A182334 -id:A182334 - cp's OEIS frontend

A001110 Square triangular numbers: numbers that are both triangular and square.

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100

Offset: 0

N. J. A. Sloane

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004
For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007
The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012
Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012
For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013
For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014
The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014
The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

			a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 204.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-259.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.

Kade Robertson, Table of n, a(n) for n = 0..600 [This replaces an earlier b-file computed by T. D. Noe]
Nikola Adžaga, Andrej Dujella, Dijana Kreso, and Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares, The Mathematical Gazette, 2002, pp. 423-431, esp pp. 424-426.
K. S. Brown, Square Triangular Numbers.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
Yong-Gao Chen and Jin-Hui Fang, Triangular numbers in geometric progression, INTEGERS 7 (2007), #A19.
F. Javier de Vega, On the parabolic partitions of a number, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
Colin Dickson et al., ratio of integers = sqrt(2), thread in newsgroup alt.math.recreational, March 7, 2004.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
Bill Gosper, The Triangular Squares, 2014.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020. Mentions this sequence.
Gillian Hatch, Pythagorean Triples and Triangular Square Numbers, Mathematical Gazette 79 (1995), 51-55.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Roger B. Nelsen, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
A. Nowicki, The numbers a^2+b^2-dc^2, J. Int. Seq. 18 (2015) # 15.2.3.
Matthew Parker, The first 5000 square triangular numbers (7-Zip compressed file).
J. L. Pietenpol, A. V. Sylwester, E. Just, and R. M. Warten, Problem E1473, Amer. Math. Monthly, Vol. 69, No. 2 (Feb. 1962), pp. 168-169. (From the editorial note on p. 169 of this source, we learn that the question about the existence of perfect squares in the sequence of triangular numbers cropped up in the Euler-Goldbach Briefwechsel of 1730; the translation into English of the relevant letters can be found at Correspondence of Leonhard Euler with Christian Goldbach (part II), pp. 614-615.) - _José Hernández_, May 24 2022
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
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
D. A. Q., Triangular square numbers: a postscript, Math. Gaz., 56 (1972), 311-314.
K. Ramsey, Re_Generalized_Proof_re_Square_Triangular_Numbers. [Broken link]
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Fabio Roman, On certain ratios regarding integer numbers which are both triangulars and squares, arXiv:1703.06701 [math.NT], 2017.
Jon E. Schoenfield, Prime factorization of a(n) for n = 1..300.
Jaap Spies, A Bit of Math, The Art of Problem Solving, Jaap Spies Publishers (2019).
Ralf Stephan, Boring proof of a nonlinearity.
UWC, Problem A, Nieuw Archief voor Wiskunde, Dec 2004; Jaap Spies, Solution.
Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Square triangular number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001108, A001109.
Other S_r type sequences are S_4=A000290, S_5=A004146, S_7=A054493, S_8=A001108, S_9=A049684, S_20=A049683, S_36=this sequence, S_49=A049682, S_144=A004191^2.
Cf. A001014; intersection of A000217 and A000290; A010052(a(n))*A010054(a(n)) = 1.
Cf. A005214, A054686, A232847 and also A233267 (reveals an interesting divisibility pattern for this sequence).
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
See A229131, A182334, A299921 for near-misses.

a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011

[n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015

a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq(A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); # Simon Plouffe in his 1992 dissertation

f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]],AppendTo[lst,f[n]]],{n,0,10!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
Transpose[NestList[Flatten[{Rest[#],34Last[#]-First[#]+2}]&, {0,1},20]][[1]]  (* Harvey P. Dale, Mar 25 2011 *)
LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *)
LinearRecurrence[{6,-1},{0,1},20]^2 (* Harvey P. Dale, Oct 22 2012 *)
(* Square = Triangular = Triangular = A001110 *)
ChebyshevU[#-1,3]^2==Binomial[ChebyshevT[#/2,3]^2,2]==Binomial[(1+ChebyshevT[#,3])/2,2]=={1,36,1225,41616,1413721}[[#]]&@Range[5]
True (* Bill Gosper, Jul 20 2015 *)
L=0;r={};Do[AppendTo[r,L];L=1+17*L+6*Sqrt[L+8*L^2],{i,1,19}];r (* Kebbaj Mohamed Reda, Aug 02 2023 *)

a=vector(100);a[1]=1;a[2]=36;for(n=3,#a,a[n]=34*a[n-1]-a[n-2]+2);a \\ Charles R Greathouse IV, Jul 25 2011

;; With memoizing definec-macro from Antti Karttunen's IntSeq-library.
(definec (A001110 n) (if (< n 2) n (+ 2 (- (* 34 (A001110 (- n 1))) (A001110 (- n 2))))))
;; Antti Karttunen, Dec 06 2013

;; For testing whether n is in this sequence:
(define (inA001110? n) (and (zero? (A068527 n)) (inA001109? (floor->exact (sqrt n)))))
(define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
;; Antti Karttunen, Dec 06 2013

a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001
a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006
a(n) = A001109(n)^2 = A001108(n)*(A001108(n)+1)/2 = (A000129(n)*A001333(n))^2 = (A000129(n)*(A000129(n) + A000129(n-1)))^2. - Henry Bottomley, Apr 19 2000
a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004
From Antonio Alberto Olivares, Nov 07 2003: (Start)
a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
a(n+1) = (17*A029547(n) - A091761(n) - 1)/16. - R. J. Mathar, Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129(2*n)^2/4 = binomial(A001108,2). - Bill Gosper, Jul 28 2008
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008
a(n) = (1/8)*(sinh(2*n*arcsinh(1)))^2. - Artur Jasinski, Feb 10 2010
a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011
a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011
a(2*n+1) = A002315(n)^2*(A002315(n)^2 + 1)/2. - Ivan N. Ianakiev, Oct 10 2012
a(2*n+1) = ((sqrt(t^2 + (t+1)^2))*(2*t+1))^2, where t = (A002315(n) - 1)/2. - Ivan N. Ianakiev, Nov 01 2012
a(2*n) = A001333(2*n)^2 * (A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * (A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013
a(n) = (A001542(n)/2)^2.
From Peter Bala, Apr 03 2014: (Start)
a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = A096979(2*n-1) for n > 0. - Ivan N. Ianakiev, Jun 21 2014
a(n) = (6*sqrt(a(n-1)) - sqrt(a(n-2)))^2. - Arkadiusz Wesolowski, Apr 06 2015
From Daniel Poveda Parrilla, Jul 16 2016 and Sep 21 2016: (Start)
a(n) = A000290(A002965(2*n)*A002965(2*n + 1)) (after Hugh Darwen).
a(n) = A000217(2*(A000129(n))^2 - (A000129(n) mod 2)).
a(n) = A000129(n)^4 + Sum_{k=0..(A000129(n)^2 - (A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers).
a(n) = A002965(2*n)^4 + Sum_{k=A002965(2*n)^2..A002965(2*n)*A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End)
E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016

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

A001110 Square triangular numbers: numbers that are both triangular and square.

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

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A299921 Squares that differ from a triangular number by 1.

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