A006451 -id:A006451 - cp's OEIS frontend

A154146 Numbers k such that 16 plus the k-th triangular number is a perfect square.

0, 14, 17, 87, 104, 510, 609, 2975, 3552, 17342, 20705, 101079, 120680, 589134, 703377, 3433727, 4099584, 20013230, 23894129, 116645655, 139265192, 679860702, 811697025

Offset: 0

R. J. Mathar, Oct 18 2009

Numbers k such that x=2*k+1 satisfies the Pell-type equation x^2 = 8*y^2 - 127. - Robert Israel, Jul 18 2019

			0, 14, 17, and 87 are terms:
   0* (0+1)/2 + 16 =  4^2,
  14*(14+1)/2 + 16 = 11^2,
  17*(17+1)/2 + 16 = 13^2,
  87*(87+1)/2 + 16 = 62^2.

Robert Israel, Table of n, a(n) for n = 0..2608
F. T. Adams-Watters, SeqFan Discussion, Oct 2009
Robert Israel, Proof of conjectured recurrence
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000217, A000290, A006451.

f:= gfun:-rectoproc({a(n+4)-6*a(n+2)+a(n)=2, a(0)=0, a(1)=14, a(2)=17, a(3)=87}, a(n), remember):
map(f, [$0..40]); # Robert Israel, Jul 18 2019

Join[{0}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 16 &]] (* G. C. Greubel, Sep 03 2016 *)

{for (n=0, 10^9, if ( issquare(n*(n+1)\2 + 16), print1(n, ", ") ) );}

{k: 16+k*(k+1)/2 in A000290}.
Conjectures: (Start)
a(n) = +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x^2*(-14-3*x+14*x^2+x^3)/((x-1) * (x^2-2*x-1) * (x^2+2*x-1)).
G.f.: ( 2 + (8+23*x)/(x^2-2*x-1) + 1/(x-1) + (-7+6*x)/(x^2+2*x-1) )/2. (End)
Conjectures confirmed: see link. - Robert Israel, Jul 18 2019

A154147 Indices k such that 19 plus the k-th triangular number is a perfect square.

Original entry on oeis.org

3, 9, 30, 60, 179, 353, 1046, 2060, 6099, 12009, 35550, 69996, 207203, 407969, 1207670, 2377820, 7038819, 13858953, 41025246, 80775900, 239112659, 470796449, 1393650710, 2744002796, 8122791603, 15993220329, 47343098910, 93215319180, 275935801859

Offset: 1

R. J. Mathar, Oct 18 2009

			3*(3+1)/2+19 = 5^2. 9*(9+1)/2+19 = 8^2. 30*(30+1)/2+19 = 22^2. 60*(60+1)/2+19 = 43^2.

Colin Barker, Table of n, a(n) for n = 1..500
F. T. Adams-Watters, SeqFan Discussion, Oct 2009
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000217, A000290, A006451.

Join[{3, 9}, Select[Range[0, 10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 19 &]] (* or *) LinearRecurrence[{1,6,-6,-1,1}, {3,9,30,60,179}, 25] (* G. C. Greubel, Sep 03 2016 *)

{for (n=0, 10^9, if ( issquare(n*(n+1)\2 + 19), print1(n, ", ") ) ); }

{k: 19+k*(k+1)/2 in A000290}.
a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x*(3 +6*x +3*x^2 -6*x^3 -4*x^4)/((1-x) * (x^2-2*x-1) * (x^2+2*x-1)).

A154149 Indices k such that 22 plus the k-th triangular number is a perfect square.

Original entry on oeis.org

2, 12, 27, 77, 162, 452, 947, 2637, 5522, 15372, 32187, 89597, 187602, 522212, 1093427, 3043677, 6372962, 17739852, 37144347, 103395437, 216493122, 602632772, 1261814387, 3512401197, 7354393202, 20471774412, 42864544827, 119318245277, 249832875762

Offset: 1

R. J. Mathar, Oct 18 2009

			2*(2+1)/2+22 = 5^2. 12*(12+1)/2+22 = 10^2. 27*(27+1)/2+22 = 20^2. 77*(77+1)/2+22 = 55^2.

