

A253804


a(n) gives the odd leg of the second of the two Pythagorean triangles with hypotenuse A080109(n) = A002144(n)^2. This is the larger of the two possible odd legs.


3



15, 119, 255, 609, 1295, 1519, 2385, 3479, 4015, 4879, 6305, 9999, 9919, 12319, 14385, 16999, 13345, 28545, 32039, 19199, 38415, 50609, 32239, 50369, 65535, 62839, 50279, 64911, 83505, 96719
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OFFSET

1,1


COMMENTS

The corresponding even legs are given in 4*A253805.
The legs of the other Pythagorean triangle with hypotenuse A080109(n) are given A253802(n) (odd) and A253803(n) (even).
Each fourth power of a prime of the form 1 (mod 4) (see A002144(n)^= A080175(n)) has exactly two representations as sum of two positive squares (Fermat). See the Dickson reference, (B) on p. 227.
This means that there are exactly two Pythagorean triangles (modulo leg exchange) for each hypotenuse A080109(n) = A002144(n)^2, n >= 1. See the Dickson reference, (A) on p. 227.
Concerning the primitivity question of these triangles see a comment on A253802.


REFERENCES

L. E. Dickson, History of the Theory of Numbers, Carnegie Institution, Publ. No. 256, Vol. II, Washington D.C., 1920, p. 227.


LINKS

Table of n, a(n) for n=1..30.


FORMULA

A080175(n) = A002144(n)^4 = a(n)^2 + (4*A253805(n))^2,
n >= 1, that is,
a(n) = sqrt(A080175(n)  (4*A253805(n))^2), n >= 1.


EXAMPLE

n = 7: A080175(7) = 7890481 = 53^4 = 2809^2; A002144(7)^4 = a(7)^2 + (4*A253805(7))^2 = 2385^2 + (4*371)^2.
The other Pythagorean triangle with hypotenuse 53^2 = 2809 has odd leg A253802(7) = 1241 and even leg 4*A253303(7) = 4*630 = 2520: 53^4 = 1241^2 + (4*630)^2.


CROSSREFS

Cf. A002144, A002972, A002973, A070079, A070151, A080109, A253303, A253802, A253805.
Sequence in context: A328725 A331211 A183475 * A161476 A162321 A161875
Adjacent sequences: A253801 A253802 A253803 * A253805 A253806 A253807


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Jan 16 2015


STATUS

approved



