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 A336019 a(n) is the smallest integer k (k>=2) such that 13...3 (1 followed by n 3's) mod k is even. 0
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. LINKS 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. EXAMPLE 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. PROG (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 (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 CROSSREFS Cf. A097166. Sequence in context: A064496 A084513 A084523 * A213886 A053673 A204910 Adjacent sequences:  A336016 A336017 A336018 * A336020 A336021 A336022 KEYWORD easy,nonn,base AUTHOR Yuan-Hao Huang, Jul 05 2020 STATUS approved

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Last modified May 10 22:16 EDT 2021. Contains 343780 sequences. (Running on oeis4.)