Colin Barker, Table of n, a(n) for n = 1..500
F. T. Adams-Watters, SeqFan Discussion, Oct 2009
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000217, A000290, A006451.

Join[{2, 12}, Select[Range[0, 10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 22 &]] (* or *) LinearRecurrence[{1,6,-6,-1,1}, {2,12,27,77,162}, 25] (* G. C. Greubel, Sep 03 2016 *)

Vec(x*(-2-10*x-3*x^2+10*x^3+3*x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^30)) \\ Colin Barker, Jul 11 2015

{k: 22+k*(k+1)/2 in A000290}
a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x*(-2-10*x-3*x^2+10*x^3+3*x^4)/((x-1) * (x^2-2*x-1) * (x^2+2*x-1)).
G.f.: ( 6 + (10+25*x)/(x^2-2*x-1) - 5/(x^2+2*x-1) + 1/(x-1) )/2.

Extended by D. S. McNeil, Dec 04 2010

A154151 Indices k such that 25 plus the k-th triangular number is a perfect square.

Original entry on oeis.org

0, 18, 21, 111, 128, 650, 749, 3791, 4368, 22098, 25461, 128799, 148400, 750698, 864941, 4375391, 5041248, 25501650, 29382549, 148634511, 171254048, 866305418, 998141741, 5049197999, 5817596400, 29428882578, 33907436661, 171524097471, 197627023568

Offset: 1

R. J. Mathar, Oct 18 2009

			0*(0+1)/2+25 = 5^2. 18*(18+1)/2+25 = 14^2. 21*(21+1)/2+25 = 16^2. 111*(111+1)/2+25 = 79^2.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
F. T. Adams-Watters, SeqFan Discussion, Oct 2009
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000217, A000290, A006451.

Join[{0}, Select[Range[0, 10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 25 &]] (* or *) LinearRecurrence[{1,6,-6,-1,1}, {0,18,21,111,128}, 25] (* G. C. Greubel_, Sep 03 2016 *)

for(n=1,10^10,if(issquare(25+n*(n+1)/2),print1(n,", ")))

{k: 25+k*(k+1)/2 in A000290}.
Conjectures: (Start)
a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x^2*(-18-3*x+18*x^2+x^3)/( (x-1) * (x^2+2*x-1) * (x^2-2*x-1)).
G.f.: ( 2 + 1/(x-1) + (10+29*x)/(x^2-2*x-1) + (-9+8*x)/(x^2+2*x-1) )/2. (End)
The first conjecture is true for the first 1000 terms of the sequence. - Harvey P. Dale, Jun 15 2013

A182334 Triangular numbers that differ from a square by 1.

Original entry on oeis.org

0, 1, 3, 10, 15, 120, 325, 528, 4095, 11026, 17955, 139128, 374545, 609960, 4726275, 12723490, 20720703, 160554240, 432224101, 703893960, 5454117903, 14682895930, 23911673955, 185279454480, 498786237505, 812293020528, 6294047334435, 16944049179226

Offset: 1

Arkadiusz Wesolowski, Apr 25 2012

From Robert G. Wilson v, Jun 20 2015: (Start)
Actually this sequence is the union of two subsequences; the triangular numbers that are less than a square by 1 and those that are greater than a square by 1.
The first sequence by index of the triangular numbers is A072221: b(n) = 6b(n-1) - b(n-2) + 2, with b(0)=1, b(1)=4.
And obviously the second sequence by index of the triangular numbers is A006451: c(n) = 6c(n-2) - c(n-4) + 2 with c(0)=0, c(1)=2, c(2)=5, c(3)=15.
(End)

			T(2) = 3 = 2^2 - 1, T(4) = 10 = 3^2 + 1,  T(5) = 15 = 4^2 - 1, and T(15) = 120 = 11^2 - 1.

