login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A335506 Start with a(0) = 1; thereafter a(n) is obtained from 5*a(n-1) by removing all 7's. 5
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 (list; graph; refs; listen; history; text; internal format)
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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified October 26 14:28 EDT 2021. Contains 348267 sequences. (Running on oeis4.)