A242219 Smallest a(n) in Pythagorean triple (a, b, c) such that c(n) - b(n) = n.

3, 4, 9, 8, 15, 12, 21, 12, 15, 20, 33, 24, 39, 28, 45, 24, 51, 24, 57, 40, 63, 44, 69, 36, 35, 52, 45, 56, 87, 60, 93, 40, 99, 68, 105, 48, 111, 76, 117, 60, 123, 84, 129, 88, 75, 92, 141, 72, 63, 60, 153, 104, 159, 72, 165, 84, 171, 116, 177, 120, 183, 124, 105, 80, 195, 132, 201, 136, 207, 140, 213, 84, 219, 148, 105, 152, 231, 156

Offset: 1

Author

V.J. Pohjola, May 07 2014

Keywords

• nonn

Comments

The local minima a(n) predominantly fluctuate, with an increasing amplitude, between the multiples of the leg lengths of the smallest primitive triple (3,4,5) and of its symmetric counterpart (4,3,5). When n grows, minima appear from higher primitive triples which further increase the amplitude.

We have a^2 = c^2 - b^2 = (c-b)(c+b) = n*(c+b). To find the least such square, use n=core(n)*f^2 with core = A007913, f = A000188(n), and look for the least c+b = c-b+2b = n+2b = core(n)*x^2 or x^2 = (n+2b)/core(n) = f^2 + 2b/core(n). The least such integer x is f+1 if core(n) is even, or else f+2. - M. F. Hasler, May 08 2014

Examples

			For n=7, a(7) = sqrt(2*7*h(12)-7^2) = 21;
for n=8, a(8) = sqrt(2*8*h(3)-8^2) = 12;
for n=9, a(9) = sqrt(2*9*h(5)-9^2) = 15;
for n=10, a(10) = sqrt(2*10*h(7)-10^2) = 20.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Programs

Mathematica

(* first do *) hypos = A009003; (* then *)
Table[ First[ Union[ Flatten[ Table[ Select[{Sqrt[2 hypos[[i]]*n - n^2]}, IntegerQ && hypos[[i]] > n], {i, 1, Length[hypos]}]]]], {n, 1, 200}]
(* view table *) ListLinePlot[%]

PARI

a(n)={ my( f=core(n,1)); sqrtint(( if( bittest( f[1],0), 4*f[2]+4, 2*f[2]+1)*f[1]+n )*n )} \\ M. F. Hasler, May 08 2014

Formula

a(n) = min(sqrt(2n*h(i)-n^2), where h(i)=A009003(i)>n.

a(n) = sqrt(( n + (x^2 - f^2)*core(n))*n ) where f = A000188(n), x = f+1 if core(n) = A007913(n) = n / f^2 is even, x = f+2 if core(n) is odd. - M. F. Hasler, May 08 2014