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%I #45 Sep 08 2022 08:45:41
%S 1,10,200,5000,140000,4200000,132000000,4290000000,143000000000,
%T 4862000000000,167960000000000,5878600000000000,208012000000000000,
%U 7429000000000000000,267444000000000000000,9694845000000000000000,353576700000000000000000
%N a(n) = 10^n*Catalan(n).
%C In general, for m >= 1, Sum_{k>=0} 1/(m^k * Catalan(k)) = 2*m*(8*m + 1) / (4*m - 1)^2 + 24 * m^2 * arcsin(1/(2*sqrt(m))) / (4*m - 1)^(5/2). - _Vaclav Kotesovec_, Nov 23 2021
%H Vincenzo Librandi, <a href="/A156275/b156275.txt">Table of n, a(n) for n = 0..200</a>
%H Vincent Pilaud, <a href="https://arxiv.org/abs/2205.06686">Pebble trees</a>, arXiv:2205.06686 [math.CO], 2022.
%F a(n) = 10^n*A000108(n).
%F From _Gary W. Adamson_, Jul 18 2011: (Start)
%F a(n) is the upper left term in M^n, M = an infinite square production matrix:
%F 10, 10, 0, 0, 0, ...
%F 10, 10, 10, 0, 0, ...
%F 10, 10, 10, 10, 0, ...
%F 10, 10, 10, 10, 10, ...
%F ... (End)
%F E.g.f.: KummerM(1/2, 2, 40*x). - _Peter Luschny_, Aug 26 2012
%F G.f.: c(10*x) with c(x) the o.g.f. of A000108 (Catalan). - _Philippe Deléham_, Nov 15 2013
%F a(n) = Sum_{k=0..n} A085880(n,k)*9^k. - _Philippe Deléham_, Nov 15 2013
%F G.f.: 1/(1 - 10*x/(1 - 10*x/(1 - 10*x/(1 - ...)))), a continued fraction. - _Ilya Gutkovskiy_, Aug 08 2017
%F Sum_{n>=0} 1/a(n) = 180/169 + 800*arctan(1/sqrt(39)) / (507*sqrt(39)). - _Vaclav Kotesovec_, Nov 23 2021
%F Sum_{n>=0} (-1)^n/a(n) = 1580/1681 - 2400*arctanh(1/sqrt(41)) / (1681*sqrt(41)). - _Amiram Eldar_, Jan 25 2022
%F D-finite with recurrence (n+1)*a(n) +20*(-2*n+1)*a(n-1)=0. - ~~~
%t Table[10^n CatalanNumber[n],{n,0,20}] (* _Harvey P. Dale_, Mar 12 2013 *)
%o (Magma) [10^n*Catalan(n): n in [0..20]]; // _Vincenzo Librandi_, Jul 19 2011
%Y Cf. A000108, A005159, A085880, A151374, A151403, A156058, A156128, A156266, A156270, A156273.
%Y Column k=10 of A290605.
%K nonn,easy
%O 0,2
%A _Philippe Deléham_, Feb 07 2009
%E a(15) corrected by _Vincenzo Librandi_, Jul 19 2011
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