A004724 Delete all 5's from the sequence of nonnegative integers.

0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 1, 16, 17, 18, 19, 20, 21, 22, 23, 24, 2, 26, 27, 28, 29, 30, 31, 32, 33, 34, 3, 36, 37, 38, 39, 40, 41, 42, 43, 44, 4, 46, 47, 48, 49, 0, 1, 2, 3, 4, 6, 7, 8, 9, 60, 61, 62, 63, 64, 6, 66, 67, 68, 69, 70, 71, 72, 73, 74, 7, 76

Offset: 0

Author

N. J. A. Sloane

Keywords

• base
• nonn

Comments

In contrast to the variant A004180 where a(n) = 0 when all the digits of n are 5's, here a number completely disappears in that case, so that subsequent indices are shifted and for n > 4, a(n) is not the result of deleting 5's from n: see formula. - M. F. Hasler, Jan 13 2020

Examples

			From  _M. F. Hasler_, Jan 13 2020: (Start)
After a(4) = 4 comes a(5) = 6, since the number 5 completely disappears.
a(48) = 49 is followed by 0, 1, 2, 3, 4 (i.e., 50, ..., 54 with the initial digit removed) and then a(54) = 6, because 55 disappears completely.
Illustration of the formula: as long as n < 5 (the first number that completely disappears) we have a(n) = A004180(n). Here n has 1 digit but n+1 does not exceed the (single repdigit) 5 (left hand side in the Iverson bracket), so m = 0. From n = 5 on, n+1 > 5, so m = 1.
Then, when n has L(n) = 2 digits, we still have n = 2 - 1 = 1 as long as n+2 <= 55 or n <= 53, but m = 3 for n > 55 - 2 = 53, i.e., from n = 54 on (where the term 55 has disappeared, see above).
Similarly, m = 3 for n > 555 - 3, i.e., from n >= 553 on, etc. (End)
		

Crossrefs

Cf. A004180 (delete digits 5 in n), A052413 (numbers with no digit 5).

Cf. A004719, A004720, A004721, A004722, A004723, A004725, A004726, A004727, A004728.

Programs

MATLAB

m=1; for u=0:76 v=dec2base(u, 10)-'0'; v = v(v~=5);  if length(v)>0; sol(m)=(str2num(strrep(num2str(v), ' ', ''))); m=m+1; end; end; sol % Marius A. Burtea, Jan 16 2020

PARI

apply( {A004724(n,L=logint(n+!n,10)+1)=A004180(n+L-(10^L\9*5-L>=n))}, [0..99])
A004724_upto(N)={[fromdigits(v)|v<-[[d|d<-digits(n+!n*50),d!=5]|n<-[0..N]],#v]} \\ M. F. Hasler, Jan 13 2020

Python

def A004724(n):
    l = len(str(n))
    m = 5*(10**l-1)//9
    k = n + l - int(n+l < m)
    return 4 if k == m else int(str(k).replace('5','')) # Chai Wah Wu, Apr 20 2021

Formula

a(n) = A004180(n + m) where m = L(n) - [ (10^L(n)-1)/9*5 >= n + L(n) ], L(n) = floor(log_10(max(n,1)) + 1), the number of digits of n, and [...] is the Iverson bracket (1 if true, 0 else). - M. F. Hasler, Jan 13 2020