OFFSET
1,5
COMMENTS
Apparently, for n>11, there seems always to be a pairing possible. Note that all primes have the 4k+1 form. By the 4k+1 theorem, such a prime has a unique representation as the sum of two squares.
LINKS
Bert Dobbelaere, Table of n, a(n) for n = 1..50
L. E. Greenfield and S. J. Greenfield, Some Problems of Combinatorial Number Theory Related to Bertrand's Postulate, J. Integer Sequences, 1998, #98.1.2.
FORMULA
a(n) = permanent(m), where the n X n matrix m is defined by m(i,j) = 1 or 0, depending on whether i^2 + (j+n)^2 is prime or composite, respectively. - T. D. Noe, Feb 10 2007
EXAMPLE
a(5) = 2 because there are two ways: (1,4,9,16,25) + (36,49,100,81,64) = (37,53,109,97,89) and (1,4,9,16,25) + (100,49,64,81,36) = (101,53,73,97,61).
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Nov 15 2002
EXTENSIONS
More terms from Bert Dobbelaere, Sep 08 2019
STATUS
approved