A268281 Numbers n such that n-tau(n), phi(n) and n form a Heronian triangle, where tau=A000005 is the number of divisors and phi=A000010 the totient.

5, 34, 53, 90, 120, 440, 780, 1954, 120994, 140453, 28813276834

Offset: 1

Author

Frank M Jackson, Jan 29 2016

Keywords

• nonn
• more

Comments

For all n, n > tau(n) and n > phi(n) and if n is prime then n-tau(n) = n-2 and phi(n) = n-1. So n = 5 gives the triangle {3, 4, 5} which is a primitive Pythagorean triangle and this is the only one. Other Pythagorean triangles are {30, 16, 34} and {756, 192, 780}, the remainder are only Heronian.

It is not known if this sequence is infinite. Prime numbers in the sequence are 5, 53 and 140453 and generate triangles {3, 4, 5}, {51, 52, 53} and {140451, 140452, 140453}.

If n = 2p where p is prime then n-tau(n) = n-4 and phi(n) = n/2-1. So n = 34 gives the triangle {16, 30, 34}. Similar numbers in this sequence are a(8), a(9) and a(11). See A272365 for generating Heronian triangles with sides n, n-4, n/2-1.

a(12) > 2*10^12. - Giovanni Resta, Apr 14 2016

Next prime value of a(n) after 140453 is > 2*10^5719. See A003500 for generating Heronian triangles with consecutive sides. - Frank M Jackson, Apr 19 2016

A003500(n)+1 is a member of this sequence iff it is prime. Also A272365(n) is a member of this sequence iff A272365(n)/2 is prime. - Frank M Jackson, Apr 29 2016

Examples

			a(2) = 34 because the triangle so formed has sides 30, 16, 34. It is Heronian with integer area 240 and is also Pythagorean. It is the second Heronian triangle.
The triangle corresponding to a(11) has sides n = 28813276834, n-tau(n) = 28813276830, phi(n) = 14406638416, and area 200960614753814018640.
		

Links

• Eric Weisstein's World of Mathematics, Divisor Function
• Eric Weisstein's World of Mathematics, Totient Function
• Eric Weisstein's World of Mathematics, Heronian Triangle.

Crossrefs

Cf. A000005, A000010, A003500, A046022, A049820, A084820, A272365.

Programs

Mathematica

triples[n_] := ({a, b, c}={n-DivisorSigma[0, n], EulerPhi[n], n}; s=(a+b+c)/2; If[a+b>c&&IntegerQ[Sqrt[s(s-a)(s-b)(s-c)]], {a, b, c}, {}]); lst={}; Do[If[triples[n]!={}, AppendTo[lst, Last[triples[n]]]], {n, 1, 200000}]; lst

Extensions

a(11) from Giovanni Resta, Apr 14 2016