cp's OEIS Frontend

a(n)=permanent(m), where the n X n matrix m is defined by m(i,j) = 1 or 0, depending on whether i+j is prime or composite respectively. - T. D. Noe, Oct 16 2007

Conjecture: Each term of the sequence is a positive square.

See also A321597 and A321610 for similar conjectures.

This conjecture fails for n = 12. The values of a(12),...,a(20) were first computed by the user MTson on Mathoverflow. Ilya Bogdanov has confirmed that a(n) is indeed a square. See answers and comments to Question 315351 on Mathoverflow. - Zhi-Wei Sun, Nov 17 2018

Clearly, a(n) is also the permanent of the matrix of order n whose (i,j)-entry is 1 or 0 according as prime(i) + prime(j) + 1 is prime or not.

Conjecture: a(n) > 0 for all n > 0.

Note that there is no permutation f of {1,...,10} such that prime(k) + prime(f(k)) - 1 is prime for every k = 1,...,10.

Though the first 27 terms are positive, we have a(50) = 0 since all the numbers 50!*k + 1, with k = 1..50, are composite.

Conjecture: a(n) > 0 for all n > 0. Moreover, for each n > 0, there is an even permutation f of {1,...,n} with prime(k)*prime(f(k)) - 2 prime for all k = 1,...,n. Also, for any integer n > 2, there is an odd permutation f of {1,...,n} with prime(k)*prime(f(k)) - 2 prime for all k = 1,...,n.

If we let b(n) denote the number of even permutations f of {1,...,n} with prime(k)*prime(f(k)) - 2 prime for all k = 1,...,n, then (b(1),...,b(11)) = (1,1,1,1,3,6,1,1,33,125,226).

In 1973 J.-R. Chen proved that there are infinitely many primes p with p + 2 a product of at most two primes, such primes p are now called Chen primes.

Conjecture 1: a(n) > 0 for all n > 0.

Conjecture 2: For any positive integer n, there is a permutation f of {1,...,n} such that k*f(k) is practical for every k = 1,...,n.

P. Bradley proved in arXiv:1809.01012 that for any positive integer n there is a permutation f of {1,...,n} such that all the numbers k + f(k) (k = 1,...,n) are prime. Modifying his proof slightly we see that for each n = 1,2,3,... there is a permutation f of {1,...,n} such that k + f(k) is practical for every k = 1,...,n.

Clearly, a(n) is the permanent of the matrix of order n whose (i,j)-entry is 1 or 0 according as 3^i + 3^j - 1 is prime or not.

Although the first 30 terms are positive, we have a(154) = 0 since it is easy to verify that 3^154 + 3^k - 1 is composite for every k = 1..154.

From Robert Israel, Dec 08 2019: (Start)

In fact 3^154 + 3^k - 1 is composite for k = 1..383, so a(n)=0 for 154 <= n <= 383.

Conjecture: for each n >= 154 there is m <= n such that 3^m + 3^k - 1 is composite for k = 1..n. This implies that a(n) = 0 for such n.

Conjecture verified for 154 <= n <= 2635. (End)

If we let b(n) denote the number of even permutations g of {1,...,n} with 3^k + 3^(g(k)) - 1 prime for all k = 1,...,n, then the values of b(1),b(2),...,b(11) are 1, 1, 1, 6, 8, 17, 30, 144, 422, 353, 111 respectively.

In the linked 2017 paper (see Conjecture 3.26), the author conjectured that for any integer a > 1 there are infinitely many primes of the form a^(2k) + a^m - 1 with k and m positive integers.