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Search: id:A001110
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| A001110 |
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a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34a(n-1) - a(n-2) + 2. (Formerly M5259 N2291)
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+0 25
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| 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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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 (r.rosenthal(AT)web.de), 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 (noe(AT)sspectra.com), Apr 30 2007
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..100
Index entries for sequences related to linear recurrences with constant coefficients
K. S. Brown, Square Triangular Numbers
Yong-Gao Chen and Jin-Hui Fang, Triangular numbers in geometric progression, INTEGERS 7 (2007), #A19.
S. 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.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
R. Stephan, Boring proof of a nonlinearity
Chris Thatcher, Square Triangular Numbers [Broken link?]
Eric Weisstein, CRC Online Dictionary, Square Triangular Number
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Wikipedia, Square triangular number
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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, circa Mar 13 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2) - Lekraj Beedassy (blekraj(AT)yahoo.com), 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 (j.spies(AT)hccnet.nl), 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 (olivares14031(AT)yahoo.com), Nov 07 2003
a(n+1)=[17*A029547(n)-A091761(n)-1]/16. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129[2n]^2/4 = binom(A001108,2). - R. W. Gosper (rwg(AT)sdf.lonestar.org), Jul 28 2008
Comment from R. W. Gosper (rwg(AT)sdf.lonestar.org), Jul 25 2008: Closed form (as square = triangular): ( (sqrt(2) + 1)^(2n)/(4 sqrt(2)) - (1 - sqrt(2))^(2n)/(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.
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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)); [S. Plouffe in his 1992 dissertation.]
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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. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 27 2008]
Sequence in context: A004294 A075760 A113938 this_sequence A064196 A060786 A063819
Adjacent sequences: A001107 A001108 A001109 this_sequence A001111 A001112 A001113
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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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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