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Smallest a(n) in Pythagorean triple (a, b, c) such that c(n) - b(n) = n.
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%I #23 May 17 2014 11:35:03

%S 3,4,9,8,15,12,21,12,15,20,33,24,39,28,45,24,51,24,57,40,63,44,69,36,

%T 35,52,45,56,87,60,93,40,99,68,105,48,111,76,117,60,123,84,129,88,75,

%U 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

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

%C 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.

%C 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

%H Charles R Greathouse IV, <a href="/A242219/b242219.txt">Table of n, a(n) for n = 1..10000</a>

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

%F 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

%e For n=7, a(7) = sqrt(2*7*h(12)-7^2) = 21;

%e for n=8, a(8) = sqrt(2*8*h(3)-8^2) = 12;

%e for n=9, a(9) = sqrt(2*9*h(5)-9^2) = 15;

%e for n=10, a(10) = sqrt(2*10*h(7)-10^2) = 20.

%t (* first do *) hypos = A009003; (* then *)

%t Table[ First[ Union[ Flatten[ Table[ Select[{Sqrt[2 hypos[[i]]*n - n^2]}, IntegerQ && hypos[[i]] > n], {i, 1, Length[hypos]}]]]], {n, 1, 200}]

%t (* view table *) ListLinePlot[%]

%o (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

%K nonn

%O 1,1

%A _V.J. Pohjola_, May 07 2014