A277838 Number of '8' 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, 22, 343, 4664, 58985, 713306, 8367628, 96021959, 1083676380, 12071331701, 133058996022, 1454046750343, 15775035404664, 170096033058985, 1824417120713306, 19478739108367627, 207133070096021958, 2194787491083676380, 23182442812071331701

Offset: 0

Author

M. F. Hasler, Nov 01 2016

Keywords

• nonn
• base

Examples

			For n=2 there is only one digit '8' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '8' in { 18, 28, ..., 78, 80, ..., 89, 98, 108, 118 }, where 88 accounts for two '8's.
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..998

Crossrefs

Cf. A277830 - A277837, 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==8,digits(k)))))

PARI

A277838(n,m=8)=if(n>m,A277838(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

Formula

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 8, a(8) = A277849(8) + 1 = A277837(8) - 9.

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