A046176 -id:A046176 - 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

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025

Offset: 0

N. J. A. Sloane

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Muniru A Asiru, Table of n, a(n) for n = 0..100
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes, and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Steven C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018

I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)

a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))

vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Entry edited by N. J. A. Sloane, Sep 14 2007

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

Original entry on oeis.org

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461

Offset: 1

Paul Weisenhorn, Feb 21 2009

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n)+1 = a(n)^2,
   (A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
  A   a-sequence    b-sequence     c-sequence
  1     A001653       A002315(n-1)   A078522
  2     A072256       A054320(n-1)   A045502(n-1)
  3     A001570       A028230        A059989(n-1)
  4     A007805       A049629(n-1)   A157459
  5  -> A157014 <-    A133283        A157460
  6     A153111       A157461        A157874
  7     A157877       A157878        A157879
  8     A077420(n-1)  A046176        A157880
  9     A097315(n-1)  A097314(n-1)   A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. similar sequences listed in A238379.

a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014

seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020

CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041049 *)
a[30, 20] (* Gerry Martens, Jun 07 2015 *)
Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)

Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012

[chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Edited by Alois P. Heinz, Sep 09 2011

A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

Original entry on oeis.org

1, 33, 1121, 38081, 1293633, 43945441, 1492851361, 50713000833, 1722749176961, 58522759015841, 1988051057361633, 67535213191279681, 2294209197446147521, 77935577499977736033, 2647515425801796877601

Offset: 0

Wolfdieter Lang, Nov 29 2002

(3*a(n))^2 - 2*(2*b(n))^2 = 1 with companion sequence b(n)= A046176(n+1), n>=0 (special solutions of Pell equation).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas Numbers, INTEGERS 6 (2006) #A15
Tanya Khovanova, Recursive Sequences
S. Vidhyalakshmi, V. Krithika, and K. Agalya, On The Negative Pell Equation y^2 = 72x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER) Volume 4, Issue 2, February (2016).
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A056771 (even part).
Row 34 of array A094954.
Row 3 of array A188646.
Cf. similar sequences listed in A238379.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329. This is the case k=3.

I:=[1,33]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

LinearRecurrence[{34,-1},{1,33},20] (* Vincenzo Librandi, Nov 22 2011 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041027 *)
a[18, 20] (* Gerry Martens, Jun 07 2015 *)

makelist(expand(((1+sqrt(2))^(4*n+2)+(1-sqrt(2))^(4*n+2))/6),n,0,14);  /* _Bruno Berselli, Nov 22 2011 */

Vec((1-x)/(1-34*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Nov 22 2011

a(n) = 34*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = T(2*n+1, 3)/3 = S(n, 34) - S(n-1, 34), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 34)= A029547(n), T(n, 3)=A001541(n).
G.f.: (1-x)/(1-34*x+x^2).
a(n) = sqrt(8*A046176(n+1)^2 + 1)/3.
a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(2*n)-a(n-1), where k = (sqrt(2)+1)^4 = 17+12*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
a(n) = a(-n-1) = A029547(n)-A029547(n-1) = ((1+sqrt(2))^(4n+2)+(1-sqrt(2))^(4n+2))/6. - Bruno Berselli, Nov 22 2011

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Joseph L. Pe, Jan 07 2003

Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176.
Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016
From Sergey Pavlov, Mar 14 2017: (Start)
The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

From Benoit Cloitre, Jan 19 2003: (Start)
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
a(n) = A008844(n) - 1. (End)
From R. J. Mathar, Sep 21 2011: (Start)
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = 24*A029546(n-2). (End)
a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020
From Amiram Eldar, Dec 02 2024: (Start)
a(n) = A001542(n-1)*A001542(n).
Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Edited by Bruno Berselli, Nov 17 2016

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844

Offset: 0

Amarnath Murthy, Jun 08 2003

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Edited and extended by Robert G. Wilson v, Jun 15 2003

A008843 Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.

Original entry on oeis.org

1, 49, 1681, 57121, 1940449, 65918161, 2239277041, 76069501249, 2584123765441, 87784138523761, 2982076586042449, 101302819786919521, 3441313796169221281, 116903366249966604049, 3971273138702695316401, 134906383349641674153601, 4582845760749114225906049, 155681849482120242006652081

Offset: 0

N. J. A. Sloane

Also indices of triangular numbers (A000217) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 16 2015

a(n)-th triangular number is a square; subsequence of A001108. - Jaroslav Krizek, Aug 05 2016

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.
P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 288.
P. F. Teilhet, Note #2094, L'Intermédiaire des Mathématiciens, 10 (1903), pp. 235-238.

M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
Index entries for sequences related to Bernoulli numbers.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A002315, A016754, A146313, A253826.

