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%I #125 Oct 06 2023 11:08:26
%S 5,10,13,15,17,20,25,26,29,30,34,35,37,39,40,41,45,50,51,52,53,55,58,
%T 60,61,65,68,70,73,74,75,78,80,82,85,87,89,90,91,95,97,100,101,102,
%U 104,105,106,109,110,111,113,115,116,117,119,120,122,123,125,130,135,136,137,140
%N Hypotenuse numbers (squares are sums of 2 nonzero squares).
%C Multiples of Pythagorean primes A002144 or of primitive Pythagorean triangles' hypotenuses A008846. - _Lekraj Beedassy_, Nov 12 2003
%C 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
%C Circumradius R of the triangles such that the area, the sides and R are integers. - _Michel Lagneau_, Mar 03 2012
%C The 2 squares summing to a(n)^2 cannot be equal because sqrt(2) is not rational. - _Jean-Christophe Hervé_, Nov 10 2013
%C 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
%C 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
%C 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
%C 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
%C 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
%D Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.
%H Alois P. Heinz, <a href="/A009003/b009003.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from T. D. Noe)
%H R. Chapman, <a href="http://empslocal.ex.ac.uk/people/staff/rjchapma/courses/nt13/sumsquares.pdf">Pythagorean triples and sums of squares</a>
%H Steven R. Finch, <a href="http://www.people.fas.harvard.edu/~sfinch/constant/lr/lr.html">Landau-Ramanujan Constant</a> [Broken link]
%H Steven R. Finch, <a href="http://web.archive.org/web/20010605004309/http://www.mathsoft.com/asolve/constant/lr/lr.html">Landau-Ramanujan Constant</a> [From the Wayback machine]
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H J. Pahikkala, <a href="http://dx.doi.org/10.4171/EM/141">On contraharmonic mean and Pythagorean triples</a>, Elemente der Mathematik, 65:2 (2010), 62-67.
%H Patrick A. Thomas, <a href="/A009003/a009003_1.txt">Solutions of x^5=x up to base 100</a>
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>
%F A005089(a(n)) > 0. - _Reinhard Zumkeller_, Jan 07 2013
%F a(n) ~ n. - _Charles R Greathouse IV_, Jan 13 2022
%p isA009003 := proc(n)
%p local p;
%p for p in numtheory[factorset](n) do
%p if modp(p,4) = 1 then
%p return true;
%p end if;
%p end do:
%p false;
%p end proc:
%p for n from 1 to 200 do
%p if isA009003(n) then
%p printf("%d,",n) ;
%p end if;
%p end do: # _R. J. Mathar_, Nov 17 2014
%t 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 *)
%t Select[Range[200], Length[PowersRepresentations[#^2, 2, 2]] > 1 &] (* _Alonso del Arte_, Feb 11 2014 *)
%o (PARI) is_A009003(n)=setsearch(Set(factor(n)[,1]%4),1) \\ _M. F. Hasler_, May 27 2012
%o (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
%o (Haskell)
%o import Data.List (findIndices)
%o a009003 n = a009003_list !! (n-1)
%o a009003_list = map (+ 1) $ findIndices (> 0) a005089_list
%o -- _Reinhard Zumkeller_, Jan 07 2013
%o (Python)
%o from itertools import count, islice
%o from sympy import primefactors
%o def A009003_gen(): # generator of terms
%o return filter(lambda n:any(map(lambda p: p % 4 == 1,primefactors(n))),count(1))
%o A009003_list = list(islice(A009003_gen(),20)) # _Chai Wah Wu_, Jun 22 2022
%Y Cf. A009000, A024507, A004431, A072592, A004613, A187811.
%Y Complement of A004144. Primes in this sequence give A002144. Same as A146984 (integer contraharmonic means) as sets - see Pahikkala 2010, Theorem 5.
%Y Cf. A083025, A084645 (primitive elements), A084646, A084647, A084648, A084649, A006339.
%K nonn
%O 1,1
%A _David W. Wilson_
%E Definition edited by _Jean-Christophe Hervé_, Nov 10 2013
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