OFFSET
1,1
COMMENTS
Conjecture: The sequence has infinitely many terms. Also, there are infinitely many Pythagorean triples (x,y,z) with x,y,z all practical and gcd(x,y,z) = 6.
It is easy to show that there are no Pythagorean triples (x,y,z) with x,y,z all practical and gcd(x,y,z) = 2.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..80
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. B. Nathanson (ed.), Combinatorial and Additive Number Theory II: CANT, New York, NY, USA, 2015 and 2016, Springer Proc. in Math. & Stat., Vol. 220, Springer, New York, 2017; arXiv:1211.1588 [math.NT], 2012-2017.
Li-Yuan Wang, Zhi-Wei Sun, On practical numbers of some special forms, arXiv:1809.01532 [math.NT], 2018.
EXAMPLE
a(1) = 20 since 20^2 = 12^2 + 16^2 with 12, 16, 20 all practical and gcd(12,16,20) = 4.
a(2) = 100 since 100^2 = 28^2 + 96^2 with 28, 96, 100 all practical and gcd(28,96,100) = 4.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
f[n_]:=f[n]=FactorInteger[n];
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2]);
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}];
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0);
n=0; Do[If[pr[4m]==False, Goto[aa]]; Do[If[SQ[(4m)^2-x^2]&&GCD[x, 4m, Sqrt[(4m)^2-x^2]]==4&&pr[x]&&pr[Sqrt[(4m)^2-x^2]], n=n+1; Print[n, " ", 4m]; Goto[aa]], {x, 1, Sqrt[8]m]}]; Label[aa], {m, 1, 9265}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 22 2017
STATUS
approved