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

0,2

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).

Links

Table of n, a(n) for n=0..28.

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).

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.

Example

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.

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 *)

Prog

(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)

Crossrefs

Cf. A335505.

Sequence in context: A206451 A291164 A216126 * A129066 A102169 A060391

Adjacent sequences: A335503 A335504 A335505 * A335507 A335508 A335509

Keyword

nonn,base

Author

Pontus von Brömssen, Jun 13 2020

Status

approved