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%I #18 Jul 31 2020 11:49:10
%S 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,
%T 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,
%U 7,17,7,7,11,9,7,17,7,7,11,9,7,17,7,7,11,9,7,17,7,7
%N a(n) is the smallest integer k (k>=2) such that 13...3 (1 followed by n 3's) mod k is even.
%C More sequences can be generated by replacing the digit 1 by any integers of the form 3x+1. However, the sequence won't be as interesting if the following digits (the 3's) are replaced by any other digits.
%F I have proved the following properties:
%F For n=12x+1, a(n)=7.
%F For n=12x+2, a(n)=7.
%F For n=12x+3, a(n)=11.
%F For n=12x+4, a(n)=9.
%F For n=12x+5, a(n)=7.
%F For n=12x+6, a(n)=17.
%F For n=12x+7, a(n)=7.
%F For n=12x+8, a(n)=7.
%F For n=12x+9, a(n)=11.
%F For n=12x+10, a(n)=9.
%F For n=12x+11, a(n)=7.
%F For n=12x, a(n) can be 17, 19, 23 or 25.
%e a(5)=7 because
%e 133333 mod 2 = 1
%e 133333 mod 3 = 1
%e 133333 mod 4 = 1
%e 133333 mod 5 = 3
%e 133333 mod 6 = 1
%e 133333 mod 7 = 4, which is the first time the result is even.
%o (Python)
%o n=1
%o a=13
%o while n<=1000:
%o c=2
%o while True:
%o if (a%c)%2==1:
%o c=c+1
%o else:
%o print(c,end=", ")
%o break
%o n=n+1
%o a=10*a+3
%o (PARI) f(n) = (4*10^n-1)/3; \\ A097166
%o a(n) = my(k=2); while ((f(n) % k) % 2, k++); k; \\ _Michel Marcus_, Jul 05 2020
%Y Cf. A097166.
%K easy,nonn,base
%O 1,1
%A _Yuan-Hao Huang_, Jul 05 2020
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