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A277849 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, ...). 10
0, 0, 1, 22, 343, 4664, 58985, 713306, 8367627, 96021949, 1083676281, 12071330713, 133058986145, 1454046651577, 15775034417009, 170096023182441, 1824417021947873, 19478738120713305, 207133060219478737, 2194787392318244180, 23182441824417009723 (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) = A083449(n) = A277830(n) - 1 for 0 < n < 9.

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

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 '9' in the sequence 0, 1, 2, ..., 12.

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.

PROG

(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)))))

(PARI) A277849(n)=(9*n-11)*(10^n+1)\729+1+(n>8) \\ Possibly incorrect for n>9. - M. F. Hasler, Nov 02 2016

CROSSREFS

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

Sequence in context: A231647 A083449 A272525 * A277838 A277635 A277837

Adjacent sequences:  A277846 A277847 A277848 * A277850 A277851 A277852

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 22 17:04 EDT 2019. Contains 327311 sequences. (Running on oeis4.)