%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

%O 0,2

%A _Reinhard Zumkeller_, Aug 18 2004