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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, ...).
12

%I #16 Nov 17 2020 14:34:54

%S 0,0,1,22,343,4664,58985,713306,8367627,96021949,1083676281,

%T 12071330713,133058986145,1454046651577,15775034417009,

%U 170096023182441,1824417021947873,19478738120713305,207133060219478737,2194787392318244180,23182441824417009723

%N 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, ...).

%H Lars Blomberg, <a href="/A277849/b277849.txt">Table of n, a(n) for n = 0..997</a>

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

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

%F 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.

%e For n = 2 there is only one digit '9' in the sequence 0, 1, 2, ..., 12.

%e 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.

%o (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)))))

%o (PARI) A014824(n)=(10^n-1)*(10/81)-n/9;

%o 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;

%o A277849(n)=A102684(A014824(n));

%o vector(50,n,A277849(n-1)) \\ _Lars Blomberg_, Nov 11 2020

%Y Cf. A277830 - A277838, A277635, A272525, A083449, A014824.

%K nonn,base

%O 0,4

%A _M. F. Hasler_, Nov 01 2016

%E More terms from _Lars Blomberg_, Nov 05 2016

%E Replaced incorrect b-file by _Lars Blomberg_, Nov 11 2020