Edward J. Barbeau, Pell's Equation (Springer 2003) at 17.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,35,0,0,-35,0,0,1).

Subsequence of A000217 and of A087279.

Cf. A001110, A006451, A072221, A229131, A299921.

I:=[0,1,3,10,15,120,325,528,4095,11026,17955]; [n le 11 select I[n] else 35*Self(n-3)-35*Self(n-6)+Self(n-9): n in [1..30]]; // Vincenzo Librandi, Jun 21 2015

lst = {}; Do[t = n*(n + 1)/2; If[IntegerQ[(t - 1)^(1/2)] || IntegerQ[(t + 1)^(1/2)], AppendTo[lst, t]], {n, 0, 10^4}]; lst (* Arkadiusz Wesolowski, Aug 06 2012 *)
b[n_] := b[n] = 6 b[n - 1] - b[n - 2] + 2; b[0] = 1; b[1] = 4; c[n_] := c[n] = 6 c[n - 2] - c[n - 4] + 2; c[0] = 0; c[1] = 2; c[2] = 5; c[3] = 15; #(# + 1)/2 & /@ Union@ Join[ Array[b, 9, 0], Array[c, 18, 0]] (* or *)
#(# + 1)/2 & /@ Join[{0, 1}, LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {2, 4, 5, 15, 25, 32, 90}, 35]] (* or *)
#(# + 1)/2 & /@ CoefficientList[ Series[x + x^2 (1 + x) (2 + x^2 - 3 x^3 + x^4)/((1 - x) (1 - 6 x^3 + x^6)), {x, 0, 36}], x] (* Robert G. Wilson v, Jun 20 2015 *)
a[n_] := a[n] = 35 a[n - 3] - 35 a[n - 6] + a[n - 9]; a[1] = 0; a[2] = 1; a[3] = 3; a[4] = 10; a[5] = 15; a[6] = 120; a[7] = 325; a[8] = 528; a[9] = 4095; a[10] = 11026; a[11] = 17955; Array[a, 36] (* Robert G. Wilson v after Charles R Greathouse IV, Apr 25 2012 *)
Select[Accumulate[Range[0,6*10^6]],AnyTrue[Sqrt[#+{1,-1}],IntegerQ]&] (* or *) LinearRecurrence[{0,0,35,0,0,-35,0,0,1},{0,1,3,10,15,120,325,528,4095,11026,17955},40] (* The first program uses the AnyTrue function from Mathematica version 10 *) (* Harvey P. Dale, Dec 24 2015 *)

concat(0, Vec(x^2*(1+3*x+10*x^2-20*x^3+15*x^4-25*x^5+38*x^6+x^8-x^9)/((1-x)*(1+x+x^2)*(1-34*x^3+x^6)) + O(x^30))) \\ Colin Barker, Sep 17 2016

a(n) = 35*a(n-3) - 35*a(n-6) + a(n-9). - Charles R Greathouse IV, Apr 25 2012

G.f.: x^2*(1+3*x+10*x^2-20*x^3+15*x^4-25*x^5+38*x^6+x^8-x^9) / ((1-x)*(1+x+x^2)*(1-34*x^3+x^6)). - Colin Barker, Sep 17 2016

A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

Original entry on oeis.org

2, 2, 1, 3, 4, 8, 9, 19, 22, 46, 53, 111, 128, 268, 309, 647, 746, 1562, 1801, 3771, 4348, 9104, 10497, 21979, 25342, 53062, 61181, 128103, 147704, 309268, 356589, 746639, 860882, 1802546, 2078353, 4351731, 5017588, 10506008, 12113529, 25363747, 29244646, 61233502

