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A083752
Minimal k > n such that (4k+3n)(4n+3k) is a square.
4
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
1,1
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
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]
FORMULA
(4a(n)+3n)(4n+3a(n)) is a square.
n < a(n) <= 393n. - Charles R Greathouse IV, Dec 13 2013
EXAMPLE
a(24)=157 because (4*157+3*24)(3*157+4*24)= 396900=630*630.
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 *)
PROG
(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
(Haskell)
a083752 n = head [k | k <- [n+1..], a010052 (12*(k+n)^2 + k*n) == 1]
-- Reinhard Zumkeller, Apr 06 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Jun 17 2003
EXTENSIONS
a(12) corrected by Charles R Greathouse IV, Dec 13 2013
STATUS
approved