A299068 Number of pairs of factors of n^2*(n^2-1) which differ by n.

3, 4, 8, 7, 11, 6, 10, 12, 11, 9, 9, 9, 13, 22, 12, 7, 7, 11, 21, 28, 9, 7, 17, 14, 13, 14, 13, 13, 11, 9, 10, 12, 17, 33, 28, 8, 7, 20, 19, 15, 9, 10, 21, 29, 10, 7, 14, 19, 18, 21, 11, 9, 16, 44, 46, 14, 7, 9, 15, 9, 9, 18, 40, 24, 18, 8, 9, 30, 18, 17, 11

Offset

2,1

Comments

The question arose when seeking triples of numbers for which the sum of the squares of any two is congruent to 1 modulo the third.

From Robert Israel, Feb 04 2018: (Start)

For n > 7, a(n)>= 7, as there are at least the following pairs:

(1,n+1), (n,2*n), (2*n,3*n), ((n^2-n)/2,(n^2+n)/2), (n^2-n,n^2), (n^2,n^2+n), and (3*n, 4*n) (if n is odd) or (n/2,3*n/2) (if n is even).

If k in A299159 is sufficiently large, then a(12*k-2)=7. Dickson's conjecture implies there are infinitely many such k, and thus infinitely many n with a(n)=7. (End)

Links

Robert Israel, Table of n, a(n) for n = 2..10000

Maple

a:= n-> (s-> add(`if`(i+n in s, 1, 0), i=s))(
        numtheory[divisors](n^2*(n^2-1))):
seq(a(n), n=2..100);  # Alois P. Heinz, Feb 01 2018

Mathematica

Array[With[{d = Divisors[# (# - 1)] &[#^2]}, Count[d + #, _?(MemberQ[d, #] &)]] &, 71, 2] (* Michael De Vlieger, Feb 01 2018 *)

Crossrefs

Cf. A299159.

Sequence in context: A255697 A019972 A064406 * A344867 A049826 A310014

Adjacent sequences: A299065 A299066 A299067 * A299069 A299070 A299071

Keyword

nonn

Author

John H Mason, Feb 01 2018

Status

approved