Offset: 0

Raphie Frank, Dec 30 2015

This sequence gives all x in N | 2*x^2 - 7(-1)^x = y^2. The companion sequence to this sequence, giving y values, is A266505.
A266505(n)/a(n) converges to sqrt(2).
Alternatively, 1/4*(3*A002203(floor[n/2]) - A002203(n-(-1)^n)), where A002203 gives the Companion Pell numbers, or, in Lucas sequence notation, V_n(2, -1).
Alternatively, bisection of A266506.
Alternatively, A048654(n -1) and A078343(n + 1) interlaced.
Alternatively, A100525(n-1), A266507(n), A038761(n) and A253811(n) interlaced.
Let b(n) = (a(n) - a(n)(mod 2))/2, that is b(n) = {1, 1, 0, 1, 2, 4, 4, 9, 11, 23, 26, 55, 64, ...}. Then:
A006452(n) = {b(4n+0) U b(4n+1)} gives n in N such that n^2 - 1 is triangular;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that n^2 + n + 1 is triangular (indices of Sophie Germain triangular numbers);
A216162(n) = {b(4n+0) U b(4n+2) U b(4n+1) U b(4n+3)}, sequences A006452 and A216134 interlaced.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[2,2,1,3]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

LinearRecurrence[{0, 2, 0, 1}, {2, 2, 1, 3}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(1 - x) (2 + 4 x + x^2)/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 41}] (* Michael De Vlieger, Dec 31 2015 *)

Vec((1-x)*(2+4*x+x^2)/(1-2*x^2-x^4) + O(x^50)) \\ Colin Barker, Dec 31 2015

a(n) = 1/sqrt(8)*(+sqrt(2)*(1+sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n - 3*(1-sqrt(2))^(floor(n/2)-(-1)^n) + sqrt(2)*(1-sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n + 3*(1+sqrt(2))^(floor(n/2)-(-1)^n)).
a(n) = 1/4*((3*((1+sqrt(2))^floor(n/2)+(1-sqrt(2))^floor(n/2))) - (-1)^n*((1+sqrt(2))^(floor(n/2)-(-1)^n)+(1-sqrt(2))^(floor(n/2)-(-1)^n))).
a(2n) = (+sqrt(2)*(1+sqrt(2))^(n-1) - 3 *(1-sqrt(2))^(n-1) + sqrt(2)*(1-sqrt(2))^(n-1) + 3*(1 + sqrt(2))^(n-1))/sqrt(8) = A048654(n -1).
a(2n) = 1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) - ((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))) = A048654(n -1).
a(2n + 1) = (-sqrt(2)*(1+sqrt(2))^(n+1) - 3 *(1-sqrt(2))^(n+1) - sqrt(2)*(1-sqrt(2))^(n+1) + 3*(1+sqrt(2))^(n+1))/sqrt(8) = A078343(n + 1).
a(2n + 1) =1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) + ((1+sqrt(2))^(n+1)+(1-sqrt(2))^(n+1))) =  A078343(n + 1).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A100525(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A266507(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038761(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A253811(n).
sqrt(2*a(n)^2 - 7(-1)^a(n))*sgn(2*n - 1) = A266505(n).
(a(2n + 1) + a(2n))/2 = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 1) - a(2n))/2 = A000129(n), where A000129 gives the Pell numbers.
(a(2n+2) + a(2n+1))*2 = A002203(n+2)
(a(2n+2) - a(2n+1))*2 = A002203(n-1).
G.f.: (1-x)*(2+4*x+x^2) / (1-2*x^2-x^4). - Colin Barker, Dec 31 2015

A285984 Numbers k such that 27*T(k)+1 is a square, where T(m) is the m-th triangular number A000217(m).

Original entry on oeis.org

0, 110, 374, 107184, 363264, 103968854, 352366190, 100849681680, 341794841520, 97824087261230, 331540643908694, 94889263793711904, 321594082796592144, 92042488055813286134, 311945928772050471470, 89281118524875093838560, 302587229314806160734240, 86602592926640785210117550

Offset: 0

Vladimir Pletser, May 01 2017

Numbers a(n) that make sqrt(27*T(a(n))+1) an integer.

