A362468 Number of distinct n-digit suffixes generated by iteratively multiplying an integer by 4, where the initial integer is 1.

3, 11, 52, 252, 1253, 6253, 31254, 156254, 781255, 3906255, 19531256, 97656256, 488281257, 2441406257, 12207031258, 61035156258, 305175781259, 1525878906259, 7629394531260, 38146972656260, 190734863281261, 953674316406261, 4768371582031262, 23841857910156262

Offset: 1

Author

Gil Moses, Apr 21 2023

Keywords

• nonn
• base
• easy

Comments

This process produces a family of similar sequences when using different constant multipliers. See crossrefs below.

Examples

			For n = 2, we begin with 1, iteratively multiply by 4 and count the number of terms before the last 2 digits begin to repeat. We obtain 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, ... . The next term is 4194304, which repeats the last 2 digits 04. Thus, the number of distinct terms is a(2) = 11.
		

Links

• Wikipedia, Multiplicative order
• Index entries for linear recurrences with constant coefficients, signature (6,-4,-6,5).

Crossrefs

Period of powers mod 10^n: A020699 (4), A216099 (3), A216164 (7), A216156 (11).

Programs

PARI

a(n)=(n+1)\2*2*5^(n-1) \\ Charles R Greathouse IV, Apr 28 2023

Python

def a(n):
    s, x, M = set(), 1, 10**n
    while x not in s: s.add(x); x = (x<<2)%M
    return len(s), x
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, Apr 22 2023

Python

def A362468(n): return (n+1>>1)+(5**(n-1)<<1) # Chai Wah Wu, Apr 24 2023

Formula

a(n) = t + k, where t = A004526(n+1) and k = A020699(n), since 4^t == 4^(t+k) (mod 10^n). Here, t is the "transient" portion and k = ord_5^n(4), the multiplicative order of 4 modulo 5^n, is the period of the orbit. - Michael S. Branicky, Apr 22 2023

Extensions

a(13) and beyond from Michael S. Branicky, Apr 22 2023