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 A001110 Square triangular numbers: numbers that are both triangular and square. (Formerly M5259 N2291) 81
 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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 REFERENCES A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193. 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). LINKS 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. 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. Nelson, 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 E 1473, 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, 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). R. Stephan, Boring proof of a nonlinearity Michel Waldschmidt, Continued fractions, Ecole 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). FORMULA 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 Problem A of Nieuw Archief voor Wiskunde (URL www.math.leidenuniv.nl/~naw/serie5/deel05/dec2004/pdf/uwc.pdf is no longer valid). Since Feb 01 2005 (submission deadline) a solution can be found at http://www.jaapspies.nl/mathfiles/problem2004-4A.pdf - 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 EXAMPLE 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. MAPLE 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 MATHEMATICA 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]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *) Transpose[NestList[Flatten[{Rest[#], 34Last[#]-First[#]+2}]&, {0, 1}, 20]][]  (* 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 True (* Bill Gosper, Jul 20 2015 *) PROG (PARI) a=vector(100); a=1; a=36; for(n=3, #a, a[n]=34*a[n-1]-a[n-2]+2); a \\ Charles R Greathouse IV, Jul 25 2011 (Haskell) a001110 n = a001110_list !! n a001110_list = 0 : 1 : (map (+ 2) \$    zipWith (-) (map (* 34) (tail a001110_list)) a001110_list) -- Reinhard Zumkeller, Oct 12 2011 (MIT/GNU Scheme, 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)))))) ;; The following two are 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 (Magma) [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 CROSSREFS 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. Sequence in context: A004294 A075760 A113938 * A331234 A278806 A064196 Adjacent sequences:  A001107 A001108 A001109 * A001111 A001112 A001113 KEYWORD nonn,easy,nice AUTHOR STATUS approved

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Last modified September 26 02:05 EDT 2022. Contains 356986 sequences. (Running on oeis4.)