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return m # Chai Wah Wu, Jan 23 2024
(Python)
from itertools import count
def A083116(n):
if not (n%10 and n%16 and n%25): return 0
for l in count(1):
k = (10**l-1)//9
for a in range(1, 10):
if not (m:=a*k)%n:
return m # Chai Wah Wu, Jan 23 2024
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1. If p is a prime > 5 then there exists a d such that a(p) = concatenation of '1' d times where p = k*d + 1 for some k. a(p)= (10^d -1)/9 < ={10^(p-1)- 1}/9. 2. a(n) = 0 if n = 10k, 16k or 25k.
2. a(n) = 0 if n = 10k, 16k or 25k.
Amarnath Murthy, " On the divisors of the Smarandache Unary sequence," Smarandache Notions Journal, Volume 11, 1-2-3, Spring 2000.
Amarnath Murthy and Meenakshi Srikanth (amarnath_murthymenakan_s(AT)yahoo.com), Apr 23 2003
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_Amarnath Murthy _ and Meenakshi Srikanth (amarnath_murthy(AT)yahoo.com), Apr 23 2003
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1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 11, 444, 111111, 222222, 555, 0, 1111111111111111, 666, 111111111111111111, 0, 777, 22, 1111111111111111111111, 888, 0, 222222, 999, 444444, 1111111111111111111111111111, 0, 111111111111111, 0, 33, 2222222222222222, 555555, 444444444
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