|
|
A235334
|
|
Numbers n such that for any positive integers (a, b), if a * b = n then a + b is a square.
|
|
1
|
|
|
3, 323, 5183, 777923, 1327103, 6718463, 12446783, 16402499, 229159043, 432972863, 1214383103, 2191925123, 4787532863, 6927565823, 10809345023, 12619826243, 22218287363, 31123310723, 32399999999, 42469790723, 79101562499, 131734154303, 151291437443
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
It seems that n is the product of twin primes of A232878 for n > 3.
Conjecture: the numbers n such that for any positive integers (a, b), a * b = n and a + b is a square are the product of twin primes, and a*b+1 is also a perfect square.
|
|
LINKS
|
|
|
EXAMPLE
|
323 is the product of two positive integers in 2 ways: 1 * 323 and 17 * 19. The sums of the pairs of multiplicands are 323+1 = 18^2 and 17+19 = 6^2 respectively. All are squares.
|
|
MATHEMATICA
|
t={}; Do[ds=Divisors[n]; If[EvenQ[Length[ds]], ok=True; k=1; While[k<=Length[ds]/2 && (ok=IntegerQ[Sqrt[ds[[k]]+ds[[ -k]]]]), k++ ]; If[ok, AppendTo[t, n]]], {n, 2, 10^8}]; t ***[Program from T.D. Noe adapted for this sequence. See A080715]***
|
|
PROG
|
(PARI) isok(n) = {d = divisors(n); if (#d % 2, return (0)); for (i = 1, #d/2, if (! issquare(d[i]+n/d[i]), return (0)); ); return (1); } \\ Michel Marcus, Jan 06 2014
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|