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A347138
Numbers k such that (100^k + 1)/101 is prime.
0
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.
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)
KEYWORD
nonn,hard,more
AUTHOR
Paul Bourdelais, Aug 19 2021
STATUS
approved