LinearRecurrence[{35,-35,1},{1,49,1681},17] (* Stefano Spezia, Aug 17 2024 *)

a(n) = 34*a(n-1) - a(n-2) + 16 = A002315(n)^2 = 2*A001653(n)^2 - 1 = 2*A008844(n) - 1 = floor(A046176(n)*sqrt(2)) = 6*A055792(n+1) - a(n-1) + 4 = (6*A055792(n+2) + a(n-1) - 20)/35. - Henry Bottomley, Nov 13 2001
a(n) = A001108(2n+1). - Ira M. Gessel, Nov 05 2014
a(n) = Sum_{k=1..2*n+1} 2^(k-1)*binomial(4*n+2, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 03 2003
O.g.f.: -(1+14*x+x^2)/((-1+x)*(1-34*x+x^2)). - R. J. Mathar, Nov 23 2007
a(n) = -(cosh((2*n - 1)*arctanh(sqrt(2))))^2 = -1 - (sinh((2*n - 1)*arctanh(sqrt(2))))^2. - Artur Jasinski, Oct 30 2008
a(n) = Sum_{k=0..4n+1} A000129(k), see Santana and Diaz-Barrero link at A002315. - Ivan N. Ianakiev, Jul 15 2022

a(14)-a(17) from Stefano Spezia, Aug 17 2024

A046177 Squares (A000290) which are also hexagonal numbers (A000384).

Original entry on oeis.org

1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481

Offset: 1

Eric W. Weisstein

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007
Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011
Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015
Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Colin Barker, Table of n, a(n) for n = 1..327
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (1155,-1155,1).

Cf. A008844, A046176, A253826.
Cf. A001110 (Numbers that are both triangular and square).
Cf. A000290, A000384, A016754, A253826.

LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)

Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015

a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007
a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016
a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016

A336624 Triangular numbers that are one-eighth of other triangular numbers; T(t) such that 8*T(t)=T(u) for some u where T(k) is the k-th triangular number.

Original entry on oeis.org

0, 15, 66, 17391, 76245, 20069280, 87986745, 23159931810, 101536627566, 26726541239541, 117173180224500, 30842405430498585, 135217748442445515, 35592109140254127630, 156041164529401899891, 41073263105447832786516, 180071368649181350028780, 47398510031577658781511915

Offset: 0

Vladimir Pletser, Aug 07 2020

The triangular numbers T(t) that are one-eighth of other triangular numbers T(u) : T(t)=T(u)/8. The t's are in A336623, the T(u)'s are in A336626 and the u's are in A336625.

Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(1)= 15 is a term because it is triangular and 8*15 = 120 is also triangular.
a(2) = 1154*a(0) - a(-2) + 81 = 0 - 15 + 81 = 66;
a(3) = 1154*a(1) - a(-1) + 81 = 1154*15 - 0 + 81 = 17391, etc.

Vladimir Pletser, Table of n, a(n) for n = 0..650
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, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Subsequence of A000217.
Cf. A336623, A336626, A336625.
Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400.

f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 81, a(1) = 15, a(0) = 0, a(-1) = 0, a(-2) = 15}, a(n), remember): map(f, [$ (0 .. 40)])[]; #

LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 15, 66, 17391, 76245}, 18] (* Amiram Eldar, Aug 08 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(4*n + 2)*(11 - 6*(-1)^n*Sqrt[2]) + (Sqrt[2] - 1)^(4*n + 2)*(11 + 6*(-1)^n*Sqrt[2]) - 18)/256, {n, 0, 17}]] (* Vaclav Kotesovec, Sep 08 2020 *)
Select[Accumulate[Range[0, 10^6]]/8, OddQ[Sqrt[8 # + 1]] &] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Jan 15 2024 *)

concat(0, Vec(3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)) + O(x^40))) \\ Colin Barker, Aug 08 2020

a(n) = 1154*a(n-2) - a(n-4) + 81, for n>=2 with a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(2)=66, a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = b(n)*(b(n)+1)/2 where b(n) is A336623(n).
G.f.: 3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)). - Colin Barker, Aug 08 2020
a(n) = ((sqrt(2) + 1)^(4*n + 2) * (11 - 6*(-1)^n*sqrt(2)) + (sqrt(2) - 1)^(4*n + 2) * (11 + 6*(-1)^n*sqrt(2)) - 18)/256. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11 - 6*sqrt(2))*(1 + sqrt(2))^(4n + 2) + (11 + 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for even n.
a(n) = ((11 + 6*sqrt(2))*(1 + sqrt(2) )^(4n + 2) + (11 - 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for odd n. (End)
128*a(n) = -9+33*A077420(n)-24*(-1)^n*A046176(n+1). - R. J. Mathar, May 05 2023

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