A006454 -id:A006454 - 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

A217278 Sequences A124174 and A006454 interlaced.

Original entry on oeis.org

0, 0, 1, 3, 10, 15, 45, 120, 351, 528, 1540, 4095, 11935, 17955, 52326, 139128, 405450, 609960, 1777555, 4726275, 13773376, 20720703, 60384555, 160554240, 467889345, 703893960, 2051297326, 5454117903, 15894464365, 23911673955, 69683724540, 185279454480

Offset: 0

Raphie Frank, Sep 29 2012

a(2n) and 2*a(2n) + 1 are triangular.

a(2n + 1) is triangular and a(2n + 1)/2 is the harmonic mean of consecutive triangular numbers (therefore, a(2n + 1) + 1 is square).

			a(18) = 35*(52326 - 1540) + 45 = 1777555,
a(19) = 35*(139128 - 4095) + 120 = 4726275.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,-34,0,-1,0,1).

Cf. (sqrt(8a(2n) + 1) - 1)/2 = A216134(n) = A216162(2n + 1).
Cf. sqrt(a(2n+1) + 1) = A006452(n + 1) = A216162(2n + 2).
Cf. (sqrt(8a(2n+1) + 1) - 1)/2 = A006451(n).

LinearRecurrence[{0,1,0,34,0,-34,0,-1,0,1},{0,0,1,3,10,15,45,120,351,528},40] (* Harvey P. Dale, Aug 04 2019 *)

concat([0,0], Vec(-x^2*(3*x^5+x^4+12*x^3+9*x^2+3*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)) + O(x^100))) \\ Colin Barker, Jun 23 2015

a(n) = 35*(a(n-4) - a(n-8)) + a(n-12).
lim n --> infinity a(2n)/a(2n - 1) = (3 + sqrt(8))/2.
From Raphie Frank, Dec 21 2015: (Start)
a(2*n + 1) = 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.
a(2*n + 2) = 1/2*(3*(a(2*n + 1)) + sqrt((a(2*n + 1)) + 1) * sqrt(8*(a(2*n + 1)) + 1) + 1).
(End)

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).

Original entry on oeis.org

0, 9, 225, 14400, 278784, 16769025, 322382025, 19356600384, 372051201600, 22337675375625, 429347532814209, 25777663981977600, 495466706924481600, 29747402099825117409, 571768151330225342025, 34328476252406392070400, 659819951198501829398784, 39615031848108328736769225, 761431651915943270106720225, 45715712424248689455481003584, 878691466491082103705616000000

Offset: 0

Vladimir Pletser, Apr 30 2017

Numbers a(n) which are the square of triangular number T(b(n)), 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 parameters K of the 3rd degree Diophantine Bachet-Mordell equation y^2=x^3+K, with x= T(b(n)) = A006454(n) and y= T(b(n))* sqrt(T(b(n))+1) = A285955(n).
Also: a(n) = (A000217(A006451(n)))^2 or a(n) = A006454(n)^2.

			For n=2, b(n)=5, a(n)=225.
For n=5, b(n)=90, a(n)= 16769025.
For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120 and (A000217(A006451(n)))^2 = (A000217(15))^2 = (120)^2 = 14400.

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..500
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. A285955, A006454, A000217, A006451, A081119, A054504.

restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print ('0,0','1,9','2,225'); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=(b*(b+1)/2)^2; 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)) (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)^2.
Empirical g.f.: 9*x*(1 + 24*x + 387*x^2 + 864*x^3 + 387*x^4 + 24*x^5 + x^6) / ((1 - x)*(1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 + 6*x + x^2)*(1 + 34*x + x^2)). - Colin Barker, Apr 30 2017

A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

Original entry on oeis.org

0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074, 6437, 17919, 37520, 104442, 218685, 608735, 1274592, 3547970, 7428869, 20679087, 43298624, 120526554, 252362877, 702480239, 1470878640, 4094354882, 8572908965, 23863649055, 49966575152

Offset: 0

N. J. A. Sloane and Jeffrey Shallit

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Problem J/A2, Jul/August 1973 pp. 73-74; solution Jan 1974 (see link).
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, 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
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000124, A006452, A006454, A098586, A098790.

Cf. numbers m such that k*A000217(m)+1 is a square: this sequence for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

a006451 n = a006451_list !! n
a006451_list = 0 : 2 : 5 : 15 : map (+ 2)
   (zipWith (-) (map (* 6) (drop 2 a006451_list)) a006451_list)
-- Reinhard Zumkeller, Jan 10 2012

N:= 100: # to get a(0) to a(N)
A[0]:= 0: A[1]:= 2: A[2]:= 5: A[3]:= 15:
for n from 4 to N do A[n]:= 6*A[n-2] - A[n-4] + 2 od:
seq(A[n],n=0..N); # Robert Israel, Aug 26 2014

LinearRecurrence[{1,6,-6,-1,1},{0,2,5,15,32},30] (* Harvey P. Dale, Jul 17 2013 *)
Select[Range[10^6], IntegerQ@ Sqrt[# (# + 1)/2 + 1] &] (* Michael De Vlieger, Apr 25 2017 *)

for(n=1,10000,t=n*(n+1)/2+1;if(issquare(t), print1(n,", "))) \\ Joerg Arndt, Oct 10 2009

