A285985 -id:A285985 - cp's OEIS frontend

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

Vladimir Pletser, Apr 29 2017

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).

			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

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

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:

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

A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

Original entry on oeis.org

0, 3, 15, 120, 528, 4095, 17955, 139128, 609960, 4726275, 20720703, 160554240, 703893960, 5454117903, 23911673955, 185279454480, 812293020528, 6294047334435, 27594051024015, 213812329916328, 937385441796000, 7263325169820735, 31843510970040003, 246739243443988680

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Alternative definition: a(n) is triangular and a(n)/2 is the harmonic average of consecutive triangular numbers. See comments and formula section of A005563, of which this sequence is a subsequence. - Raphie Frank, Sep 28 2012
As with the Sophie Germain triangular numbers (A124174), 35 = (a(n) - a(n-6))/(a(n-2) - a(n-4)). - Raphie Frank, Sep 28 2012
Sophie Germain triangular numbers of the second kind as defined in A217278. - Raphie Frank, Feb 02 2013
Triangular numbers m such that m+1 is a square. - Bruno Berselli, Jul 15 2014
From Vladimir Pletser, Apr 30 2017: (Start)
Numbers a(n) which are the triangular number T(b(n)), where b(n) is the sequence A006451(n) of numbers n such that T(n)+1 is a square.
a(n) also gives the x solutions of the 3rd-degree Diophantine Bachet-Mordell equation y^2 = x^3 + K, with y = T(b(n))*sqrt(T(b(n))+1) = A285955(n) and K = T(b(n))^2 = A285985(n), the square of the triangular number of b(n) = A006451(n).
Also: This sequence is a subsequence of A000217(n), namely A000217(A006451(n)). (End)

			From _Raphie Frank_, Sep 28 2012: (Start)
35*(528 - 15) + 0 = 17955 = a(6),
35*(4095 - 120) + 3 = 139128 = a(7),
35*(17955 - 528) + 15 = 609960 = a(8),
35*(139128 - 4095) + 120 = 4726275 = a(9). (End)
From _Raphie Frank_, Feb 02 2013: (Start)
a(7) = 139128 and a(9) = 4726275.
a(9) = (2*(sqrt(8*a(7) + 1) - 1)/2 + 3*sqrt(a(7) + 1) + 1)^2 - 1 = (2*(sqrt(8*139128 + 1) - 1)/2 + 3*sqrt(139128 + 1) + 1)^2 - 1 = 4726275.
a(9) = 1/2*((3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)^2 + (3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)) = 1/2*((3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)^2 + (3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)) = 4726275. (End)
From _Vladimir Pletser_, Apr 30 2017: (Start)
For n=2, b(n)=5, a(n)=15
For n=5, b(n)=90, a(n)= 4095
For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. (End)

Edward J. Barbeau, Pell's Equation, New York: Springer-Verlag, 2003, p. 17, Exercise 1.2.
Allan J. Gottlieb, How four dogs meet in a field, and other problems, Technology Review, Jul/August 1973, pp. 73-74.
Vladimir Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vladimir Pletser, Table of n, a(n) for n = 0..1000 (first 60 terms from Vincenzo Librandi)
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.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Jeffrey Shallit, Letter to N. J. A. Sloane, Oct. 1975.
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Eric Weisstein's World of Mathematics, Mordell Curve.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. sqrt(a(n) + 1) = A006452(n + 1) = A216162(2n + 2) and (sqrt(8a(n) + 1) - 1)/2 = A006451.
Cf. A217278, A124174, A216134. - Raphie Frank, Feb 02 2013
Subsequence of A182334.
Cf. A001109, A245031.
Cf. A285955, A285985, A000217, A006451, A081119, A054504. - Vladimir Pletser, Apr 30 2017

I:=[0,3,15,120,528,4095]; [n le 6 select I[n] else 35*(Self(n-2) - Self(n-4)) + Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 21 2015

A006454:=-3*z*(1+4*z+z**2)/(z-1)/(z**2-6*z+1)/(z**2+6*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print ('0,0','1,3','2,15'); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=b*(b+1)/2; print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do: # Vladimir Pletser, Apr 30 2017

Clear[a]; a[0] = a[1] = 1; a[2] = 2; a[3] = 4; a[n_] := 6a[n - 2] - a[n - 4]; Array[a, 40]^2 - 1 (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)
LinearRecurrence[{1,34,-34,-1,1},{0,3,15,120,528},30] (* Harvey P. Dale, Feb 18 2023 *)

concat(0, Vec(3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30))) \\ Colin Barker, Apr 30 2017

a(n) = A006451(n)*(A006451(n)+1)/2.
a(n) = A006452(n)^2 - 1. - Joerg Arndt, Mar 04 2011
a(n) = 35*(a(n-2) - a(n-4)) + a(n-6). - Raphie Frank, Sep 28 2012
From Raphie Frank, Feb 01 2013: (Start)
a(0) = 0, a(1) = 3, and a(n+2) = (2x + 3y + 1)^2 - 1  = 1/2*((3x + 4y + 1)^2 + (3x + 4y + 1)) where x = (sqrt(8*a(n) + 1) - 1)/2 = A006451(n) =  1/2*(A216134(n + 1) + A216134(n - 1)) and y = sqrt(a(n) + 1) = A006452(n + 1) = 1/2*(A216134(n + 1) - A216134(n - 1)).
Note that A216134(n + 1) = x + y, and A216134(n + 3) = (2x + 3y + 1) + (3x + 4y + 1) = (5x + 7y + 2), where A216134 gives the indices of the Sophie Germain triangular numbers. (End)
a(n) = (1/64)*(((4 + sqrt(2))*(1 -(-1)^(n+1)*sqrt(2))^(2* floor((n+1)/2)) + (4 - sqrt(2))*(1 + (-1)^(n+1)*sqrt(2))^(2*floor((n+1)/2))))^2 - 1. - Raphie Frank, Dec 20 2015
From Vladimir Pletser, Apr 30 2017: (Start)
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)) (this sequence), we have:
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). (End)
From Colin Barker, Apr 30 2017: (Start)
G.f.: 3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) for n > 4.
(End)
a(n)  = (A001109(n/2+1) - 2*A001109(n/2))^2 - 1 if n is even, and (A001109((n+3)/2) - 4*A001109((n+1)/2))^2 - 1 if n is odd (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Better description from Harvey P. Dale, Jan 28 2001
More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001
Minor edits by N. J. A. Sloane, Oct 24 2009

cp's OEIS Frontend

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).

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