cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A253802 a(n) gives the odd leg of one of the two Pythagorean triangles with hypotenuse A080109(n) = A002144(n)^2. This is the smaller of the two possible odd legs.

Original entry on oeis.org

7, 65, 161, 41, 1081, 369, 1241, 671, 721, 3471, 959, 9401, 4681, 1695, 3281, 7599, 10199, 24521, 3439, 18335, 37241, 45241, 24465, 29281, 64001, 18561, 31855, 27761, 76601, 7825
Offset: 1

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Author

Wolfdieter Lang, Jan 14 2015

Keywords

Comments

The corresponding even legs are given in 4*A253803.
The legs of the other Pythagorean triangle with hypotenuse A080109(n) are given A253804(n) (odd) and A253805(n) (even).
Each fourth power of a prime of the form 1 (mod 4) (see A002144(n)^2 = 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.
Note that the Pythagorean triangles are not always primitive. E.g., n = 2: (65, 4*39, 13^2) = 13*(5, 4*3, 13). For each prime congruent 1 (mod 4) (A002144) there is one and only one such non-primitive triangle with hypotenuse p^2 (just scale the unique primitive triangle with hypotenuse p with the factor p). Therefore, one of the two existing Pythagorean triangles with hypotenuse from A080109 is primitive and the other is imprimitive.

Examples

			n = 7: A080175(7) = 7890481 = 53^4 = 2809^2; A002144(7)^4  =  a(7)^2 + (4*A253803(7))^2 = 1241^2 + (4*630)^2.
The other Pythagorean triangle with hypotenuse 53^2 = 2809 has odd leg A253804(7) = 2385 and even leg 4*A253305(7) = 4*371 = 1484: 53^4 = 2385^2 + (4*371)^2.
		

References

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

Crossrefs

Formula

A080175(n) = A002144(n)^4 = a(n)^2 + (4*A253803(n))^2,
n >= 1, that is,
a(n) = sqrt(A080175(n) - (4*A253803(n))^2), n >= 1.

A253803 a(n) gives one fourth of the even leg of one of the two Pythagorean triangles with hypotenuse A080109(n) = A002144(n)^2. The odd leg is given in A253802(n).

Original entry on oeis.org

6, 39, 60, 210, 210, 410, 630, 915, 1320, 1780, 2340, 990, 2730, 3164, 4620, 5215, 5610, 4290, 8145, 8106, 2730, 6630, 12116, 12540, 4080, 17485, 17451, 18480, 9690, 24414
Offset: 1

Views

Author

Wolfdieter Lang, Jan 14 2015

Keywords

Comments

See A253802 for comments and the Dickson reference.

Examples

			n = 7: A080175(7) = 7890481 = 53^4 = 2809^2; A002144(7)^4 = A253802(7)^2 + (4*a(7))^2 = 1241^2 + (4*630)^2.
The other Pythagorean triangle with hypotenuse
53^2 = 2809 has odd leg A253804(7) = 2385 and even leg 4*A253305(7) = 4*371 = 1484: 53^4 = 2385^2 + (4*371)^2.
		

References

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

Crossrefs

Formula

a(n) = sqrt(A080109(n)^2 - A253802(n)^2)/4, n >= 1.

A253304 Numbers n such that the sum of the heptagonal numbers H(n) and H(n+1) is equal to the octagonal number O(m) for some m.

Original entry on oeis.org

1, 22, 77, 1376, 4785, 85302, 296605, 5287360, 18384737, 327731030, 1139557101, 20314036512, 70634155537, 1259142532726, 4378178086205, 78046522992512, 271376407189185, 4837625283003030, 16820959067643277, 299854721023195360, 1042628085786694001
Offset: 1

Views

Author

Colin Barker, Dec 30 2014

Keywords

Comments

Also positive integers x in the solutions to 5*x^2-3*y^2+2*x+2*y+1 = 0, the corresponding values of y being A253305.

Examples

			1 is in the sequence because H(1)+H(2) = 1+7 = 8 = O(2).
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1,62,-62,-1,1},{1,22,77,1376,4785},30] (* Harvey P. Dale, Nov 05 2024 *)
  • PARI
    Vec(x*(3*x^3+7*x^2-21*x-1)/((x-1)*(x^2-8*x+1)*(x^2+8*x+1)) + O(x^100))

Formula

a(n) = a(n-1)+62*a(n-2)-62*a(n-3)-a(n-4)+a(n-5).
G.f.: x*(3*x^3+7*x^2-21*x-1) / ((x-1)*(x^2-8*x+1)*(x^2+8*x+1)).
Showing 1-3 of 3 results.