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A277834
Number of '4' 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, ...).
2
0, 0, 1, 22, 344, 4671, 59053, 713985, 8374417, 96089849, 1084355281, 12078120713, 133126886145, 1454725651577, 15781824417015, 170163923182508, 1825096021948551, 19485528120720094, 207200960219546637, 2195466392318923180, 23189231824423799723
OFFSET
0,4
LINKS
FORMULA
a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 4,
a(n) = A277833(n) - 5*10^(n-4) for n >= 4, a(n) = A277835(n) + 6*10^(n-5) for n >= 5.
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 '4' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '4' in { 14, 24, 34, 40, ..., 49, 54, ..., 114 }, where 44 accounts for two '4's.
PROG
(PARI) print1(c=N=0); for(n=1, 8, print1(", "c+=sum(k=N+1, N=N*10+n, #select(d->d==4, digits(k)))))
(PARI) A277834(n, m=4)=if(n>m, A277833(n, m+1)+(m+2)*10^(n-m-1), A277830(n)-(m>n))
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Nov 01 2016
EXTENSIONS
More terms from Lars Blomberg, Nov 05 2016
Removed incorrect b-file. - David A. Corneth, Dec 31 2020
STATUS
approved