A197040 Occurrences of edge-lengths of Euler bricks in every 100 consecutive integers.

3, 8, 9, 8, 9, 9, 6, 9, 10, 8, 7, 9, 6, 8, 7, 8, 11, 6, 7, 8, 9, 8, 7, 6, 8, 10, 6, 6, 6, 8, 8, 8, 8, 9, 6, 9, 7, 6, 7, 8, 8, 9, 7, 11, 7, 8, 5, 9, 8, 9, 9, 7, 6, 7, 9, 6, 7, 9, 7, 8, 10, 5, 9, 7, 7, 7, 7, 6, 9, 9, 6, 8, 7, 9, 8, 6, 9, 5, 9, 9, 8, 6, 6, 7, 7

Offset: 1

Author

Christopher Monckton of Brenchley, Oct 08 2011

Keywords

• nonn
• base

Comments

Distribution of edge-length occurrences for Euler bricks is remarkably near-uniform.

Examples

			For n=1 (i.e., the integers 1..100), there are only 3 possible edge-lengths for Euler bricks: 44, 85, 88.
		

References

• L. E. Dickson, History of the Theory of Numbers, vol. 2, Diophantine Analysis, Dover, New York, 2005.
• P. Halcke, Deliciae Mathematicae; oder, Mathematisches sinnen-confect., N. Sauer, Hamburg, Germany, 1719, page 265.

Links

• Robin Visser, Table of n, a(n) for n = 1..10000
• E. W. Weisstein, MathWorld: Euler brick

Crossrefs

cf. A195816, A196943, A031173, A031174, A031175. Edge lengths of Euler bricks are A195816; face diagonals are A196943.

Programs

Sage

def a(n):
    ans = set()
    for x in range(100*(n-1)+1, 100*n+1):
        divs = Integer(x^2).divisors()
        for d in divs:
            if (d <= x^2/d): continue
            if (d-x^2/d)%2==0:
                y = (d-x^2/d)/2
                for e in divs:
                    if (e <= x^2/e): continue
                    if (e-x^2/e)%2==0:
                        z = (e-x^2/e)/2
                        if (y^2+z^2).is_square(): ans.add(x)
    return len(ans)  # Robin Visser, Jan 02 2024