A006451 -id:A006451 - cp's OEIS frontend

A001652 a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.

0, 3, 20, 119, 696, 4059, 23660, 137903, 803760, 4684659, 27304196, 159140519, 927538920, 5406093003, 31509019100, 183648021599, 1070379110496, 6238626641379, 36361380737780, 211929657785303, 1235216565974040, 7199369738058939, 41961001862379596, 244566641436218639

Offset: 0

N. J. A. Sloane

Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.
Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571).
Members of Diophantine pairs. Solution to a*(a+1) = 2*b*(b+1) in natural numbers including 0; a = a(n), b = b(n) = A053141(n); The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.
The three sequences x (A001652), y (A046090) and z (A001653) may be obtained by setting u and v equal to the Pell numbers (A000129) in the formulas x = 2uv, y = u^2 - v^2, z = u^2 + v^2 [Joseph Wiener and Donald Skow]. - Antonio Alberto Olivares, Dec 22 2003
All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy, Aug 14 2006
Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n). - Kenneth J Ramsey, Sep 22 2007
In general, if b(n) = A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g., 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870 = 5741*14+84. - Charlie Marion, Nov 19 2007
Limit_{n -> oo} a(n)/a(n-1) = 3+2*sqrt(2) = A156035. - Klaus Brockhaus, Feb 17 2009
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with pMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = y^2 with pMohamed Bouhamida, Sep 02 2009
a(n+k) = A001541(k)*a(n) + A001542(k)*A001653(n+1) + A001108(k). - Charlie Marion, Dec 10 2010
The numbers 3*A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in A075841). - Bruno Berselli, Aug 26 2010
Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,
  T(a(n) + 2*k*A001653(n+1)) = 2*T(A053141(n-1) + k*A002315(n)) + k^2 and
  T(a(n) + (2*k+1)*A001653(n+1)) = (A001109(n+1) + k*A002315(n))^2 + k*(k+1).
  Also (a(n) + k*A001653(n))^2 + (a(n) + k*A001653(n) + 1)^2 = (A001653(n+1) + k*A002315(n))^2 + k^2. - Charlie Marion, Dec 09 2010
For n>0, A143608(n) divides a(n). - Kenneth J Ramsey, Jun 28 2012
Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - Carmine Suriano, Dec 17 2013
The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1) = 1 and c(1) = 2, then b(n) = c(n-1) and c(n) = 2*c(n-1) + b(n-1). Alternatively, b(n) = c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - J. M. Bergot, Oct 09 2014
Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016
Numbers m such that m^2+(m+1)^2 is a square. - César Aguilera, Aug 14 2017
For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - Charlie Marion, Dec 08 2018
The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - Bernard Schott, Mar 10 2019
Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. - Steven Blasberg, Mar 04 2021

			The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Christian Aebi and Grant Cairns, Lattice equable quadrilaterals III: tangential and extangential cases, Integers (2023) Vol. 23, #A48.
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See p. 5.
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
Thibault Gauthier, Deep Reinforcement Learning for Synthesizing Functions in Higher-Order Logic, arXiv:1910.11797 [cs.AI], 2019. See also EPiC Series in Computing, (2020) Vol. 73, 230-248.
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
Ron Knott, Pythagorean Triples and Online Calculators
A. Martin, Table of prime rational right-angled triangles, The Mathematical Magazine, 2 (1910), 297-324.
A. Martin, Table of prime rational right-angled triangles (annotated scans of a few pages)
A. Martin, On rational right-angled triangles, Proceedings of the Fifth International Congress of Mathematicians (Cambridge, 22-28 August 1912).
S. P. Mohanty, Which triangular numbers are products of three consecutive integers, Acta Math. Hungar., 58 (1991), 31-36.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
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.
Burkard Polster, Nice merging together, Mathologer video (2015).
B. Polster and M. Ross, Marching in squares, arXiv:1503.04658 [math.HO], 2015.
Zhang Zaiming, Problem #502, Pell's Equation - Once Again, The College Mathematics Journal, 25 (1994), 241-243.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A046090(n) = -a(-1-n).
Cf. A001108, A143608, A089950 (partial sums), A156035.
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; this sequence for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013
Cf. A053141, A001541, A001109, A274307.
Cf. A002315, A001653 (solutions of x^2 - 2*y^2 = -1).
Cf. A000129, A001333, A001542, A002024, A002378, A029549, A048739, A053142, A055997, A075841, A084159, A084703, A097571, A123737.

