12) and beyond from _Andrew Howroyd_, Nov 08 2019
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