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

1,1

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

Table of n, a(n) for n=1..80.

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