Talk:Deceptive numbers

Since the repunit ends in 1 5 can not be a factor of n--Nigel Galloway (talk) 11:44, 12 February 2022 (UTC)

Since the repunit is odd, 2 cannot be a factor of n. And because the digit sum of the repunit is n - 1, 3 cannot be a factor of n (an integer is divisible by 3, if its digit sum is divisible by 3, but n - 1 and n cannot both be divisible by 3).
All non-primes below 7² are divisible by 2, 3, or 5. So it is sufficient to start the search at 49. --Querfeld (talk) 14:01, 24 September 2023 (UTC)

Regarding the repunit R_n-1, the equation R_n-1 * 9 = 10^n-1 - 1 is fulfilled. And if it is ensured, that n is not divisible by 3, then if R_n-1 * 9 is divisible by n, also R_n-1 is divisible by n. In that case, the check of R_n-1 mod n = 0 can be simplified to 10^n-1 mod n = 1, which can be calculated via modular exponentiation, so that big integers are not required for this task. --Querfeld (talk) 19:09, 26 September 2023 (UTC)

Yep, that certainly works! (actually nine entries were already doing just that last year) --Petelomax (talk) 23:52, 26 September 2023 (UTC)

Indeed (actually, the OCaml, Python, and C implementations have been written by me; translations followed), I just wanted to explain, how this works — to encourage implementations in languages, that don't provide native BigInt support (32-bit precision is sufficient for the first 50 deceptive numbers). And the principle has already been shown on the OEIS page (just that they mod by n * 9, instead of ruling out factor 3 in advance). But it is often faster to do the Fermat primality test before the full primality check. --Querfeld (talk) 10:59, 27 September 2023 (UTC)

With the wp:Fermat primality test being a^n-1 mod n = 1, where gcd(a, n) = 1, it becomes obvious, that the deceptive numbers are a subset of the Fermat pseudoprimes to base a = 10, with the multiples of 3 filtered out (10^x already rules out the multiples of 2 and 5). --Querfeld (talk) 19:09, 26 September 2023 (UTC)