A277833 Number of '3' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).
0, 0, 1, 23, 349, 4721, 59553, 718985, 8424417, 96589849, 1089355281, 12128120713, 133626886145, 1459725651582, 15831824417065, 170663923183008, 1830096021953551, 19535528120770094, 207700960220046637, 2200466392323923180, 23239231824473799723
Offset: 0
Examples
For n=2 there is only one digit '3' in the sequence 0, 1, 2, *3*, 4, ..., 12.
For n=3 there are 12 + 10 = 22 more digits '3' in { 13, 23, 30, ..., 39, 43, 53, ..., 123 }, where 33 accounts for two '3's.
Links
- David A. Corneth, Table of n, a(n) for n = 0..998
Programs
-
PARI
print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==3,digits(k))))) -
PARI
A277833(n,m=3)=if(n>12, error("not yet implemented"), n>m, A277833(n,m+1)+(m+2)*10^(n-m-1), (9*n-11)*(10^n+1)\729+2-(m>n)) \\ M. F. Hasler, Nov 02 2016, edited Dec 29 2020
Formula
Extensions
More terms from Lars Blomberg, Nov 05 2016
Removed incorrect b-file. - David A. Corneth, Dec 31 2020