A210503 Numbers k that form a primitive Pythagorean triple with k' and sqrt(k^2 + k'^2), where k' is the arithmetic derivative of k.

15, 35, 143, 323, 899, 1763, 3599, 4641, 5183, 10403, 11663, 13585, 19043, 22499, 32399, 35581, 36863, 39203, 51983, 57599, 72899, 79523, 97343, 121103, 176399, 186623, 213443, 272483, 324899, 359999, 381923, 412163, 435599, 446641, 622081, 656099, 675683

Offset: 1

Author

Paolo P. Lava, Jan 25 2013

Keywords

• nonn

Comments

A037074 is a subsequence of this sequence.

If k is the product of a pair of twin primes we have k=p(p+2), k'=2(p+1) and sqrt(k^2+k'^2)=(p+1)^2+1=p(p+2)+2=k+2. These numbers are relatively prime and therefore they form a primitive Pythagorean triple.

Also in the sequence are the following numbers with four distinct prime factors:

4641 = 3*7*13*17 [form p(p+4)*q(q+4)],

13585 = 5*11*13*19 [form p(p+6)*q(q+6)],

35581 = 7*13*17*23 [form p(p+6)*q(q+6)],

446641 = 13*17*43*47 [form p(p+4)*q(q+4)],

622081 = 17*23*37*43 [form p(p+6)*q(q+6)],

700321 = 19*29*31*41 [form p(p+10)*q(q+10)],

From Ray Chandler, Jan 25 2017: (Start)

24887581 = 47*53*97*103 [form p(p+6)*q(q+6)],

43518577 = 59*67*101*109 [form p(p+8)*q(q+8)],

115539901 = 83*97*113*127 [form p(p+14)*q(q+14)],

158682817 = 89*101*127*139 [form p(p+12)*q(q+12)],

305162941 = 103*113*157*167 [form p(p+10)*q(q+10)],

1093514641 = 103*107*313*317 [form p(p+4)*q(q+4)],

1415940061 = 167*193*197*223 [form p(p+26)*q(q+26)].

And one term with six distinct prime factors:

650344079 = 7*11*37*53*59*73. (End)

Examples

			m=57599, m'=480, sqrt(57599^2 + 480^2) = 57601.
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..500 (terms 1..100 from Paolo P. Lava)

Crossrefs

Cf. A003415, A009003, A009004, A037074.

Programs

Maple

with(numtheory);
A210503:= proc(q)
local a,n,p;
for n from 1 to q do
  a:=n*add(op(2,p)/op(1,p),p=ifactors(n)[2]);
  if trunc(sqrt(n^2+a^2))=sqrt(n^2+a^2) and gcd(n,gcd(a,n^2+a^2))=1 then print(n); fi;
od; end:
A210503(100000);

Python

from math import sqrt
from sympy import factorint
from gmpy2 import mpz, is_square, gcd
A210503 = []
for n in range(2, 10**5):
    nd = sum([mpz(n*e/p) for p, e in factorint(n).items()])
    if is_square(nd**2+n**2) and gcd(gcd(n, nd), mpz(sqrt(nd**2+n**2))) == 1:
        A210503.append(n) # Chai Wah Wu, Aug 21 2014