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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.

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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

Views

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

Cf. A002144, A002972, A002973, A070079, A070151, A080109, A253305, A253803, A253804.

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.

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.

Original entry on oeis.org

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
Offset: 1

Views

Author

Wolfdieter Lang, Jan 16 2015

Keywords

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.

Examples

			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.

References

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

Crossrefs

Cf. A002144, A002972, A002973, A070079, A070151, A080109, A253303, A253802, A253805.

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.

A267859 The p-defect p - N(p) of the elliptic curve y^2 = x^3 + x for primes p congruent to 1 modulo 4 (A002144).

Original entry on oeis.org

2, -6, 2, 10, 2, 10, -14, 10, -6, 10, 18, 2, -6, -14, -22, -14, -22, 26, 18, -14, 2, -30, 26, -30, 2, 26, 18, 10, 34, 26, -22, 18, 10, 34, -14, 34, -38, 2, -6, -30, 34, -14, 42, -38, 10, -22, 42, -38, 26, 2, -46, 10, 34, -38, 50, 26, 50, -46, 2, 10, -30, -54, 18, -38, 50, 34, -22, 10, 50, -54
Offset: 1

Views

Author

Wolfdieter Lang, Feb 06 2016

Keywords

Comments

See A002172 for a differently signed sequence.
The number N(p) of solutions modulo a prime p of the elliptic curve y^2 = x^3 + x (of discriminant -4) is given for all p in A095978.
The p-defect a_p = p - N(p) for prime 2 and primes congruent to 3 modulo 4 vanishes.
A002144(n) - (a(n)/2)^2 = (2*A002973(n))^2, n >= 1. See the formula for A095978 for primes 1 (mod 4).
This sequence gives also the non-vanishing p-defects of the elliptic curve y^2 = x^3 - 4*x. See a comment on A138515 with the Martin and Ono link for the modularity series for these two elliptic curves. - Wolfdieter Lang, May 26 2016

Examples

			n = 2: p = A002144(2) = 13 = A000040(6), m = 6, a(2) = 13 - A095978(6) = 13 - 19  = -6.
n = 2:  -6 = A138515((A002144(2) - 1)/4) =
A138515(3) = -6. - _Wolfdieter Lang_, May 26 2016

References

  • J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 398. In the 4th ed., 2014, p. 371.

Links

Crossrefs

Cf. A000040, A002144, A002172, A095978, A138515.

Programs

  • Mathematica
    terms = 100; A002144 = Select[Range[5, 20*terms, 4], PrimeQ]; A095978[n_] := Module[{p, xy, x}, p = Prime[n]; If[n==1 || Mod[p, 4]==3, Return[p]]; xy = {Re[#], Im[#]}& @ FactorInteger[p, GaussianIntegers -> True][[2, 1]]; x = SelectFirst[xy, OddQ]; If[Mod[x, 4]==1, p - 2*x, p + 2*x]]; a[n_] := (p = A002144[[n]]; m = PrimePi[p]; p - A095978[m]); Array[a, terms] (* Jean-François Alcover, Feb 26 2016, after Robert Israel (A095978) *)

Formula

a(n) = A002144(n) - A095978(m) with A002144(n) = A000040(m), n >= 1.
a(n) = A138515((A002144(n) - 1)/4), n >= 1. - Wolfdieter Lang, May 26 2016

A330487 Sum of those x(p)*y(p) with p <= n, where p is a prime congruent to 1 modulo 4, and p = x(p)^2 + y(p)^2 with 1 <= x(p) <= y(p).

Original entry on oeis.org

0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, 8, 8, 8, 8, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 22, 22, 22, 22, 22, 22, 22, 22, 28, 28, 28, 28, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 62, 62, 62, 62, 62, 62, 62, 62, 92, 92, 92, 92, 92, 92, 92, 92, 92, 92, 92, 92, 116, 116, 116, 116, 116, 116, 116, 116
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 15 2019

Keywords

Comments

Conjecture: Let b(n) be the sum of all primes p <= n with p == 1 (mod 4). Then a(n)/b(n) = 1/Pi + O(1/sqrt(n)).
A classical theorem of Euler (conjectured by Fermat) states that any prime p == 1 (mod 4) can be written uniquely as x^2 + y^2 with 1 <= x <= y.
For any prime p == 1 (mod 4), we obviously have a(p) > a(p-1). Also, b(n) >= 2*a(n) for all n > 0 since x^2 + y^2 >= 2*x*y.
Via computation we find that a(10^10) = 353452066546904620, b(10^10) = 1110397615780409147, and 3.14157907 < b(10^10)/a(10^10) < 3.14157908.

Examples

			a(5) = 2 since 5 is the first prime congruent to 1 mod 4 and 5 = 1^2 + 2^2 with 1*2 = 2.
a(13) = 8 since 13 = 2^2 + 3^2 is the second prime congruent to 1 mod 4 and 1*2 + 2*3 = 8.

Links

Crossrefs

Cf. A000796, A002144, A002972, A002973.

Programs

  • Mathematica
    tab={};Do[m=0;Do[If[PrimeQ[(2x+1)^2+(2y)^2],m=m+(2x+1)*(2y)],{x,0,(Sqrt[n]-1)/2},{y,1,Sqrt[n-(2x+1)^2]/2}];tab=Append[tab,m],{n,1,80}];Print[tab]
Previous Showing 11-14 of 14 results.

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