A009003 - OEIS
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
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.
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;
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 *)
PROG
(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
(Haskell)
import Data.List (findIndices)
a009003 n = a009003_list !! (n-1)
a009003_list = map (+ 1) $ findIndices (> 0) a005089_list
(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))