A009003 Hypotenuse numbers (squares are sums of 2 nonzero squares).

5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 80, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120, 122, 123, 125, 130, 135, 136, 137, 140

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

Multiples of Pythagorean primes A002144 or of primitive Pythagorean triangles' hypotenuses A008846. - Lekraj Beedassy, Nov 12 2003

This is exactly the sequence of positive integers with at least one prime divisor of the form 4k + 1. Compare A072592. - John W. Layman, Mar 12 2008 and Franklin T. Adams-Watters, Apr 26 2009

Circumradius R of the triangles such that the area, the sides and R are integers. - Michel Lagneau, Mar 03 2012

The 2 squares summing to a(n)^2 cannot be equal because sqrt(2) is not rational. - Jean-Christophe Hervé, Nov 10 2013

Closed under multiplication. The primitive elements are those with exactly one prime divisor of the form 4k + 1 with multiplicity one, which are also those for which there exists a unique integer triangle = A084645. - Jean-Christophe Hervé, Nov 11 2013

a(n) are numbers whose square is the mean of two distinct nonzero squares. This creates 1-to-1 mapping between a Pythagorean triple and a "Mean" triple. If the Pythagorean triple is written, abnormally, as {j, k, h} where j^2 +(j+k)^2 = h^2, and h = a(n), then the corresponding "Mean" triple with the same h is {k, 2j, h} where (k^2 + (k+2j)^2)/2 = h^2. For example for h = 5, the Pythagorean triple is {3, 1, 5} and the Mean triple is {1, 6, 5}. - Richard R. Forberg, Mar 01 2015

Integral side lengths of rhombuses with integral diagonals p and q (therefore also with integral areas A because A = pq/2 is some multiple of 24). No such rhombuses are squares. - Rick L. Shepherd, Apr 09 2017

Conjecture: these are bases n in which exists an n-adic integer x satisfying x^5 = x, and 5 is the smallest k>1 such that x^k =x (so x^2, x^3 and x^4 are not x). Example: the 10-adic integer x = ...499879186432 (A120817) satisfies x^5 = x, and x^2, x^3, and x^4 are not x, so 10 is in this sequence. See also A120817, A210850 and A331548. - Patrick A. Thomas, Mar 01 2020

Didactic comment: When students solve a quadratic equation a*x^2 + b*x + c = 0 (a, b, c: integers) with the solution formula, they often make the mistake of calculating b^2 + 4*a*c instead of b^2 - 4*a*c (especially if a or c is negative). If the root then turns out to be an integer, they feel safe. This sequence lists the absolute values of b for which this error can happen. Reasoning: With p^2 = b^2 - 4*a*c and q^2 = b^2 + 4*a*c it follows by addition immediately that p^2 + q^2 = 2*b^2. If 4*a*c < 0, let p = x + y and q = x - y. If 4*a*c > 0, let p = x - y and q = x + y. In both cases follows that y^2 + x^2 = b^2. So every Pythagorean triple gives an absolute value of b for which this error can occur. Example: From (y, x, b) = (3, 4, 5) follows (q^2, b^2, p^2) = (1, 25, 49) or (p^2, b^2, q^2) = (1, 25, 49) with abs(4*a*c) = 24. - Felix Huber, Jul 22 2023

Conjecture: Numbers m such that the limit: Limit_{s->1} zeta(s)*Sum_{k=1..m} [k|m]*A008683(k)*(i^k)/(k^(s - 1)) exists, which is equivalent to numbers m such that abs(Sum_{k=1..m} [k|m]*A008683(k)*(i^k)) = 0. - Mats Granvik, Jul 06 2024

References

• Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
• R. Chapman, Pythagorean triples and sums of squares
• Steven R. Finch, Landau-Ramanujan Constant [Broken link]
• Steven R. Finch, Landau-Ramanujan Constant [From the Wayback machine]
• Ron Knott, Pythagorean Triples and Online Calculators
• J. Pahikkala, On contraharmonic mean and Pythagorean triples, Elemente der Mathematik, 65:2 (2010), 62-67.
• Patrick A. Thomas, Solutions of x^5=x up to base 100
• Index entries for sequences related to sums of squares

Crossrefs

Cf. A000404 (sums of 2 squares), A004431 (sums of 2 distinct squares), A009000 (hypotenuse numbers with repetition), A072592, A004613, A187811.

Complement of A004144. Primes in this sequence give A002144. Same as A146984 (integer contraharmonic means) as sets - see Pahikkala 2010, Theorem 5.

Cf. A083025, A084645 (primitive elements), A084646, A084647, A084648, A084649, A006339.

Programs

Haskell

import Data.List (findIndices)
a009003 n = a009003_list !! (n-1)
a009003_list = map (+ 1) $ findIndices (> 0) a005089_list
-- Reinhard Zumkeller, Jan 07 2013

Maple

isA009003 := proc(n)
    local p;
    for p in numtheory[factorset](n) do
        if modp(p,4) = 1 then
            return true;
        end if;
    end do:
    false;
end proc:
for n from 1 to 200 do
    if isA009003(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Nov 17 2014

Mathematica

f[n_] := Module[{k = 1}, While[(n - k^2)^(1/2) != IntegerPart[(n - k^2)^(1/2)], k++; If[2 * k^2 >= n, k = 0; Break[]]]; k]; A009003 = {}; Do[If[f[n^2] > 0, AppendTo[A009003, n]], {n, 3, 100}]; A009003 (* Vladimir Joseph Stephan Orlovsky, Jun 15 2009 *)
Select[Range[200], Length[PowersRepresentations[#^2, 2, 2]] > 1 &] (* Alonso del Arte, Feb 11 2014 *)

PARI

is_A009003(n)=setsearch(Set(factor(n)[,1]%4),1)  \\ M. F. Hasler, May 27 2012

PARI

list(lim)=my(v=List(),u=vectorsmall(lim\=1)); forprimestep(p=5,lim,4, forstep(n=p,lim,p, u[n]=1)); for(i=5,lim, if(u[i], listput(v,i))); u=0; Vec(v) \\ Charles R Greathouse IV, Jan 13 2022

Python

from itertools import count, islice
from sympy import primefactors
def A009003_gen(): # generator of terms
    return filter(lambda n:any(map(lambda p: p % 4 == 1,primefactors(n))),count(1))
A009003_list = list(islice(A009003_gen(),20)) # Chai Wah Wu, Jun 22 2022

Formula

A005089(a(n)) > 0. - Reinhard Zumkeller, Jan 07 2013

a(n) ~ n. - Charles R Greathouse IV, Jan 13 2022

a(n) = sqrt(n-th square in A000404), where A000404 lists the sums of two nonzero squares. - M. F. Hasler, Jun 20 2025

Extensions

Definition edited by Jean-Christophe Hervé, Nov 10 2013