G.f.: x*(-2-3*x+2*x^2+x^3)/(x-1)/(x^2+2*x-1)/(x^2-2*x-1). Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
a(n) = 6*a(n-2) - a(n-4) + 2 with a(0)=0, a(1)=2, a(2)=5, a(3)=15. - Zak Seidov, Apr 15 2008
a(n) = 3*a(n-2) + 4*sqrt((a(n-2)^2 + a(n-2))/2 + 1) + 1 with a(0) = 0, a(1) = 2. - Raphie Frank, Feb 02 2013
a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5); a(0)=0, a(1)=2, a(2)=5, a(3)=15, a(4)=32. - Harvey P. Dale, Jul 17 2013
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n+1) = A098790(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n) = A098586(2*n-1), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(n) = 8*sqrt(T(a(n-2)) + 1) + a(n-4) where T(a(n)) = A000217(a(n)), and a(-1) = -1, a(0)=0, a(1)=2, a(2)=5. - Vladimir Pletser, Apr 29 2017

More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001

Edited by N. J. A. Sloane, Oct 24 2009, following discussions by several correspondents in the Sequence Fans Mailing List, Oct 10 2009

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

A245031 Numbers m such that 3m+1 and 8m+1 are both squares.

Original entry on oeis.org

0, 1, 21, 120, 2080, 11781, 203841, 1154440, 19974360, 113123361, 1957283461, 11084934960, 191793804840, 1086210502741, 18793835590881, 106437544333680, 1841604094101520, 10429793134197921, 180458407386358101, 1022013289607062600

Offset: 1

Bruno Berselli, Jul 15 2014

Naturally, all terms are triangular numbers.
Numbers m such that k*m+1 and 8*m+1 are both squares:
k=1: A006454;
k=3: this sequence;
k=4: A029549;
k=5: 0, 3, 231, 4560, 333336, 6575751, ...
k=6: A200999;
k=7: A157879.
Numbers m such that 3*m+1 and k*m+1 are both squares:
k=1: A045899;
k=2: A045502;
k=4: A059989;
k=5: A159683;
k=6: 8*A029546;
k=7: A160695;
k=8: this sequence.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000217.

Cf. A006454, A029546, A029549, A045502, A045899, A059989, A157879, A159683, A160695, A200999.

I:=[0,1,21,120,2080]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..20]];

LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 21, 120, 2080}, 20] (* or *) CoefficientList[Series[x (1 + 20 x + x^2)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 20}], x]

a[1]:0$ a[2]:1$ a[3]:21$ a[4]:120$ a[5]:2080$ a[n]:=a[n-1]+98*a[n-2]-98*a[n-3]-a[n-4]+a[n-5]$ makelist(a[n], n, 1, 20);

a=vector(20); a[1]=0; a[2]=1; a[3]=21; a[4]=120; a[5]=2080; for(i=6, #a, a[i]=a[i-1]+98*a[i-2]-98*a[i-3]-a[i-4]+a[i-5]); a

G.f.: x^2*(1 + 20*x + x^2)/((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)).
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
G.f. of the quadrisections:
a(4k+1):   40*x*(52 + 3*x)/((1 - x)*(1 - 9602*x + x^2));
a(4k+2): (1 + 2178*x + 21*x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+3): (21 + 2178*x + x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+4): 40*(3 + 52*x)/((1 - x)*(1 - 9602*x + x^2)).

Changed offset from 0 to 1 and adapted formulas by Bruno Berselli, Mar 03 2016

A214838 Triangular numbers of the form k^2 + 2.

Original entry on oeis.org

3, 6, 66, 171, 2211, 5778, 75078, 196251, 2550411, 6666726, 86638866, 226472403, 2943171003, 7693394946, 99981175206, 261348955731, 3396416785971, 8878171099878, 115378189547778, 301596468440091, 3919462027838451, 10245401755863186, 133146330756959526, 348042063230908203

Offset: 1

Alex Ratushnyak, Mar 07 2013

Corresponding k values are in A077241.

Except 3, all terms are in A089982: in fact, a(2) = 3+3 and a(n) = (k-2)*(k-1)/2+(k+1)*(k+2)/2, where k = sqrt(a(n)-2) > 2 for n > 2. [Bruno Berselli, Mar 08 2013]

			2211 is in the sequence because 2211 = 47^2 + 2.

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A000217, A001110, A001219, A006454, A029549, A069017, A077241, A089982, A164055, A165892.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)))); // Bruno Berselli, Mar 08 2013

LinearRecurrence[{1, 34, -34, -1, 1}, {3, 6, 66, 171, 2211}, 25] (* Bruno Berselli, Mar 08 2013 *)

t[n]:=((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4$
makelist(expand(t[n]*(t[n]+1)/2), n, 1, 25); /* Bruno Berselli, Mar 08 2013 */

for(n=1, 10^9, t=n*(n+1)/2; if(issquare(t-2), print1(t,", "))); \\ Joerg Arndt, Mar 08 2013

import math
for i in range(2, 1<<32):
      t = i*(i+1)//2 - 2
      sr = int(math.sqrt(t))
      if sr*sr == t:
          print(f'{sr:10} {i:10} {t+2}')

G.f.: -3*x*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)). - Joerg Arndt, Mar 08 2013

a(n) = A000217(t), where t = ((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4. - Bruno Berselli, Mar 08 2013

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

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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A217278 Sequences A124174 and A006454 interlaced.

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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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A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

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

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A245031 Numbers m such that 3*m+1 and 8*m+1 are both squares.

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A245031 Numbers m such that 3m+1 and 8m+1 are both squares.