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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, ...).
11

%I #17 Jan 01 2021 11:53:57

%S 0,0,1,22,343,4664,58985,713306,8367628,96021959,1083676380,

%T 12071331701,133058996022,1454046750343,15775035404664,

%U 170096033058985,1824417120713306,19478739108367627,207133070096021958,2194787491083676380,23182442812071331701

%N 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, ...).

%H David A. Corneth, <a href="/A277838/b277838.txt">Table of n, a(n) for n = 0..998</a>

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

%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 '8' in the sequence 0, 1, 2, ..., 12.

%e 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.

%o (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)))))

%o (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

%Y Cf. A277830 - A277837, A277849, 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 Removed incorrect b-file. - _David A. Corneth_, Dec 31 2020