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A277849
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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, ...).
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12
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0, 0, 1, 22, 343, 4664, 58985, 713306, 8367627, 96021949, 1083676281, 12071330713, 133058986145, 1454046651577, 15775034417009, 170096023182441, 1824417021947873, 19478738120713305, 207133060219478737, 2194787392318244180, 23182441824417009723
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OFFSET
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0,4
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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PROG
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(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) A014824(n)=(10^n-1)*(10/81)-n/9;
A102684(n)=my(pow, f, g, h); sum(j=1, #Str(n), pow=10^j; f=floor(n/pow); g=floor(n/pow+1/10); h=(4/5+g)*pow; g*(2*n+2-h)-f*(2*n+2-(1+f)*pow))/2;
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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