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%I #29 Apr 09 2024 11:49:27
%S 1,1001,1002001,1003003001,1004006004001,1005010010005001,
%T 1006015020015006001,1007021035035021007001,1008028056070056028008001,
%U 1009036084126126084036009001,1010045120210252210120045010001
%N a(n) = 1001^n.
%C 694 is the smallest exponent e such that 1001^e begins with a digit greater than 1: A000030(a(694)) = 2, A000030(a(693)) = 1. - _Reinhard Zumkeller_, Nov 05 2010
%C a(n) gives the n-th row of Pascal's triangle (A007318) as long as all the binomial coefficients have at most three digits, otherwise the binomial coefficients with more than three digits overlap. - _Daniel Forgues_, Aug 12 2012
%D Rozsa Peter, Playing with Infinity, New York, Dover Publications, 1957.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (1001).
%F From _Philippe Deléham_, Nov 30 2008: (Start)
%F a(n) = 1001*a(n-1), n > 0; a(0)=1.
%F G.f.: 1/(1-1001*x). (End)
%o (PARI) 1001^n \\ _Charles R Greathouse IV_, Jan 30 2012
%Y Cf. A097660, A001020, A096884, A007318.
%K nonn,easy,changed
%O 0,2
%A _Reinhard Zumkeller_, Aug 18 2004
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