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Kalmynin conjectures that this sequence is infinite.

I have some doubts about terms 17497, 52489, 164617 that are == 1 (mod 8).

Numbers k such that A068494(k) = A002322(k).

If k is in the sequence, then k*gpf(k) is in the sequence.

Are there infinitely many terms of the form (p-1)*p, where p is a prime?

The terms are squarefree.

It seems that this sequence is {2, 6, 42} union {squarefree kernels of A345765} in natural order.

k is a term of A124240 if and only if Clausen(k, 0) divides Clausen(k, 1).

Composite numbers k such that lambda(k) divides k-1 and lambda(k-1) divides k-1, where lambda(m) = A002322(m).

These are composites k such that b^(k-1) == 1 (mod (k-1)*k) for every b coprime to (k-1)*k.

Composites k such that B^(b^(k-1)-1) == 1 (mod k^2) for every B coprime to k and for every b coprime to (k-1)*k.

There are exactly 78 terms in this sequence.

k is a term in this sequence iff k is divisible by A002322(k) (k is in A124240) and k is a divisor of 131040.

From Jianing Song, Nov 17 2018: (Start)

Proof. Let psi = A002322. For every x coprime to k we must have x^x == (x+k)^(x+k) == x^(x+k) (mod k), that is, x^k == 1 (mod k), so psi(k) must be divisible by k, and the condition is equivalent to x^x^(x+1) == x (mod k) for every x coprime to k.

The case k = 1, 2 is obvious. Now let k > 2, then k must be even.

Let g be a primitive lambda-root modulo k and psi(k), 0 < g < k, then g must be odd. By the condition, we have g^g^(g+1) == g (mod k), so g^(g+1) == 1 (mod psi(k)), g + 1 == 0 (mod psi(psi(k)). Also, (k-g)^(k-g)^(k-g+1) == k - g (mod k). Let d be the multiplicative order of k - g modulo k, then (k-g)^d == 1 (mod k) and (k-g)^(k-g+1) == 1 (mod d).

Case (i). d is even, then we have g^d == 1 (mod k) => d = psi(k). So (k-g)^(k-g+1) == 1 (mod psi(k)). Since k - g + 1 is an even number, we have g^(k-g+1) == 1 (mod psi(k)) => k - g + 1 == 0 (mod psi(psi(k))). psi(k) | k implies that psi(psi(k)) | psi(k) (and also divides k), so -g + 1 == 0 (mod psi(psi(k)) => psi(psi(k)) | 2.

Case (ii). d is odd, then g^d == -1 (mod k) => d = psi(k)/2. So (k-g)^(k-g+1) == 1 (mod psi(k)/2). Since k - g is an odd number, we get (k-g)^(k-g+1) == 1 (mod psi(k)) again. The same as above, psi(psi(k)) | 2.

So psi(k) | 24, k divides 131040.

Note (i). Now we show that there does exist such g being a primitive lambda-root modulo k and psi(k). Let k = 2^(e_0)*Product_{i=1..m} (p_i)^(e_i)*s, psi(k) = 2^(f_0)*Product_{i=1..m} (p_i)^(f_i)*t, where p_i are distinct odd primes, gcd(t, k) = gcd(s, psi(k)) = 1, then let g == 3 (mod 8), g be a primitive root modulo (p_i)^2 and a primitive lambda-root modulo s and t.

Note (ii). We show that k divides 131040 and psi(k) divides k suffice. Suppose k > 1, then gcd(x, k) = 1 implies that x is odd.

(a) k is not divisible by 3, by psi(k) | k we have psi(k) is also not divisible by 3, so k divides 2^5*5 = 160. x^(x+1) == 1 (mod 8) => x^x^(x+1) == x (mod 160).

(b) k is divisible by 3, then gcd(x, 6) = 1 => x^(x+1) == 1 (mod 24) => x^x^(x+1) == x (mod 131040). (End) [Revised by Jianing Song, Feb 04 2019]