A277849 Number of digits '9' 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, 8367627, 96021949, 1083676281, 12071330713, 133058986145, 1454046651577, 15775034417009, 170096023182441, 1824417021947873, 19478738120713305, 207133060219478737, 2194787392318244180, 23182441824417009723

Offset: 0

Author

M. F. Hasler, Nov 01 2016

Keywords

• nonn
• base

Examples

			For n = 2 there is only one digit '9' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '9' in { 19, 29, ..., 89, 90, ..., 99, 109, 119 }, where 99 accounts for two '9's.
		

Links

• Lars Blomberg, Table of n, a(n) for n = 0..997

Crossrefs

Cf. A277830 - A277838, 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==9,digits(k)))))

PARI

A014824(n)=(10^n-1)*(10/81)-n/9;
A102684(n)=my(pow,f,g,h);sum(j=1,#Str(n),pow=10^j;f=floor(n/pow);g=floor(n/pow+1/10);h=(4/5+g)*pow;g*(2*n+2-h)-f*(2*n+2-(1+f)*pow))/2;
A277849(n)=A102684(A014824(n));
vector(50,n,A277849(n-1)) \\ Lars Blomberg, Nov 11 2020

Formula

a(n) = A083449(n) = A277830(n) - 1 for 0 < n < 9.

a(n) = A277838(n) for n < 8, and a(8) = A277838(8) - 1.

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

Replaced incorrect b-file by Lars Blomberg, Nov 11 2020