A083752 Minimal k > n such that (4k+3n)(4n+3k) is a square.

393, 786, 1179, 109, 1965, 2358, 2751, 218, 3537, 3930, 4323, 327, 132, 5502, 5895, 436, 6681, 7074, 7467, 545, 8253, 8646, 9039, 157, 9825, 264, 10611, 763, 11397, 11790, 12183, 872, 481, 13362, 13755, 981, 184, 14934, 396, 1090, 16113, 16506, 16899, 1199

Offset: 1

Author

Zak Seidov, Jun 17 2003

Keywords

• nonn

Comments

A problem of elementary geometry lead to the search for squares of the form (4*a^2+3*b^2)(4*b^2+3*a^2). I could not find any such squares except when a=b. See link to ZS.

Letting j := 24k+25n in (4k+3n)(4n+3k)=x^2 yields the Pell-like equation j^2 - 48 x^2 = 49 n^2. The recurrence relationship for solutions to Pell equations implies that if k,x is a solution for n, then so is k1=18817k+19600n-5432x, x1=18817x-65184k-67900n. As a result, if there is a solution with 109/4n < k < 393n, then there is also one with n < k < 109/4n, so either n < a(n) <= 109/4n or a(n)=393n. - David Applegate, Jan 09 2014

Examples

			a(24)=157 because (4*157+3*24)(3*157+4*24)= 396900=630*630.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Zak Seidov, Two "triangles" in a right triangle [Cached copy, pdf file, with permission]

Crossrefs

Cf. A085018, A085019.

Cf. A010052, A235188.

Programs

Haskell

a083752 n = head [k | k <- [n+1..], a010052 (12*(k+n)^2 + k*n) == 1]
-- Reinhard Zumkeller, Apr 06 2015

Maple

a:= proc(n) local k; for k from n+1
      while not issqr((4*k+3*n)*(4*n+3*k)) do od; k
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Dec 13 2013

Mathematica

a[n_] := For[k = n + 1, True, k++, If[IntegerQ[Sqrt[(4k+3n)(4n+3k)]], Return[k]]]; Table[an = a[n]; Print[an]; an, {n, 1, 50}] (* Jean-François Alcover, Oct 31 2016 *)

PARI

a(n)=my(k=n+1); while(!issquare((4*k+3*n)*(4*n+3*k)), k++); k \\ Charles R Greathouse IV, Dec 13 2013

PARI

diff(v)=vector(#v-1,i,v[i+1]-v[i])
a(n)=my(v=select(k->issquare(12*Mod(k,n)^2),[0..n-1])); forstep(k=n+v[1], 393*n, diff(concat(v,n)), if(issquare((4*k+3*n)*(4*n+3*k)) && k>n, return(k))) \\ Charles R Greathouse IV, Dec 13 2013

PARI

a(n)=for(k=n+1, 109*n\4, if(issquare((4*k+3*n)*(4*n+3*k)), return(k))); 393*n \\ Charles R Greathouse IV, Jan 09 2014

Sage

def a(n):
    k = n + 1
    while not is_square((4*k+3*n)*(4*n+3*k)):
        k += 1
    return k
[a(n) for n in (1..44)] # Peter Luschny, Jun 25 2014

Formula

(4a(n)+3n)(4n+3a(n)) is a square.

n < a(n) <= 393n. - Charles R Greathouse IV, Dec 13 2013

Extensions

a(12) corrected by Charles R Greathouse IV, Dec 13 2013