A347138 Numbers k such that (100^k + 1)/101 is prime.

3, 293, 461, 11867, 90089

Offset

1,1

Comments

These are the repunit primes in base -100. It is unusual to represent numbers in a negative base, but it follows the same formulation as any base: numbers are represented as a sum of powers in that base, i.e., a0*1 + a1*b^1 + a2*b^2 + a3*b^3 ... Since the base is negative, the terms will be alternating positive/negative. For repunits the coefficients are all ones so the sum reduces to 1 + b + b^2 + b^3 + ... + b^(k-1) = (b^k-1)/(b-1). Since b is negative and k is an odd prime, the sum equals (|b|^k+1)/(|b|+1). For k=3, the sum is 9901, which is prime. As with all repunits, we only need to PRP test the prime exponents. The factors of repunits base -100 will be of the form p=2*k*m+1 where m must be even, which is common for (negative) bases that are squares.

Links

Table of n, a(n) for n=1..5.

Paul Bourdelais, A Generalized Repunit Conjecture

Example

3 is a term since (100^3 + 1)/101 = 9901 is a prime.

Mathematica

Do[ If[ PrimeQ[ (100^n + 1)/101], Print[n]], {n, 0, 18000}]

Prog

(PARI) is(n)=isprime((100^n+1)/101)

Crossrefs

Cf. A309533, A309532, A237052, A229145, A057191, A057182, A057175.

Sequence in context: A058455 A116532 A199644 * A124357 A212798 A157579

Adjacent sequences: A347135 A347136 A347137 * A347139 A347140 A347141

Keyword

nonn,hard,more

Author

Paul Bourdelais, Aug 19 2021

Status

approved