login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A350588
a(n) is the number of distinct numbers of steps required for the last n digits of integers to repeat themselves by iterating the map m -> m^5.
1
1, 2, 3, 4, 6, 9, 14, 23, 33, 45, 59, 75, 93, 113, 135, 159, 184, 211, 240, 271, 304, 339, 376, 415, 456, 499, 544, 591, 640, 691, 744, 799, 855, 913, 973, 1035, 1099, 1165, 1233, 1303, 1375, 1449, 1525, 1603, 1683, 1765, 1849, 1935, 2023, 2113, 2205, 2299
OFFSET
1,2
FORMULA
For n >= 9, a(n) = a(n-1) + 2*n - 4 - ceiling(log_2 (n)) or a(n) = n^2 - 3*n - 17 - Sum_{i=9..n} ceiling(log_2 (i)).
EXAMPLE
a(1) = 1. It takes one step to repeat the last digit by iterating the map on an integer. For example, 2^5 = 32 and 9^5 = 59049. Thus, the distinct number of steps for n = 1 is {1} and a(1) = 1.
a(2) = 2. It takes 1 or 2 steps for an integer to repeat its last two digits. For example, 24 -> 7962624; 27 -> 14348907 -> 608266787713357709119683992618861307. Thus, a(2) = 2: {1, 2}.
a(3) = 3: {1..3}.
a(4) = 4: {1..4}.
a(5) = 6: {1..6}.
a(6) = 9: {1..9}.
a(7) = 14: {1..14}.
a(8) = 23: {1..23}.
a(9) = 33: {1..24, 32..40}.
a(10) = 45: {1..25, 32..41, 64..73}.
a(11) = 59: {1..26, 32..42, 64..74, 128..138}.
PROG
(Python)
from math import log, ceil
def A350588(n):
if n <= 8:
b, S = 10**n, set()
for i in range(b):
t, s, T = i, 0, set()
while t not in T: T.add(t); t = (t**5)%b; s += 1
S.add(s)
return(len(S))
else: return n*n - 3*n - 17 - sum(ceil(log(i, 2)) for i in range(9, n+1))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Ya-Ping Lu, Jan 07 2022
STATUS
approved