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 A277838 Number of digits '8' 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, ...). 10
 0, 0, 1, 22, 343, 4664, 58985, 713306, 8367628, 96021959, 1083676380, 12071331701, 133058996022, 1454046750343, 15775035404664, 170096033058985, 1824417120713306, 19478739108367627, 207133070096021958, 2194787491083676380, 23182442812071331701 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 LINKS Lars Blomberg, Table of n, a(n) for n = 0..998 FORMULA a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 8, a(8) = A277849(8) + 1 = A277837(8) - 9. 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. EXAMPLE For n=2 there is only one digit '8' in the sequence 0, 1, 2, ..., 12. 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. PROG (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))))) (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 CROSSREFS Cf. A277830 - A277837, A277849, A277635, A272525, A083449, A014824. Sequence in context: A083449 A272525 A277849 * A277635 A277837 A277836 Adjacent sequences:  A277835 A277836 A277837 * A277839 A277840 A277841 KEYWORD nonn,base AUTHOR M. F. Hasler, Nov 01 2016 EXTENSIONS More terms from Lars Blomberg, Nov 05 2016 STATUS approved

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Last modified September 19 15:21 EDT 2019. Contains 327198 sequences. (Running on oeis4.)