A277835 Number of '5' 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, ...).

0, 0, 1, 22, 343, 4665, 58993, 713385, 8368417, 96029849, 1083755281, 12072120713, 133066886145, 1454125651577, 15775824417009, 170103923182448, 1824496021947951, 19479528120714094, 207140960219486637, 2194866392318323180, 23183231824417799723

Offset: 0

Author

M. F. Hasler, Nov 01 2016

Keywords

• nonn
• base

Examples

			For n = 2 there is only one digit '5' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '5' in { 15, 25, ..., 45, 50, ..., 59, 65, ..., 115 }, where 55 accounts for two '5's.
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..998

Crossrefs

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Programs

PARI

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==5,digits(k)))))

PARI

A277835(n,m=5)=if(n>m,A277835(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

Formula

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 5,

a(5) = A277836(5) + 1, a(n) = A277836(n) + 7*10^(n-6) for n >= 6.

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.

Extensions

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020