a:=[0,3];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]+2; od; a; # Muniru A Asiru, Dec 08 2018

a001652 n = a001652_list !! n
a001652_list = 0 : 3 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
-- Reinhard Zumkeller, Jan 10 2012

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ (-2+(r2+1)*(3+2*r2)^n-(r2-1)*(3-2*r2)^n)/4: n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Feb 17 2009

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018

A001652 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,3]) ;
    else
        6*procname(n-1)-procname(n-2)+2 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

LinearRecurrence[{7,-7,1}, {0,3,20}, 30] (* Harvey P. Dale, Aug 19 2011 *)
With[{c=3+2*Sqrt[2]},NestList[Floor[c*#]+3&,3,30]] (* Harvey P. Dale, Oct 22 2012 *)
CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 21 2014 *)
Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* G. C. Greubel, Jul 15 2018 *)

{a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* Michael Somos, Aug 11 2006 */

concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ Altug Alkan, Nov 08 2015

{a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ G. C. Greubel, Jul 15 2018

(x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 08 2019

G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Simon Plouffe in his 1992 dissertation
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). a_{n} = -1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n. - Antonio Alberto Olivares, Oct 13 2003
a(n) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - Pierre CAMI, Mar 30 2005
a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - Bill Gosper, Feb 07 2010
Binomial(a(n)+1,2) = 2*binomial(A053141(n)+1,2) = A029549(n). See A053141. - Bill Gosper, Feb 07 2010
Let b(n) = A046090(n) and c(n) = A001653(n). Then for k>j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n<0, a(n) = -b(-n-1). Also a(n)*a(n+2*k+1) + b(n)*b(n+2*k+1) + c(n)*c(n+2*k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2*k) + b(n)*b(n+2*k) + c(n)*c(n+2*k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion, Jul 01 2003
For n and j >= 1, Sum_{k=0..j} A001653(k)*a(n) - Sum_{k=0...j-1} A001653(k)*a(n-1) + A053141(j) = A001109(j+1)*a(n) - A001109(j)*a(n-1) + A053141(j) = a(n+j). - Charlie Marion, Jul 07 2003
Sum_{k=0...n} (2*k+1)*a(n-k) = A001109(n+1) - A000217(n+1). - Charlie Marion, Jul 18 2003
a(n) = A055997(n) - 1 + sqrt(2*A055997(n)*A001108(n)). - Charlie Marion, Jul 21 2003
a(n) = {A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003
a(2*n+k) + a(k) + 1 = A001541(n)*A002315(n+k). For k>0, a(2*n+k) - a(k-1) = A001541(n+k)*A002315(n); e.g., 803760-119 = 19601*41. - Charlie Marion, Mar 17 2003
a(n) = (A001653(n+1) - 3*A001653(n) - 2)/4. - Lekraj Beedassy, Jul 13 2004
a(n) = {2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy, Jul 21 2004
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8*a(n) +4) + 1, a(1)=0. - Richard Choulet, Sep 18 2007
As noted (Sep 20 2006), a(n) = 5*(a(n-1) + a(n-2)) - a(n-3) + 4. In general, for n > 2*k, a(n) = A001653(k)*(a(n-k) + a(n-k-1) + 1) - a(n-2*k-1) - 1. Also a(n) = 7*(a(n-1) - a(n-2)) + a(n-3). In general, for n > 2*k, A002378(k)*(a(n-k)-a(n-k-1)) + a(n-2*k-1). - Charlie Marion, Dec 26 2007
In general, for n >= k >0, a(n) = (A001653(n+k) - A001541(k) * A001653(n) - 2*A001109(k-1))/(4*A001109(k-1)); e.g., 4059 = (33461-3*5741-2*1)/(4*1); 4059 = (195025-17*5741-2*6)/(4*6). - Charlie Marion, Jan 21 2008
From Charlie Marion, Jan 04 2010: (Start)
a(n) = ( (1 + sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 2)/4 = (A001333(2n+1) - 1)/2.
a(2*n+k-1) = Pell(2*n-1)*Pell(2*n+2*k) + Pell(2*n-2)*Pell(2*n+2*k+1) + A001108(k+1);
a(2*n+k) = Pell(2*n)*Pell(2*n+2*k+1) + Pell(2*n-1)*Pell(2*n+2*k+2) - A055997(k+2). (End)
a(n) = A048739(2*n-1) for n > 0. - Richard R. Forberg, Aug 31 2013
a(n+1) = 3*a(n) + 2*A001653(n) + 1 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 + (a(n)+1)^2 = A001653(n+1)^2. - Pierre CAMI, Mar 30 2005; clarified by Hermann Stamm-Wilbrandt, Aug 31 2014
a(n+1) = 3*A001541(n) + 10*A001109(n) + A001108(n). - Hermann Stamm-Wilbrandt, Sep 09 2014
For n>0, a(n) = Sum_{k=1..2*n} A000129(k). - Charlie Marion, Nov 07 2015
a(n) = 3*A053142(n) - A053142(n-1). - R. J. Mathar, Feb 05 2016
E.g.f.: (1/4)*(-2*exp(x) - (sqrt(2) - 1)*exp((3-2*sqrt(2))*x) + (1 + sqrt(2))*exp((3+2*sqrt(2))*x)). - Ilya Gutkovskiy, Apr 11 2016
a(n) = A001108(n) + 2*sqrt(A000217(A001108(n))). - Dimitri Papadopoulos, Jul 06 2017
a(A000217(n-1)) = ((A001653(n)+1)/2) * ((A001653(n)-1)/2), n > 1. - Ezhilarasu Velayutham, Mar 10 2019
a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
In general, for each k >= 0, a(n) = ((a(n-k)+a(k-1)+1)*(a(n-k)-a(k)))/a(n-2*k) for n > 2*k. - Charlie Marion, Dec 27 2020
A generalization of the identity a(n)^2 + A046090(n)^2 = A001653(n+1)^2 follows. Let P(k,n) be the n-th k-gonal number. Then P(k,a(n)) + P(k,A046090(n)) = P(k,A001653(n+1)) + (4-k)*A001109(n). - Charlie Marion, Dec 07 2021
a(n) = A046090(n)-1 = A002024(A029549(n)). - Pontus von Brömssen, Sep 11 2024

