A336019 a(n) is the smallest integer k (k>=2) such that 13...3 (1 followed by n 3's) mod k is even.
7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 23, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 23, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7, 11, 9, 7, 17, 7, 7
Offset: 1
Examples
a(5)=7 because 133333 mod 2 = 1 133333 mod 3 = 1 133333 mod 4 = 1 133333 mod 5 = 3 133333 mod 6 = 1 133333 mod 7 = 4, which is the first time the result is even.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..20004
Crossrefs
Cf. A097166.
Programs
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PARI
f(n) = (4*10^n-1)/3; \\ A097166 a(n) = my(k=2); while ((f(n) % k) % 2, k++); k; \\ Michel Marcus, Jul 05 2020
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Python
n=1 a=13 while n<=1000: c=2 while True: if (a%c)%2==1: c=c+1 else: print(c,end=", ") break n=n+1 a=10*a+3
Formula
I have proved the following properties:
For n=12x+1, a(n)=7.
For n=12x+2, a(n)=7.
For n=12x+3, a(n)=11.
For n=12x+4, a(n)=9.
For n=12x+5, a(n)=7.
For n=12x+6, a(n)=17.
For n=12x+7, a(n)=7.
For n=12x+8, a(n)=7.
For n=12x+9, a(n)=11.
For n=12x+10, a(n)=9.
For n=12x+11, a(n)=7.
For n=12x, a(n) can be 17, 19, 23 or 25.
Comments