

A299068


Number of pairs of factors of n^2*(n^21) which differ by n.


2



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
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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^2n)/2,(n^2+n)/2), (n^2n,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*k2)=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^21))):
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 * A049826 A310014 A310015
Adjacent sequences: A299065 A299066 A299067 * A299069 A299070 A299071


KEYWORD

nonn


AUTHOR

John H Mason, Feb 01 2018


STATUS

approved



