login
This site is supported by donations to The OEIS Foundation.

 

Logo

Many excellent designs for a new banner were submitted. We will use the best of them in rotation.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A001110 a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
(Formerly M5259 N2291)
43
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

These are the numbers that are both triangular and square.

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

It appears that 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 4-th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

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, pp. 16-17, Freeman 1988

J. H. Silverman, A Friendly Introduction to Number Theory, p 196, Prentice Hall 2001

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

T. D. Noe, Table of n, a(n) for n=0..100

Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares<:a>, The Mathematical Gazette, 2002, pp. 423--431, esp pp. 424--426.

K. S. Brown, Square Triangular Numbers

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.

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.

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

D. A. Q., Triangular square numbers - a postscript, Math. Gaz., 56 (1972), 311-314.

K. Ramsey, Re_Generalized_Proof_re_Square_Triangular_Numbers

R. Stephan, Boring proof of a nonlinearity

Eric Weisstein's World of Mathematics, Square Triangular Number

Eric Weisstein's World of Mathematics, Triangular Number

Wikipedia, Square triangular number

Index to sequences with linear recurrences with constant coefficients, signature (35,-35,1).

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

FORMULA

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

As n goes to infinity the ratio a(n+1)/a(n) goes to 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 more 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

a(n) = 35(a(n-1)-a(n-2)) + a(n-3); a(n) = -1/16 +((-24+17*2^(1/2))/2^(11/2))*(17-12*2^(1/2))^(n-1) +((24+17*2^(1/2))/2^(11/2))*(17+12*2^(1/2))^(n-1). - Antonio Alberto Olivares, Nov 07 2003

a(n+1)=[17*A029547(n)-A091761(n)-1]/16. - R. J. Mathar, Nov 16 2007

a(n) = A001333^2 * A000129^2 = A000129[2n]^2/4 = binom(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[2n 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(2n+1) = (A002315(n))^2*(((A002315(n))^2)+1)/2. - Ivan N. Ianakiev, Oct 10 2012

a(2n+1) = ((sqrt(a^2 + (a+1)^2))*(2a+1))^2, where a = (A002315(n)-1)(1/2). - Ivan N. Ianakiev, Nov 01 2012

a(2n) = A001333(2n)^2 * (A001333(2n)^2 - 1)/2, and  a(2n+1) = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2.  The later 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 2X2 matrix T(n, M), where M is the 2X2 matrix [0, -17; 1, 18].

See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4-th order linear divisibility sequences. (End)

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 sixth 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[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 *)

PROG

(PARI) 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

(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

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.

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.

Sequence in context: A004294 A075760 A113938 * A064196 A060786 A162850

Adjacent sequences:  A001107 A001108 A001109 * A001111 A001112 A001113

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified April 17 20:01 EDT 2014. Contains 240655 sequences.