A096884 - OEIS

%I #40 Aug 29 2024 05:57:41

%S 1,101,10201,1030301,104060401,10510100501,1061520150601,

%T 107213535210701,10828567056280801,1093685272684360901,

%U 110462212541120451001,11156683466653165551101,1126825030131969720661201,113809328043328941786781301,11494742132376223120464911401,1160968955369998535166956051501

%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