This sequence a(n) gives also the indices of the triangular numbers T(a(n)) such that the 3rd degree Diophantine Bachet-Mordell equation y^2 = x^3+K holds with x = 3*T(a(n)) = A286035(n), y = T(a(n))* sqrt(27*T(a(n))+1) = A286036(n) and K = T(a(n))^2 = A286037(n).

			k = 110 is a term because 27*(T(110) + 1) = 27 * (110*111/2 + 1) is a square. - _David A. Corneth_, May 02 2017
For n = 2, a(2) = 264*sqrt(27*(a(0)*(a(0)+1)/2)+1)+ a(-2) = 264*sqrt(27*(0*(0+1)/2)+1) + 110 = 374.
For n = 6, a(6) = 264*sqrt(27*(a(4)*(a(4)+1)/2)+1)+ a(2) = 264*sqrt(27*(363264*(363264+1)/2)+1) + 374 = 352366190.

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

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 and Amir Ghadermarzi, Mordell's equation : a classical approach, arXiv:1311.7077 [math.NT], 2013.
Eric Weisstein's World of Mathematics, Mordell Curve

Cf. A286035, A286036, A286037, A285955, A006454, A000217, A006451, A081119, A054504.

restart: am2:=110: am1:=0: a0:=0: ap1:=110: print ('0,0','1,110'); for n from 2 to 1000 do a:= 264*sqrt(27* (a0^2+a0)/2+1)+am2; print(n,a); am2:=am1; am1:=a0; a0:=ap1; ap1:=a; end do:

nxt[{a_,b_}]:={b,485*a+242+66*Sqrt[54a^2+54*a+4]}; NestList[nxt,{0,110},20][[All,1]] (* Harvey P. Dale, May 30 2018 *)

is(n) = issquare(27*binomial(n+1, 2)+1) \\ David A. Corneth, May 02 2017

a(n) = 264*sqrt(27*T(a(n-2))+1)+ a(n-4) = 264*sqrt(27*(a(n-2)*(a(n-2)+1)/2)+1)+ a(n-4), with a(-2)=110, a(-1)=0, a(0)=0, a(1)=110.
Empirical g.f.: 22*x*(5 + 12*x + 5*x^2) / ((1 - x)*(1 - 970*x^2 + x^4)). - Colin Barker, May 01 2017, verified by Robert Israel, May 03 2017
a(n) = 485*a(n-2)+242+66*sqrt(54*a(n-2)^2+54*a(n-2)+4). - Robert Israel, May 03 2017

A286035 a(n) = 3*T(A285984(n)), where T(m) is the m-th triangular number A000217(m).

Original entry on oeis.org

0, 18315, 210375, 17232775560, 197941645440, 16214284059063255, 186242898311223435, 15255987442587265956120, 175235570535035566127880, 14354328072739259079522561195, 164878797845087651200279041495, 13505958574968967401962031517525680, 155134131134672045268505114018663320

Offset: 0

Vladimir Pletser, May 01 2017

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).

			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.

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

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

Cf. A285984, A286036, A286037, A285955, A006454, A000217, A006451, A081119, A054504

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:

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

A286036 a(n) is the solution y to the Bachet Mordell equation y^2=x^3+K, with x = 3*T(b(n)) and K = (T(b(n)))^2, where T(b(n)) is the triangular number of b(n)= A285984(n).

Original entry on oeis.org

0, 2478630, 96492000, 2262209634604920, 88065491686677120, 2064651070850763887750940, 80374740223699340246041830, 1884345278651963087653858708518360, 73355621393690297028946986338029560, 1719785575058362227821108881720941727234290, 66949481579385248741161156467886515267346140

Offset: 0

Vladimir Pletser, May 01 2017

a(n) is the producs of the triangular number T(b(n)) and the square root of 27 times this triangular number plus one, sqrt(27*T(b(n))+1), where b(n) is the sequence A285984(n) of numbers n such that (27*T(n)+1) is a square.

