%I A029549
%S A029549 0,6,210,7140,242556,8239770,279909630,9508687656,323015470680,
%T A029549 10973017315470,372759573255306,12662852473364940,430164224521152660,
%U A029549 14612920781245825506,496409142337836914550
%N A029549 a(0) = 0, a(1) = 6, a(2) = 210; for n >= 0, a(n+3) = 35*a(n+2) - 35*a(n+1) 
               + a(n).
%C A029549 Triangular numbers that are twice other triangular numbers. - Don N. 
               Page
%C A029549 Triangular numbers that are also pronics. These will be shown to have 
               a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein 
               (ellerstein(AT)aol.com), Mar 09 2002
%C A029549 In other words, triangular numbers which are products of two consecutive 
               numbers. E.g. a(2)=210: 210 is a triangular number which is the product 
               of two consecutive numbers: 14*15.- Shyam Sunder Gupta (guptass(AT)rediffmail.com), 
               Oct 26 2002
%C A029549 Coefficients of the series giving the best rational approximations to 
               sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) 
               - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), 
               which constitute every second convergent of the continued fraction. 
               The corresponding continued fractions are [2;1,4,1], [2;1,4,1,4,1], 
               [2;1,4,1,4,1,4,1], [2;1,4,1,4,1,4,1,4,1] and so forth. - Gene Ward 
               Smith (genewardsmith(AT)gmail.com), Sep 30 2006
%H A029549 Shyam Sunder Gupta <a href="http://www.shyamsundergupta.com/triangle.htm">
               Fascinating Triangular Numbers</a>
%F A029549 G.f.: 6x/(1-35*x+35*x^2-x^3).
%F A029549 a(n) = -3/16 + (3/32+1/16*2^(1/2))*(17+12*2^(1/2))^n + (3/32-1/16*2^(1/
               2))*(17-12*2^(1/2))^n. - Gene Ward Smith (genewardsmith(AT)gmail.com), 
               Sep 30 2006
%F A029549 a(n) = Binomial(A001652(n), 2) = A000217(A001652(n)). - Mitch Harris, 
               Apr 19 2007, R. J. Mathar, Jun 26 2009
%t A029549 CoefficientList[Series[6/(1 - 35*x + 35*x^2 - x^3), {x, 0, 14}], x]
%t A029549 Table[ Floor[ N[ (Sqrt[ 2 ]+1)^(4n+2)/32 ]], {n, 0, 20} ],
%Y A029549 Cf. A123478, A123479, A123480, A123482.
%Y A029549 Cf. A075528 = triangular numbers that are half other triangular numbers.
%Y A029549 Equals A029546/6.
%Y A029549 a(n) = 6*A029546(n-1) = 2*A075528(n).
%Y A029549 Sequence in context: A084694 A065945 A076715 this_sequence A087639 A028350 
               A099788
%Y A029549 Adjacent sequences: A029546 A029547 A029548 this_sequence A029550 A029551 
               A029552
%K A029549 nonn
%O A029549 0,2
%A A029549 Don N. Page (don(AT)phys.ualberta.ca)
%E A029549 Additional comments from Christian G. Bower (bowerc(AT)usa.net), Sep 
               19 2002; Shyam Sunder Gupta (guptass(AT)rediffmail.com), Oct 26 2002; 
               T. D. Noe (noe(AT)sspectra.com), Nov 07 2006; and others.
%E A029549 Edited by N. J. A. Sloane (njas(AT)research.att.com), Apr 18 2007, following 
               suggestion from Andrew Plewe and Tanya Khovanova.