Additional comments from Wolfdieter Lang, Feb 10 2000

A006452 a(n) = 6*a(n-2) - a(n-4).

Original entry on oeis.org

1, 1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552, 12671, 26531, 73852, 154634, 430441, 901273, 2508794, 5253004, 14622323, 30616751, 85225144, 178447502, 496728541, 1040068261, 2895146102, 6061962064, 16874148071, 35331704123

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Solution to a Diophantine equation.
Integers k such that k^2-1 is a triangular number. - Benoit Cloitre, Apr 05 2002
For all elements "x" of the sequence, 8*x^2 - 7 is a square. - Gregory V. Richardson, Oct 07 2002
a(n) mod 10 is a sequence of period 12: repeat (1, 1, 2, 4, 1, 3, 4, 4, 3, 1, 4, 2). - Paul Curtz, Dec 07 2012
a(n)^2 - 1 = A006454(n - 1) is a Sophie Germain triangular number of the second kind as defined in A217278. - Raphie Frank, Feb 08 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6xy + y^2 + 7 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 34xy + y^2 + 252 = 0. - Colin Barker, Mar 04 2014
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n+1), for n >= 0, gives one half of all positive y solutions of the Pell equation x^2 - 2*y^2 = -7. The corresponding x-solutions are x(n) = A077446(n+1).
See a comment on A077446 for the first and second class solutions separately, and the connection to the Pell equation X^2 - 2*Y^2 = 14. (End)
For n > 0, a(n) is the n-th almost balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: 11^2 - 2*(2*4)^2 = -7 (see the Pell comment above);
(4*4)^2 - 2*11^2 = +14. - _Wolfdieter Lang_, Feb 26 2015

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Jul/Aug 1973 pp. 73-74.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
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
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001333, A006451, A006452, A006454, A038723, A077446, A216134, A217278.

