A335506 Start with a(0) = 1; thereafter a(n) is obtained from 5*a(n-1) by removing all 7's.

1, 5, 25, 125, 625, 3125, 15625, 8125, 40625, 203125, 1015625, 508125, 2540625, 1203125, 6015625, 3008125, 15040625, 5203125, 26015625, 13008125, 65040625, 325203125, 1626015625, 813008125, 4065040625, 20325203125, 101626015625, 50813008125, 254065040625

Offset: 0

Author

Pontus von Brömssen, Jun 13 2020

Keywords

• nonn
• base

Comments

This sequence is a rare non-periodic case of the recurrence where x(0)=1 and x(n+1) is obtained from m*x(n) by removing all digits k and all trailing zeros in base b. In fact, except for (m, b, k) = (5, 10, 7) (this sequence), x is eventually periodic whenever m <= 5 and 2 <= b <= 32, or m <= 16 and 2 <= b <= 16. However, for negative b it seems that x is non-periodic more frequently, for example when (m, b, k) is (2, -5, 1) or (2, -8, 1).

Examples

			For n <= 6, no 7's occur in 5^n, so a(n) = 5^n. Deleting the 7 in 5*a(6) = 78125, we obtain a(7) = 8125.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,100000,0,0,0,-100000).

Crossrefs

Cf. A335505.

Programs

Mathematica

Remove7[n_] := FromDigits[Select[IntegerDigits[n], # != 7 &]]; a[0] = 1; a[n_] := a[n] = Remove7[5 * a[n - 1]]; Array[a, 29, 0] (* Amiram Eldar, Jun 20 2020 *)

Python

from sympy.ntheory.factor_ import digits
from functools import reduce
def drop(x,n,k):
  # Drop all digits k from x in base n.
  return reduce(lambda x,j:n*x+j if j!=k else x,digits(x, n)[1:],0)
def A335506(n):
  if n==0:
    return 1
  else:
    return drop(5*A335506(n-1),10,7)

Formula

a(n) = a(n-4) + 100000*(a(n-16) - a(n-20)) for n > 22.

Recurrence (for n > 2):

a(n+1) = 5*a(n) if n is 1, 3, 4, 5, 7, 8, 9, 11, 13, or 15 mod 16 (these are the cases where 5*a(n) doesn't contain any 7's);

a(n+1) = (a(n) + 625)/2 if n is 2, 6, 10, or 14 mod 16;

a(n+1) = 5*a(n) - 7*10^(5*n/16 + 2) if n is 0 mod 16;

a(n+1) = 5*a(n) - 115*10^(5*(n + 4)/16) if n is 12 mod 16.

Explicit expressions:

a(16*n) = (37*2^4*10^(5*n - 2) - 1)*5^5/123 for n >= 1;

a(16*n + 1) = (2^12*10^(5*n - 4) - 1)*5^6/123 for n >= 1;

a(16*n + 2) = (2^12*10^(5*n - 4) - 1)*5^7/123 for n >= 1;

a(16*n + 3) = (2^8*10^(5*n - 1) - 1)*5^4/123 for n >= 0;

a(16*n + 4) = (2^8*10^(5*n - 1) - 1)*5^5/123 for n >= 0;

a(16*n + 5) = (2^8*10^(5*n - 1) - 1)*5^6/123 for n >= 0;

a(16*n + 6) = (2^8*10^(5*n - 1) - 1)*5^7/123 for n >= 0;

a(16*n + 7) = (2^4*10^(5*n + 2) - 1)*5^4/123 for n >= 0;

a(16*n + 8) = (2^4*10^(5*n + 2) - 1)*5^5/123 for n >= 0;

a(16*n + 9) = (2^4*10^(5*n + 2) - 1)*5^6/123 for n >= 0;

a(16*n + 10) = (2^4*10^(5*n + 2) - 1)*5^7/123 for n >= 0;

a(16*n + 11) = (10^(5*n + 5) - 1)*5^4/123 for n >= 0;

a(16*n + 12) = (10^(5*n + 5) - 1)*5^5/123 for n >= 0;

a(16*n + 13) = (37*2^8*10^(5*n) - 1)*5^6/123 for n >= 0;

a(16*n + 14) = (37*2^8*10^(5*n) - 1)*5^7/123 for n >= 0;

a(16*n + 15) = (37*2^4*10^(5*n + 3) - 1)*5^4/123 for n >= 0.