			For n = 2, b(n) = 374, a(n)= 96492000.
For n = 3, b(n) = A285984(n) =107184. Therefore, a(n) = T(b(n))* sqrt(27*T(b(n))+1) = A000217(A285984(n))* sqrt(27*A000217(A285984(n))+1) = A000217(107184)* sqrt(27*A000217(107184)+1) =5744258520* sqrt(27*5744258520 +1) = 2262209634604920.

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

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

Cf. A285984, A286035, A286037, A285955, A006454, A000217, A006451, A081119, A054504.

restart: bm2:=110: bm1:=0: b0:=0: bp1:=110: print ('0,0','1,2478630’); for n from 2 to 1000 do b:= 264*sqrt(27* (b0^2+b0)/2+1)+bm2; T:=b*(b+1)/2; a:= T*sqrt(27*T+1); print(n,a); bm2:=bm1; bm1:=b0; b0:=bp1; bp1:=b; end do:

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) = T(b(n))*sqrt(27*T(b(n))+1) (this sequence), one has :
a(n) = ([264*sqrt(27*T(b(n-2))+1)+ b(n-4)]*[ 264*sqrt(27*T(b(n-2))+1)+ b(n-4)+1]/2) *sqrt(27*([264*sqrt(27*T(b(n-2))+1)+ b(n-4)]*[ 264*sqrt(27*T(b(n-2))+1)+ b(n-4)+1]/2)+1).
Empirical g.f.: 330*x*(1 + x)*(7511 + 284889*x + 108094375*x^2 + 284889*x^3 + 7511*x^4) / ((1 - 912670090*x^2 + x^4)*(1 - 970*x^2 + x^4)). - Colin Barker, May 01 2017

A286037 a(n) = T(A285984(n))^2, where T(m) is the m-th triangular number A000217(m).

Original entry on oeis.org

0, 37271025, 4917515625, 32996505944592590400, 4353432777721630310400, 29211445283110309395256454577225, 3854046352373857001854365165911025, 25860572538708927496411840821477504196161600, 3411945020082158343071838489442339152945921600, 22894081602203374655543296113789919615194083223613314225

Offset: 0

Vladimir Pletser, May 01 2017

a(n) =(T(b(n)))^2, parameters K=a(n) of the Bachet Mordell equation y^2=x^3+K, with x= 3*T(b(n)) and y= T(b(n))*sqrt(27*T(b(n))+1), where T(b(n)) is the triangular number of b(n)= A285984(n).

			For n = 2, b(n) = 374, a(n)= 4917515625.
For n = 3, b(n) = A285984(n) =107184. Therefore, a(n) = (T(b(n)))^2 = (A000217(A285984(n)))^2 = (A000217(107184))^2 = (5744258520)^2=32996505944592590400.

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

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

Cf. A285984, A286035, A286036, A285955, A006454, A000217, A006451, A081119, A054504.

restart: bm2:=110: bm1:=0: b0:=0: bp1:=110: print ('0,0','1,4917515625’); for n from 2 to 1000 do b:= 264*sqrt(27*(b0^2+b0)/2+1)+bm2; a:=(b*(b+1)/2)^2; print(n,a); bm2:=bm1; bm1:=b0; b0:=bp1; bp1:=b; end do:

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) = (T(b(n)))^2 (this sequence), one has :
a(n) = ([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)^2.
Empirical g.f.: 27225*x*(1369 + 179256*x + 30879367019*x^2 + 168661970400*x^3 + 30879367019*x^4 + 179256*x^5 + 1369*x^6) / ((1 - x)*(1 - 940898*x + x^2)*(1 - 970*x + x^2)*(1 + 970*x + x^2)*(1 + 940898*x + x^2)). - Colin Barker, May 01 2017

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