I:=[1,1,2,4]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 09 2013

A006452:=-(z-1)*(z**2+3*z+1)/(z**2+2*z-1)/(z**2-2*z-1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's

s=0;lst={1}; Do[s+=n;If[Sqrt[s+1]==Floor[Sqrt[s+1]],AppendTo[lst, Sqrt[s+1]]], {n,0,8!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
a[0]=a[1]= 1; a[2]=2; a[3]=4; a[n_]:= 6*a[n-2] -a[n-4]; Array[a, 30, 0] (* Robert G. Wilson v, Jun 11 2010 *)
CoefficientList[Series[(1+x-4x^2-2x^3)/((1-2x-x^2)(1+2x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;1;2;4])[1,1] \\ Charles R Greathouse IV, May 10 2016

def A001333(n): return lucas_number2(n, 2, -1)/2
def A006452(n): return (A001333(n+1) + (-1)^n *A001333(n-2))/4
[A006452(n) for n in range(41)] # G. C. Greubel, Jan 22 2023

Bisection: a(2n) = A006452(n). a(2n+1) = A038723(n).
G.f.: ( 1+x-4*x^2-2*x^3 ) / ( (1-2*x-x^2)*(1+2*x-x^2) ).
From Gregory V. Richardson, Oct 07 2002: (Start)
For n (even), a(n) = ( ((3 + sqrt(8))^((n/2)+1) - (3 - sqrt(8))^((n/2)+1)) - 2*((3 + sqrt(8))^((n/2)-1) - (3 - sqrt(8))^((n/2)-1)) ) / (6*sqrt(8)).
For n (odd), a(n) = ( ((3 + sqrt(8))^((n+1)/2) - (3 - sqrt(8))^((n+1)/2)) - 2*((3 + sqrt(8))^((n-1)/2) - (3 - sqrt(8))^((n-1)/2)) ) / (2*sqrt(8)).
Limit_{n->oo} a(n)/a(n-2) = 3 + sqrt(8).
If n is odd, lim_{n->oo} a(n)/a(n-1) = (9 + 2*sqrt(8))/7.
If n is even, lim_{n->oo} a(n)/a(n-1) = (11 + 3*sqrt(8))/7. (End)
a(n+2) = (A001333(n+3) + (-1)^n *A001333(n))/4. - Paul Curtz, Dec 06 2012
a(n+2) = sqrt(17*a(n)^2 + 6*(sqrt(8*a(n)^2 - 7))*a(n)*sgn(2*n - 1) - 7) with a(0) = 1, a(1) = 1. - Raphie Frank, Feb 08 2013
a(n+2) = (A216134(n+2) - A216134(n))/2. - Raphie Frank, Feb 11 2013
E.g.f.: (2*cosh(sqrt(2)*x)*(2*cosh(x) - sinh(x)) + sqrt(2)*(3*cosh(x) - sinh(x))*sinh(sqrt(2)*x))/4. - Stefano Spezia, Nov 26 2022

More terms from James Sellers, May 03 2000

A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

Original entry on oeis.org

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, 15050244140475527879, 209622859053806301480

Offset: 0

N. J. A. Sloane

(a(n)+1)^3 - a(n)^3 is a square (that of A001570(n)).
The ratio A001570(n)/a(n) tends to sqrt(3) = 1.73205... as n increases. - Pierre CAMI, Apr 21 2005
Define a(1)=0 a(2)=7 such that 3*(a(1)^2) + 3*a(1) + 1 = j(1)^2 = 1^2 and 3*(a(2)^2) + 3*a(2) + 1 = j(2)^2 = 13^2. Then a(n) = a(n-2) + 8*sqrt(3*(a(n-1)^2) + 3*a(n-1) + 1). Another definition : a(n) such that 3*(a(n)^2) + 3*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
a(n) = A001353(n)*A001075(n+1). For n>0, the triple {a(n), a(n)+1=A001922(n), A001570(n)} forms a near-isosceles triangle with angle 2*Pi/3 bounded by the consecutive sides. - Lekraj Beedassy, Jul 21 2006
Numbers n such that A003215(n) is a square, cf. A006051. - Joerg Arndt, Jan 02 2017

			G.f. = 7*x + 104*x^2 + 1455*x^3 + 20272*x^4 + 282359*x^5 + 3932760*x^6 + ... - _Michael Somos_, Aug 17 2018

J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
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
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001570, A001922, A006051.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; A233450 for k=3; A001652 for k=4; A129556 for k=5; this sequence for k=6. - Bruno Berselli, Dec 16 2013

[Round(-1/2 - (1/6)*Sqrt(3)*(7-4*Sqrt(3))^n + (1/6)*Sqrt(3)*(7+4*Sqrt(3))^n + (1/4)*(7+4*Sqrt(3))^n + (1/4)*(7-4*Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

