A356187 Number of permutations f of {1,...,n} with f(1) = 1 such that those k*f(k) + 1 (k = 1..n) are n distinct primes.

1, 1, 0, 0, 0, 2, 2, 6, 4, 24, 6, 162, 330, 1428, 1632

Offset: 1

Author

Zhi-Wei Sun, Jul 28 2022

Keywords

• nonn
• more

Comments

Conjecture: a(n) > 0 except for n = 3,4,5. Also, for any integer n > 5, there is a permutation f of {1,...,n} with f(1) = 3 such that those k*f(k) - 1 (k = 1..n) are n distinct primes.

This is stronger than part (i) of the conjecture in A321597.

Examples

			a(7) = 2 since the only permutations f of {1,...,7} with f(1) = 1 such that {k*f(k) + 1: k = 1..7} is a set of 7 primes, are (f(1),...,f(7)) = (1,3,4,7,2,5,6) and (1,5,2,3,6,7,4). Note that 1*1 + 1 = 2, 2*3 + 1 = 7, 3*4 + 1 = 13, 4*7 + 1 = 29, 5*2 + 1 = 11, 6*5 + 1 = 31, 7*6+1 = 43 are distinct primes. Also, 1*1 + 1 = 2, 2*5 + 1 = 11, 3*2 + 1 = 7, 4*3 + 1 = 13, 5*6 + 1 = 31, 6*7 + 1 = 43, 7*4 + 1 = 29 are distinct primes.
a(10) > 0 since for (f(1),...,f(10)) = (1,3,4,7,8,5,6,9,2,10) the set {k*f(k) + 1: k = 1..10} is a set of 10 distinct primes.
		

Links

• Zhi-Wei Sun, On permutations of {1,...,n} and related topics, J. Algebraic Combin., 2021.

Crossrefs

Cf. A000040, A321597, A333083.

Programs

Mathematica

(* A program to find all the permutations f of {1,...,9} with f(1) = 1 such that U = {k*f(k)+1: k = 1..9} is a set of 9 distinct primes. *)
V[i_]:=V[i]=Part[Permutations[{2,3,4,5,6,7,8,9}],i]
m=0;Do[U={2};Do[p=j*V[i][[j-1]]+1;If[PrimeQ[p],U=Append[U,p]],{j,2,9}];
If[Length[Union[U]]==9,m=m+1;Print[m," ",V[i]," ",U]],{i,1,8!}]

Python

from itertools import permutations as perm
from itertools import islice
from sympy import isprime
from math import factorial as fact
import collections
def consume(iterator, n=None):
    "Advance the iterator n-steps ahead. If n is None, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)
for x in range(2,20):
    mult = range(1,x)
    count = 0
    q = perm(range(1,x))
    for y in q:
        keeppos = 0
        keepflag = False
        if y[0] != 1:#stop when the first digit is not 1
            break
        z = [mult[a] * y[a] + 1 for a in range(x-1)]
        for b in z[0:-2]:
            if not isprime(b):
                keeppos = z.index(b)
                keepflag = True
                break
        if keepflag:#skip ahead to advance the next non-prime term
            consume(q,fact(x-keeppos-2)-1)
        elif len(set(z)) == len(z) and all(isprime(b) for b in set(z)):#no duplicates and all primes
            count += 1
    print(x-1,count)
# David Consiglio, Jr., Aug 04 2022

Extensions

a(11)-a(13) from Jinyuan Wang, Jul 29 2022

a(14)-a(15) from David Consiglio, Jr., Aug 04 2022