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A001110
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a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
(Formerly M5259 N2291)
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38
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0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625
(list;
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refs;
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history;
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internal format)
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OFFSET
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0,3
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COMMENTS
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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 March 7th 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
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
Tom Beldon and Tony Gardiner, ``Triangular numbers and perfect squares'', The Mathematical Gazette, 2002, pp. 423--431, esp pp. 424--426.
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.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
Martin Gardner, Time Travel and other Mathematical Bewilderments, pp. 16-17, Freeman 1988
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
D. A. Q., Triangular square numbers - a postscript, Math. Gaz., 56 (1972), 311-314.
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).
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..100
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
_Simon Plouffe_, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
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).
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FORMULA
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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 7 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)=A001109(2n-1). - 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 http://www.math.leidenuniv.nl/~naw/serie5/deel05/dec2004/pdf/uwc.pdf After 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 A. 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
Comment from Bill Gosper, Jul 25 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.
a(n)=(1/8) (Sinh[2n ArcSinh[1]])^2 [From 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 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
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EXAMPLE
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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
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MAPLE
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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); (Spies)
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); [Simon Plouffe in his 1992 dissertation.]
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MATHEMATICA
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f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]], AppendTo[lst, f[n]]], {n, 0, 10!}]; lst (* Vladimir 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 *)
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PROG
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(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
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CROSSREFS
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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.
Sequence in context: A004294 A075760 A113938 * A064196 A060786 A162850
Adjacent sequences: A001107 A001108 A001109 * A001111 A001112 A001113
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KEYWORD
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nonn,easy,nice,changed
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000
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STATUS
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approved
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