A096884 a(n) = 101^n.

1, 101, 10201, 1030301, 104060401, 10510100501, 1061520150601, 107213535210701, 10828567056280801, 1093685272684360901, 110462212541120451001, 11156683466653165551101, 1126825030131969720661201, 113809328043328941786781301, 11494742132376223120464911401, 1160968955369998535166956051501

Offset

0,2

Comments

A185817(n) = smallest m such that in decimal representation n is a prefix of a(m).

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

From Peter M. Chema, Apr 10 2016: (Start)

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.

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)

Links

Table of n, a(n) for n=0..15.

Tanya Khovanova, Recursive Sequences

Index entries for linear recurrences with constant coefficients, signature (101).

Formula

a(n) = Sum_{k=0..n} binomial(n, k)*10^(n-k).

a(n) = A096883(2n).

a(n) = 101^n. a(n) = Sum_{k=0..n,} binomial(n, k)*100^k. - Paul Barry, Aug 24 2004

G.f.: 1/(1-101*x). - Philippe Deléham, Nov 25 2008

E.g.f.: exp(101*x). - Ilya Gutkovskiy, Apr 10 2016

Mathematica

Table[101^n, {n, 0, 11}] \\ Ilya Gutkovskiy, Apr 10 2016

Prog

(PARI) a(n)=101^n \\ Charles R Greathouse IV, Oct 16 2015
(PARI) x='x+O('x^99); Vec(1/(1-101*x)) \\ Altug Alkan, Apr 10 2016

Crossrefs

Cf. A007318, A003590, A001020, A097659.

Sequence in context: A135375 A267622 A172162 * A055474 A071783 A082808

Adjacent sequences: A096881 A096882 A096883 * A096885 A096886 A096887

Keyword

easy,nonn

Author

Paul Barry, Jul 14 2004

Status

approved