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%I #35 Apr 10 2016 23:03:03
%S 1,101,10201,1030301,104060401,10510100501,1061520150601,
%T 107213535210701,10828567056280801,1093685272684360901,
%U 110462212541120451001,11156683466653165551101
%N a(n) = 101^n.
%C A185817(n) = smallest m such that in decimal representation n is a prefix of a(m).
%C a(n) gives the n-th row of Pascals' triangle (A007318) as long as all the binomial coefficients have at most two digits, otherwise the binomial coefficients with more than two digits overlap. - _Daniel Forgues_, Aug 12 2012
%C From _Peter M. Chema_, Apr 10 2016: (Start)
%C One percent growth applied n times increases a value by factor of a(n)/10^(2n), since 1% increases using "1.01". Therefore (a(n)/10^(2n) - 1)*100 = the percentage increase of one percent growth applied n times.
%C For instance, 432 increasing by 1% three times gives 445.090032 (i.e., 432*1.01^3), which is 1.030301 ((a(3)/10^(2*3)) times 432 or a 3.0301% increase from the original 432 ((a(3)/10^(2*3)-1)*100 = 3.0301). (End)
%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 (101).
%F a(n) = Sum_{k=0..n} binomial(n, k)*10^(n-k).
%F a(n) = A096883(2n).
%F a(n) = 101^n. a(n) = Sum_{k=0..n,} binomial(n, k)*100^k. - _Paul Barry_, Aug 24 2004
%F G.f.: 1/(1-101*x). - _Philippe Deléham_, Nov 25 2008
%F E.g.f.: exp(101*x). - _Ilya Gutkovskiy_, Apr 10 2016
%t Table[101^n, {n, 0, 11}] \\ _Ilya Gutkovskiy_, Apr 10 2016
%o (PARI) a(n)=101^n \\ _Charles R Greathouse IV_, Oct 16 2015
%o (PARI) x='x+O('x^99); Vec(1/(1-101*x)) \\ _Altug Alkan_, Apr 10 2016
%Y Cf. A007318, A003590, A001020, A097659.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jul 14 2004
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