A001921:=z*(-7+z)/(z-1)/(z**2-14*z+1); # Conjectured by Simon Plouffe in his 1992 dissertation.

t = {0, 7}; Do[AppendTo[t, 14*t[[-1]] - t[[-2]] + 6], {20}]; t (* T. D. Noe, Aug 17 2012 *)
LinearRecurrence[{15, -15, 1}, {0, 7, 104}, 19] (* Michael De Vlieger, Jan 02 2017 *)
a[ n_] := -1/2 + (ChebyshevT[n + 1, 7] - ChebyshevT[n, 7]) / 12; (* Michael Somos, Aug 17 2018 *)

concat(0, Vec(x*(x-7)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Jan 06 2015

{a(n) = -1/2 + (polchebyshev(n + 1, 1, 7) - polchebyshev(n, 1, 7)) / 12}; /* Michael Somos, Aug 17 2018 */

G.f.: x*(-7 + x)/(x - 1)/(x^2 - 14*x + 1) (see Simon Plouffe in Maple section).
a(n) = (A028230(n+1)-1)/2. - R. J. Mathar, Mar 19 2009
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). - Colin Barker, Jan 06 2015
a(n) = -1 - a(-1-n) for all n in Z. - Michael Somos, Aug 17 2018

More terms from James Sellers, Jul 04 2000

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

A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

Original entry on oeis.org

0, 1, 4, 9, 26, 55, 154, 323, 900, 1885, 5248, 10989, 30590, 64051, 178294, 373319, 1039176, 2175865, 6056764, 12681873, 35301410, 73915375, 205751698, 430810379, 1199208780, 2510946901, 6989500984, 14634871029, 40737797126, 85298279275, 237437281774

Offset: 0

Raphie Frank, Sep 01 2012

Numbers n such that 2*triangular(n) + 1 is a triangular number. Equivalently, numbers n such that n^2 + n + 1 is a triangular number. - Alex Ratushnyak, Apr 18 2013

For n > 0, a(n) is the n-th almost cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Colin Barker, Table of n, a(n) for n = 0..1000
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Wikipedia, Pell numbers
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A124174, A000129, A001333, A006451, A006452, A124124, A079496.

Cf. A000217, A069017 (triangular numbers of the form k^2 + k + 1).

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 4, 9, 26}, 40] (* T. D. Noe, Sep 03 2012 *)

Vec( x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 13 2014

isok(n) = ispolygonal(n*(n+1) + 1, 3); \\ Michel Marcus, Aug 13 2014

G.f.: x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)). - R. J. Mathar, Sep 08 2012
sqrt(2) = lim_{k->infinity} ((a(2k+1) + a(2k) + 1)/2)/(a(2k+1) - a(2k)) = lim_{k->infinity} A001333(2k + 1)/A000129(2k + 1).
1 + (sqrt 2) = lim_{k->infinity} (a(2k + 1) - a(2k))/(a(2k + 1) - 2*a(2k) +  a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A000129(2k).
1 + 1/(sqrt 2) = lim_{k->infinity} (a(2k+1) - a(2k))/(a(2k) - a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A001333(2k).
a(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013
From Raphie Frank, Jan 04 2013: (Start)
A124174(n) = a(n)*(a(n) + 1)/2.
A079496(n) = a(n + 1) - a(n).
A000129(2n) = a(2n) - 2*a(2n - 1) +  a(2n - 2).
A000129(2n) = a(2n + 1) - 2*a(2n) +  a(2n - 1).
A000129(2n + 1) = a(2n + 1) - a(2n).
A001333(2n) = a(2n) - a(2n - 1).
A001333(2n + 1) = (a(2n + 1) + a(2n) + 1)/2.
A006451(n + 1) = (a(n + 2) + a(n))/2.
A006452(n + 2) = (a(n + 2) - a(n))/2.
A124124(n + 2) = (a(n + 2) + a(n))/2 + (a(n + 2) - a(n)).
(End)
a(n + 2) = sqrt(8*a(n)^2 + 8*a(n) + 9) + 3*a(n) + 1; a(0) = 0, a(1) = 1. - Raphie Frank, Feb 02 2013
a(n) = (3/8 + sqrt(2)/4)*(1 + sqrt(2))^n + (-1/8 - sqrt(2)/8)*(-1 + sqrt(2))^n + (3/8 - sqrt(2)/4)*(1 - sqrt(2))^n + (-1/8 + sqrt(2)/8)*(-1 - sqrt(2))^n - 1/2. - Robert Israel, Aug 13 2014
E.g.f.: (1/4)*(-2*cosh(x) - 2*sinh(x) + 2*cosh(sqrt(2)*x)*(cosh(x) + 2*sinh(x)) + sqrt(2)*(cosh(x) + 3*sinh(x))*sinh(sqrt(2)*x)). - Stefano Spezia, Dec 10 2019

