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A285955 Numbers a(n) = T(b(n))*sqrt(T(b(n))+1), where T(b(n)) is the triangular number of b(n)= A000217(b(n)) and b(n)=A006451(n). Also a(n) = y solutions of the Bachet Mordell equation y^2=x^3+K, where x= T(b(n)) = A006454(n) and K = (T(b(n)))^2= A285985(n).

%I A285955 #15 Feb 16 2025 08:33:44
%S A285955 0,6,60,1320,12144,262080,2405970,51894744,476378760,10274921850,
%T A285955 94320640056,2034382775040,18675010652760,402797515372356,
%U A285955 3697557790357470,79751873665825680,732097767490332144,15790468188346521390,144951660405354891060,3126432949419110989944
%N A285955 Numbers a(n) = T(b(n))*sqrt(T(b(n))+1), where T(b(n)) is the triangular number of b(n)= A000217(b(n)) and b(n)=A006451(n). Also a(n) = y solutions of the Bachet Mordell equation y^2=x^3+K, where x= T(b(n)) = A006454(n) and K = (T(b(n)))^2= A285985(n).
%C A285955 Numbers a(n) which are the products of the triangular number T(b(n)) and the square root of this triangular number plus one, sqrt(T(b(n))+1), where  b(n) is the sequence A006451(n) of numbers n such that T(n)+1 is a square.
%C A285955 This sequence a(n) gives also the y solutions of the 3rd degree Diophantine Bachet-Mordell equation y^2=x^3+K, with x= T(b(n)) = A006454(n) and K = (T(b(n)))^2 = A285985(n), where T(b(n)) is the triangular number of b(n)= A006451(n).
%C A285955 Also: A000217(A006451(n)) * sqrt(A000217(A006451(n))+1).
%D A285955 V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
%H A285955 Vladimir Pletser, <a href="/A285955/b285955.txt">Table of n, a(n) for n = 0..1000</a>
%H A285955 M.A. Bennett and A. Ghadermarzi, <a href="http://www.math.ubc.ca/~bennett/BeGa-data.htm">Data on Mordell's curve</a>.
%H A285955 Michael A. Bennett, Amir Ghadermarzi, <a href="https://arxiv.org/abs/1311.7077">Mordell's equation : a classical approach</a>, arXiv:1311.7077 [math.NT], 2013.
%H A285955 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/MordellCurve.html">Mordell Curve</a>
%F A285955 Since b(n) = 8*sqrt(T(b(n-2))+1)+ b(n-4) = 8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-1)=-1, b(0)=0, b(1)=2, b(2)=5 (see A006451) and a(n) = T(b(n))*sqrt(T(b(n))+1) (this sequence), one has :
%F A285955 a(n) = ([8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)]*[ 8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1]/2)* sqrt(([8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)]*[ 8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1]/2)+1).
%F A285955 Empirical g.f.: 6*x*(1 - x)*(1 + 11*x + 27*x^2 + 11*x^3 + x^4) / ((1 + 14*x - x^2)*(1 + 2*x - x^2)*(1 - 2*x - x^2)*(1 - 14*x - x^2)). - _Colin Barker_, Apr 30 2017
%e A285955 For n=2, b(n)=5, a(n)=60.
%e A285955 For n=5, b(n)=90, a(n)= 262080.
%e A285955 For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. This gives A000217(A006451(n)) * sqrt(A000217(A006451(n)) + 1) = 120 * sqrt(120 + 1) = 1320. - _David A. Corneth_, Apr 29 2017
%p A285955 restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print (‘0,0’,’1,6’,’2,60’); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=(b*(b+1)/2)* sqrt((b*(b+1)/2)+1); print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do:
%Y A285955 Cf. A285985, A006454, A000217, A006451, A081119, A054504.
%K A285955 nonn,easy
%O A285955 0,2
%A A285955 _Vladimir Pletser_, Apr 29 2017