A277831 Number of '1' 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, 1, 5, 57, 689, 8121, 93553, 1058985, 11824417, 130589849, 1429355281, 15528120716, 167626886179, 1799725651922, 19231824420465, 204663923217008, 2170096022293551, 22935528124170094, 241700960254046637, 2540466392663923180, 26639231827873799724

Offset: 0

Author

M. F. Hasler, Nov 01 2016

Keywords

• nonn
• base

Examples

			For n=2 are counted the same '1' as for n=1, plus the 4 additional digits '1' in 10, 11 and 12.
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..998
• M. F. Hasler, "Digits d in 0 through 123...n", OEIS Wiki, Nov. 2016.

Crossrefs

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Programs

PARI

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==1,digits(k)))))

PARI

A277831(n)=if(n<2,n, n<11, A277832(n)+3*10^(n-2), error("n > 10 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

Formula

a(n) = A277832(n) + 3*10^(n-2), for 2 <= n <= 10.

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

Extensions

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020