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A286035
a(n) = 3*T(A285984(n)), where T(m) is the m-th triangular number A000217(m).
4
0, 18315, 210375, 17232775560, 197941645440, 16214284059063255, 186242898311223435, 15255987442587265956120, 175235570535035566127880, 14354328072739259079522561195, 164878797845087651200279041495, 13505958574968967401962031517525680, 155134131134672045268505114018663320
OFFSET
0,2
COMMENTS
This sequence a(n) gives the solutions x of the 3rd degree Diophantine Bachet-Mordell equation y^2=x^3+K, with y = T(b(n))*sqrt(27*T(b(n))+1) = A286036(n) and K = (T(b(n)))^2 = A286037(n), the square of the triangular number of b(n)= A285984(n).
REFERENCES
V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
LINKS
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
FORMULA
Since b(n) = 264*sqrt(27*T(b(n-2))+1)+ b(n-4) = 264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-2)=110, b(-1)=0, b(0)=0, b(1)=110 (see A285984) and a(n) = 3*T(b(n)) (this sequence), one has :
a(n) = 3*[264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4) ]*[ 264*sqrt(27*(b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1]/2.
Empirical g.f.: 495*x*(37 + 388*x + 37*x^2) / ((1 - x)*(1 - 970*x + x^2)*(1 + 970*x + x^2)). - Colin Barker, May 01 2017
EXAMPLE
For n = 2, b(n) = 374, a(n)= 210375.
For n = 3, b(n) = A285984(n) =107184. Therefore, a(n) = 3*T(b(n)) = 3*A000217(A285984(n)) = 3*A000217(107184) = 3*5744258520=17232775560.
MAPLE
restart: bm2:=110: bm1:=0: b0:=0: bp1:=110: print ('0, 0', '1, 18315'); for n from 2 to 1000 do b:= 264*sqrt(27* (b0^2+b0)/2+1)+bm2; a:=3*b*(b+1)/2; print(n, a); bm2:=bm1; bm1:=b0; b0:=bp1; bp1:=b; end do:
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, May 01 2017
STATUS
approved