cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A346669 Numbers r such that the number of nonnegative m < r such that m^k == m (mod r) is equal to k*(the number of nonnegative m < r such that -m^k == m (mod r)), where k = 2^A007814(r-1) + 1.

Original entry on oeis.org

3, 5, 7, 11, 13, 15, 17, 19, 23, 27, 29, 31, 35, 37, 39, 41, 43, 47, 51, 53, 55, 57, 59, 61, 67, 71, 73, 75, 79, 83, 85, 87, 89, 91, 95, 97, 101, 103, 107, 109, 111, 113, 115, 119, 123, 125, 127, 131, 135, 137, 139, 143, 149, 151, 155, 157, 159, 163, 167, 173, 175, 179, 181, 183, 187
Offset: 1

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Author

Juri-Stepan Gerasimov, Jul 28 2021

Keywords

Comments

Conjecture: this sequence contains odd prime numbers A065091, but does not contain Carmichael numbers A002997.
Proof from Jianing Song, Jun 14 2021: (Start)
Conjecture: Let v(n) = A007814(n) be the 2-adic valuation of n. Define
A(n) = A182816(n) as the number of nonnegative m < n such that m^n == m (mod n),
B(n) = A333570(n) as the number of nonnegative m < n such that m^n == -m (mod n),
C(n) as the number of nonnegative m < n such that m^(2^v(n-1)+1) == m (mod n), and
D(n) as the number of nonnegative m < n such that m^(2^v(n-1)+1) == -m (mod n).
Then A(n)/B(n) = n and C(n)/D(n) = 2^v(n-1) + 1 if and only if n is an odd prime.
The conjecture is correct.
"<=": If n is an odd prime, then A(n) = n, B(n) = 1, C(n) = 2^v(n) + 1, D(n) = 1.
"=>": If A(n)/B(n) = n, since A(n) <= n, we must have A(n) = n, so n is either prime or a Carmichael number.
Let n = (p_1)*(p_2)*...*(p_k) be a Carmichael number. Write d = v(n-1), s = (n-1)/2^d be odd.
If m^(2^d+1) == -m (mod n), then m^(2^d) == -1 (mod n/gcd(m,n)). Since m^(s+1) == 1 (mod n), we have (-1)^s == 1 (mod n/gcd(m,n)), so n/gcd(m,n) = 1, m = 0. This shows that D(n) = 1.
For gcd(m,n) = Product_{i in S} p_i, m^(2^v(n-1)+1) == m (mod n) is equivalent to m == 0 (mod Product_{i in S} p_i), m^(2^d) == 1 (mod Product_{i not in S} p_i). The number of solutions modulo n is Product_{i not in S} gcd(2^d, p_i-1). Hence we have C(n) = Sum_{S subset of {1,2,...,k}} Product_{i not in S} gcd(2^d, p_i - 1) = Product_{i=1..k} (1 + gcd(2^d, p_i-1)).
If n is a Carmichael number such that C(n)/D(n) = 2^d + 1, then Product_{i=1..k} (1 + gcd(2^d, p_i-1)) = 2^d + 1. Hence gcd(2^d, p_i-1) < 2^d for 1 <= i <= k. By Zsigmondy's Theorem, unless d = 1 or 3, 2^d + 1 has a primitive prime factor p such that p does not divide 2^e + 1 for all e < d. So we must have d = 1 or 3 (otherwise p does not divide Product_{i=1..k} (1 + gcd(2^d, p_i-1))), then k = 1 or 2. But a Carmichael number must have at least 3 prime factors, a contradiction! (End)
In the case k = r, this sequence contains odd prime and Carmichael numbers, but does not contain any other numbers.

Crossrefs

Programs

  • Magma
    [r: r in [2..190] | #[m: m in [0..r-1] | m^k mod r eq m] eq #[m: m in [0..r-1] | -m^k mod r eq m]*k where k is 2^Valuation(r-1, 2) + 1];
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