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A277833 Number of '3' 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, ...).

%I A277833 #21 Jan 01 2021 11:51:41
%S A277833 0,0,1,23,349,4721,59553,718985,8424417,96589849,1089355281,
%T A277833 12128120713,133626886145,1459725651582,15831824417065,
%U A277833 170663923183008,1830096021953551,19535528120770094,207700960220046637,2200466392323923180,23239231824473799723
%N A277833 Number of '3' 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 A277833 David A. Corneth, <a href="/A277833/b277833.txt">Table of n, a(n) for n = 0..998</a>
%F A277833 a(n) = A277832(n) - 4*10^(n-3) [for n >= 3] = A277834(n) + 5*10^(n-4) [for n >= 4].
%F A277833 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 A277833 For n=2 there is only one digit '3' in the sequence 0, 1, 2, *3*, 4, ..., 12.
%e A277833 For n=3 there are 12 + 10 = 22 more digits '3' in { 13, 23, 30, ..., 39, 43, 53, ..., 123 }, where 33 accounts for two '3's.
%o A277833 (PARI) print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==3,digits(k)))))
%o A277833 (PARI) A277833(n,m=3)=if(n>12, error("not yet implemented"), n>m, A277833(n,m+1)+(m+2)*10^(n-m-1), (9*n-11)*(10^n+1)\729+2-(m>n)) \\ _M. F. Hasler_, Nov 02 2016, edited Dec 29 2020
%Y A277833 Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.
%K A277833 nonn,base
%O A277833 0,4
%A A277833 _M. F. Hasler_, Nov 01 2016
%E A277833 More terms from _Lars Blomberg_, Nov 05 2016
%E A277833 Removed incorrect b-file. - _David A. Corneth_, Dec 31 2020