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%I #19 Jan 01 2021 11:52:39
%S 0,0,1,22,343,4665,58993,713385,8368417,96029849,1083755281,
%T 12072120713,133066886145,1454125651577,15775824417009,
%U 170103923182448,1824496021947951,19479528120714094,207140960219486637,2194866392318323180,23183231824417799723
%N Number of '5' 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="/A277835/b277835.txt">Table of n, a(n) for n = 0..998</a>
%F a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 5,
%F a(5) = A277836(5) + 1, a(n) = A277836(n) + 7*10^(n-6) for n >= 6.
%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 '5' in the sequence 0, 1, 2, ..., 12.
%e For n = 3 there are 11 + 10 = 21 more digits '5' in { 15, 25, ..., 45, 50, ..., 59, 65, ..., 115 }, where 55 accounts for two '5's.
%o (PARI) print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==5,digits(k)))))
%o (PARI) A277835(n,m=5)=if(n>m,A277835(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ _M. F. Hasler_, Nov 02 2016
%Y Cf. A277830 - A277838, 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