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).

0, 6, 60, 1320, 12144, 262080, 2405970, 51894744, 476378760, 10274921850, 94320640056, 2034382775040, 18675010652760, 402797515372356, 3697557790357470, 79751873665825680, 732097767490332144, 15790468188346521390, 144951660405354891060, 3126432949419110989944

Offset: 0

Author

Vladimir Pletser, Apr 29 2017

Keywords

• nonn
• easy

Comments

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.

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).

Also: A000217(A006451(n)) * sqrt(A000217(A006451(n))+1).

Examples

			For n=2, b(n)=5, a(n)=60.
For n=5, b(n)=90, a(n)= 262080.
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
		

References

• V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.

Links

• Vladimir Pletser, Table of n, a(n) for n = 0..1000
• M.A. Bennett and A. Ghadermarzi, Data on Mordell's curve.
• Michael A. Bennett, Amir Ghadermarzi, Mordell's equation : a classical approach, arXiv:1311.7077 [math.NT], 2013.
• Eric Weisstein's World of Mathematics, Mordell Curve

Crossrefs

Cf. A285985, A006454, A000217, A006451, A081119, A054504.

Programs

Maple

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:

Formula

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 :

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).

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