%I A046090
%S A046090 1,4,21,120,697,4060,23661,137904,803761,4684660,27304197,159140520,
%T A046090 927538921,5406093004,31509019101,183648021600,1070379110497,
%U A046090 6238626641380,36361380737781,211929657785304,1235216565974041
%N A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence 
               gives X+1 values.
%C A046090 Solution to a(a-1) = 2b(b-1) in natural numbers: a = a(n), b = b(n) = 
               A011900(n).
%C A046090 n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - Benoit 
               Cloitre (benoit7848c(AT)orange.fr), Apr 15 2003
%D A046090 A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 
               pp. 122-125, 1964.
%D A046090 T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, 
               Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 
               94-104.
%D A046090 L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 
               (2005), 205-213.
%H A046090 <a href="Sindx_Tu.html#2wis">Index entries for two-way infinite sequences</
               a>
%H A046090 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A046090 Ron Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/
               pythag.html">Pythagorean Triples and Online Calculators</a>
%H A046090 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               PythagoreanTriple.html">Link to a section of The World of Mathematics.</
               a>
%F A046090 a(n) = (1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n).
%F A046090 a(n) = 6*a(n-1)-a(n-2)-2, n >= 2, a(0) = 1, a(1) = 4. a(n) = (A(n+1)-3*A(n)+2)/
               4 with A(n) = A001653(n).
%F A046090 G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)). a(n) = partial sums of A001541(n). 
               - Barry Williams, May 03 2000
%F A046090 A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1) 
               - Charlie Marion (charliem(AT)bestweb.net), Jul 01 2003
%F A046090 a(n) = 1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/
               2})^n. - Antonio Olivares (olivares14031(AT)yahoo.com), Oct 13, 2003
%F A046090 Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then 
               for k>j, c(i)*(c(k) - c(j)) = a(k+i)+...+a(i+j+1) + a(k-i-1)+...+a(j-i) 
               + k - j. For n<0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) 
               + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) 
               + c(n)*c(n+2k) = 2*c(n+k)^2 - Charlie Marion (charliem(AT)bestweb.net), 
               Jul 01 2003
%F A046090 2*a(n)=2*A084159(n) + 1 + (-1)^(n+1)=2*A046729(n) + 1 - (-1)^(n+1). - 
               Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 16 2004
%o A046090 (PARI) a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)),x,3))/4
%Y A046090 Other 2 sides are A001652 and A001653.
%Y A046090 Cf. A011900, A001541. A001652(n)=-a(-1-n).
%Y A046090 Sequence in context: A024051 A020048 A093426 this_sequence A045721 A101810 
               A001888
%Y A046090 Adjacent sequences: A046087 A046088 A046089 this_sequence A046091 A046092 
               A046093
%K A046090 nonn,easy,nice
%O A046090 0,2
%A A046090 Eric Weisstein (eric(AT)weisstein.com)
%E A046090 Additional comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de)