OFFSET
1,1
COMMENTS
Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives areas A*B/2.
By Theorem 2 of Mohanty and Mohanty, all these numbers are primitive Pythagorean. - T. D. Noe, Sep 24 2013
This sequence also gives Fibonacci's congruous numbers (without multiplicity, in increasing order) divided by 4. See A258150. - Wolfdieter Lang, Jun 14 2015
The same as A024406 with duplicates removed. All terms are multiples of 6, cf. A258151. - M. F. Hasler, Jan 20 2019
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000 (corrected by Giovanni Resta, Jan 21 2019)
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.
FORMULA
Positive integers of the form u*v*(u^2 - v^2) where 2uv and u^2 - v^2 are coprime or, alternatively, where u, v are coprime and one of them is even.
a(n) = 6*A258151(n). - M. F. Hasler, Jan 20 2019
EXAMPLE
6 is in the sequence because it is the area of the 3-4-5 triangle.
a(7) = 210 corresponds to the two primitive Pythagorean triangles (21, 20, 29) and (35, 12, 37). See A024406. - Wolfdieter Lang, Jun 14 2015
MATHEMATICA
nn = 22; (* nn must be even *) t = Union[Flatten[Table[If[GCD[u, v] == 1 && Mod[u, 2] + Mod[v, 2] == 1, u v (u^2 - v^2), 0], {u, nn}, {v, u - 1}]]]; Select[Rest[t], # < nn (nn^2 - 1) &] (* T. D. Noe, Sep 19 2013 *)
PROG
(PARI) select( {is_A024365(n)=my(N=1+#n=divisors(2*n)); for(i=1, N\2, gcd(n[i], n[N-i])==1 && issquare(n[i]^2+n[N-i]^2) && return(n[i]))}, [1..10^4]) \\ is_A024365 returns the smaller leg if n is a term, else 0. - M. F. Hasler, Jun 06 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Additional comments James R. Buddenhagen, Aug 10 2008 and from Max Alekseyev, Nov 12 2008
Edited by N. J. A. Sloane, Nov 20 2008 at the suggestion of R. J. Mathar
STATUS
approved