A222314 - OEIS

%I #23 Aug 29 2015 08:06:36

%S 1,3,5,7,9,13,16,19,22,27,35,39,43,51,58,65,73,81,87,95,108,117,124,

%T 137,149,159,168,181,194,207,221,233,247,263,278,291,304,325,341,355,

%U 374,397,414,431,455,473,493,517,538,555,579,605,625,643,672,697,716,743,769,795,820,847,870,899,929,957,981,1011,1046,1077,1107,1139,1173

%N Number of characters when n-th row of Pascal's triangle is written in base 10 (including the n spaces).

%C Inspired by Fig. 1 of Cobeli and Zaharescu.

%H N. J. A. Sloane, <a href="/A222314/b222314.txt">Table of n, a(n) for n = 0..2000</a>

%H C. Cobeli and A. Zaharescu, <a href="http://ssmr.ro/bulletin/volumes/56-1/node9.html">Promenade around Pascal Triangle-Number Motives</a>, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104) No. 1, 2013, 73-98.

%H N. J. A. Sloane, <a href="/A222314/a222314_1.pdf">Triangle showing silhouette of first 30 rows of Pascal's triangle (after Cobeli and Zaharescu)</a>

%F Stirling's formula shows that a(n) ~ n^2/(2 log 10) = 0.217... n^2.

%e "1 6 15 20 15 6 1" contains 16 characters, so a(6) = 16.

%p lis:=[];

%p M:=100;

%p f1:=n->[seq(binomial(n,k),k=0..n)];

%p for n from 0 to M do

%p t1:=f1(n);

%p t2:=convert(t1,string);

%p t3:=length(t2)-2-n;

%p lis:=[op(lis),t3];

%p od:

%p [seq(lis[i],i=1..M)];

%o (PARI) a(n) = n + sum(k=0, n, #digits(binomial(n,k))); \\ _Michel Marcus_, Aug 29 2015

%Y Cf. A007318.

%K nonn,base

%O 0,2

%A _N. J. A. Sloane_, Feb 16 2013