# Improving do loop speed with IntegerQ conditions

The following program runs rather slow when the upper bound for n gets large (~6 secs for 10^5).

triangleA[a_, b_, c_] := Sqrt[(a + b + c)/2 ((a + b + c)/2 - c)];
perimeterTot = 0.; Do[{If[IntegerQ@triangleA[n, n, n + 1] == True,
perimeterTot = perimeterTot + 3 n + 1],
If[IntegerQ@triangleA[n, n, n - 1] == True,
perimeterTot = perimeterTot + 3 n - 1]}, {n, 2, 333333333}]


I'm not sure what's causing this--the IntegerQ function or just the inefficiency of do loops in Mathematica? I'd appreciate any suggestion

• See here for a discussion of fast ways to check if an integer is a perfect square. I think this is the part that limits your speed. Jun 13 '19 at 21:29
• There is no need for If[IntegerQ[...] == True, ...]. Just If[IntegerQ[...], ...] will suffice. In addition, === (SameQ) is usually the appropriate check for equality in If statements, not == (Equals). Jun 14 '19 at 8:24

There is an analytic solution.

For the (n,n,n+1) triangle, the area is Sqrt[(n-1)(3n+1)]/2.

Area -> FullSimplify[triangleA[n, n, n + 1]]


The valid values of n are Sloane's A103974, the smaller sides in (n,n,n+1)-integer triangle with integer area.

Table[Simplify[((2 + Sqrt)^(2 k) + (2 - Sqrt)^(2 k) + 1)/3], {k,0,10}]


{1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985, 6575588101, 91586127425}

For the (n,n,n-1) triangle, the area is Sqrt[(n+1)(3n-1)]/2.

Area -> FullSimplify[triangleA[n, n, n - 1]]


The valid values of n are Sloane's A103772, the larger sides in (n,n,n-1)-integer triangle with integer area.

Table[
Round[(-1 + (7-4*Sqrt)^k*(2+Sqrt) - (-2+Sqrt)*(7+4*Sqrt)^k)/3],
{k,2,10}]


{17, 241, 3361, 46817, 652081, 9082321, 126500417, 1761923521, 24540428881}

For all valid values of n less than or equal to a maximum m, sum 3n+1 or 3n-1.

perimeterSum[m_] :=
Block[{area = 0, k, n},
k = 1;
While[(n = Simplify[((2 + Sqrt)^(2 k) + (2 - Sqrt)^(2 k) + 1)/3]) <= m,
k += 1;
area += (3 n + 1)];
k = 2;
While[(n = Round[(-1 + (7-4*Sqrt)^k*(2+Sqrt) - (-2+Sqrt)*(7+4*Sqrt)^k)/3]) <= m,
k += 1;
area += (3 n - 1)];
area
]


The solution is very fast.

AbsoluteTiming[perimeterSum[10^6]]


{0.001005, 2672274}

This is a partial answer only.

At the core you need a fast way of checking whether or not an integer is a perfect square. Here is a version that is moderately fast, to be optimized:

squareQ[n_] := MemberQ[{0, 1, 4, 9}, Mod[n, 16]] && IntegerQ[Sqrt[n]]


The triangle is simplified for the two specific cases:

triangleA[a_, b_, c_] = Sqrt[(a + b + c)/2 ((a + b + c)/2 - c)];
t1[n_] = triangleA[n, n, n + 1]^2 // FullSimplify;
t2[n_] = triangleA[n, n, n - 1]^2 // FullSimplify;


The search is now about 3 times faster:

perimeterTot = 0;
Do[If[squareQ[t1[n]], perimeterTot += 3 n + 1];
If[squareQ[t2[n]], perimeterTot += 3 n - 1], {n, 2, 10^5}]


To improve on this, look into superfast implementations of squareQ, inspired for example by this question and this Java thread. Probably compiling, or even hand-writing a C subroutine, would help.