A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

Original entry on oeis.org

0, 2, 21, 95, 816, 3626, 31005, 137711, 1177392, 5229410, 44709909, 198579887, 1697799168, 7540806314, 64471658493, 286352060063, 2448225223584, 10873837476098, 92968086837717, 412919472031679, 3530339074609680, 15680066099727722, 134059916748330141

Offset: 1

Alexander Adamchuk, Apr 20 2007

Corresponding numbers m > 0 such that m^2 is a centered pentagonal number are listed in A129557 = {1, 4, 34, 151, 1291, 5734, 49024, ...}.
From Andrea Pinos, Nov 02 2022: (Start)
By definition: 5*T(a(n)) = A129557(n)^2 - 1 where triangular number T(j) = j*(j+1)/2. This implies:
Every odd prime factor of a(n) and d(n)=a(n)+1 is present in b(n)=A129557(n)+1 or in c(n)=A129557(n)-1. (End)
From the law of cosines the non-Pythagorean triple {a(n), a(n)+1=A254332(n), A129557(n+1)} forms a near-isosceles triangle whose angle between the consecutive integer sides is equal to the central angle of the regular pentachoron polytope (4-simplex) (see A140244 and A140245). This implies that the terms {a(n)} are also those numbers k such that 1 + 5*A000217(k) is a square. - Federico Provvedi, Apr 04 2023

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A005891 (centered pentagonal numbers), A129557 (numbers k>0 such that k^2 is a centered pentagonal number), A221874.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; this sequence for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

A005891 := proc(n) (5*n^2+5*n+2)/2 ; end: n := 0 : while true do if issqr(A005891(n)) then print(n) ; fi ; n := n+1 ; od : # R. J. Mathar, Jun 06 2007

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[n] ], {n,1,40000} ]
LinearRecurrence[{1,38,-38,-1,1},{0,2,21,95,816},30] (* Harvey P. Dale, Nov 09 2017 *)
Table[(((x^(n+2))+(((-1)^n*(x^(2*n+1)+1)-x)/(x^n)))/(x^2+1)-1)/2/.x->3+Sqrt[10],{n,0,50}]//Round (* Federico Provvedi, Apr 04 2023 *)

a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; 1,-1,-38,38,1]^(n-1)*[0;2;21;95;816])[1,1] \\ Charles R Greathouse IV, Feb 11 2019

For n >= 5, a(n) = 38*a(n-2) - a(n-4) + 18. - Max Alekseyev, May 08 2009
G.f.: x^2*(x^3+2*x^2-19*x-2) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)). - Colin Barker, Feb 21 2013
a(n) = (A221874(n) - 1) / 2. - Bruno Berselli, Feb 21 2013
From Andrea Pinos, Oct 24 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7+2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13+4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).  (End)
a(n) = (x^(2*(n+1)) + (-1)^n*(x^(2*n+1)+1) - x) / (2*x^n*(x^2 + 1)) - (1/2), with x=3+sqrt(10). - Federico Provvedi, Apr 04 2023

More terms from R. J. Mathar, Jun 06 2007
Further terms from Max Alekseyev, May 08 2009
a(22)-a(23) from Colin Barker, Feb 21 2013

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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A285985 Numbers a(n) = (T(b(n)))^2, where T(b(n)) is the triangular number of b(n)= A000217(b(n)) and b(n)=A006451(n). Also a(n) = parameters K of the Bachet Mordell equation y^2=x^3+K, where x= T(b(n)) = A006454(n) and y= T(b(n))* sqrt(T(b(n))+1) = A285955(n).

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A001652 a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.

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A006452 a(n) = 6*a(n-2) - a(n-4).

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A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

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