A024411 Short leg of more than one primitive Pythagorean triangle.

20, 28, 33, 36, 39, 44, 48, 51, 52, 57, 60, 65, 68, 69, 75, 76, 84, 85, 87, 88, 92, 93, 95, 96, 100, 104, 105, 108, 111, 115, 116, 119, 120, 123, 124, 129, 132, 133, 135, 136, 140, 141, 145, 147, 148, 152, 155, 156, 159, 160, 161, 164, 165, 168, 172, 175, 177, 180, 183, 184

Offset

1,1

Comments

Every term is composite. - Clark Kimberling, Feb 04 2024

Proof by contradiction: let p prime be the short leg. Then p^2 + b^2 = c^2 i.e., p^2 = (c - b) * (c + b). Then (c - b, c + b) in {(1, p^2), (p, p)}. If (c - b, c + b) = (p, p) then c = p and b = 0 which is impossible. Hence there is at most one solution for (c - b, c + b). A contradiction. - David A. Corneth, Feb 04 2024

Links

Ray Chandler, Table of n, a(n) for n = 1..10000

Ron Knott, Pythagorean Triples and Online Calculators

Mathematica

aa=1; s=""; For[a=1, a<=10^2, For[b=a+1, ((b+1)^2-b^2)<=a^2, c=(a^2+b^2)^0.5; If[c==Round[c]&&GCD[a, b]==1, If[a==aa, s=s<>ToString[a]<>", "]; If[a!=aa, aa=a, aa=1]]; b++ ]; a++ ]; s (* Vladimir Joseph Stephan Orlovsky, Apr 29 2008 *)

Prog

(PARI)
is(n) = {
	my(d = divisors(n^2), q = 0, b, c);
	for(i = 1, #d\2,
		if(!bitand(d[#d + 1 - i] - d[i], 1),
			c = (d[i] + d[#d + 1 - i])/2;
			b = d[#d + 1 - i] - c;
			if(gcd(n, b) == 1 && n < b,
				q++;
				if(q >= 2,
					return(1)
				)
			)
		)
	); 0
} \\ David A. Corneth, Feb 04 2024

Crossrefs

Cf. A020884, A024352.

Sequence in context: A059617 A122146 A032482 * A269986 A139805 A141823

Adjacent sequences: A024408 A024409 A024410 * A024412 A024413 A024414

Keyword

nonn

Author

David W. Wilson

Status

approved