A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square. (Formerly M3004)

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

0,2

Comments

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)

References

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).

Links

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).

Formula

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

Example

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)

Maple

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

Mathematica

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 *)

Prog

(Magma) 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
(PARI) 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

Crossrefs

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

Sequence in context: A060639 A068052 A068859 * A225115 A303353 A112228

Adjacent sequences: A006451 A006452 A006453 * A006455 A006456 A006457

Keyword

nonn,easy

Author

N. J. A. Sloane, Jeffrey Shallit

Extensions

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

Status

approved