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A029549
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a(0) = 0, a(1) = 6, a(2) = 210; for n >= 0, a(n+3) = 35*a(n+2) - 35*a(n+1) + a(n).
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17
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0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Triangular numbers that are twice other triangular numbers. - Don N. Page
Triangular numbers that are also pronics. These will be shown to have a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Mar 09 2002
In other words, triangular numbers which are products of two consecutive numbers. E.g. a(2)=210: 210 is a triangular number which is the product of two consecutive numbers: 14*15.- Shyam Sunder Gupta (guptass(AT)rediffmail.com), Oct 26 2002
Coefficients of the series giving the best rational approximations to sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2;1,4,1], [2;1,4,1,4,1], [2;1,4,1,4,1,4,1], [2;1,4,1,4,1,4,1,4,1] and so forth. - Gene Ward Smith (genewardsmith(AT)gmail.com), Sep 30 2006
The a(n) satisfy the same recurrence relation that defines the terms of A165518 [Ant King 13 Dec 2010].
Intersection of A000217 and A002378.
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LINKS
| Reinhard Zumkeller, Table of n, a(n) for n = 0..100
Shyam Sunder Gupta Fascinating Triangular Numbers
Index to sequences with linear recurrences with constant coefficients, signature (35,-35,1).
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FORMULA
| G.f.: 6x/(1-35*x+35*x^2-x^3) = -6*x / ( (x-1)*(x^2-34*x+1) ).
a(n) = 6*A029546(n-1) = 2*A075528(n).
a(n) = -3/16 + (3/32+1/16*2^(1/2)) *(17+12*2^(1/2))^n + (3/32-1/16*2^(1/2)) *(17-12*2^(1/2))^n. - Gene Ward Smith (genewardsmith(AT)gmail.com), Sep 30 2006
a(n) = cosh((4*n+2)*log(1+sqrt(2)))-3)/16. a(n) = binomial(A001652+1,2) = 2*binomial(A053141+1,2). - Bill Gosper, Feb 07 2010
a(n) = Binomial(A001652(n), 2) = A000217(A001652(n)). - Mitch Harris, Apr 19 2007, R. J. Mathar, Jun 26 2009
a(n)= ceil((3+2sqrt(2))^(2n+1)-6)/32 =floor(1/32 (1+sqrt(2))^(4n+2)) [Ant King 13 Dec 2010].
sum_{n>=1} 1/a(n) = 3-2*sqrt(2) = A157259-4. [Ant King 13 Dec 2010].
a(n) = a(n-1)+A001109(2n). - Charlie Marion, Feb 10 2011
a(n+2)=34*a(n+1)-a(n)+6. - Charlie Marion, Feb 11 2011
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MATHEMATICA
| CoefficientList[Series[6/(1 - 35*x + 35*x^2 - x^3), {x, 0, 14}], x]
Intersection[ #, 2*# ]&@Table[ Binomial[ n, 2 ], {n, 999999} ] - Bill Gosper, Feb 07 2010
Table[ Floor[ N[ (Sqrt[ 2 ]+1)^(4n+2)/32 ]], {n, 0, 20} ]
LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* From Harvey P. Dale, June 06 2011 *)
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PROG
| (Macsyma) (makelist(binom(n, 2), n, 1, 999999), intersection(%%, 2*%%)) - Bill Gosper, Feb 07 2010
(Haskell)
a029549 n = a029549_list !! n
a029549_list = [0, 6, 210] ++
zipWith (+) a029549_list
(map (* 35) $ tail delta)
where delta = zipWith (-) (tail a029549_list) a029549_list
-- Reinhard Zumkeller, Sep 19 2011
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CROSSREFS
| Cf. A123478, A123479, A123480, A123482.
Cf. A075528. Cf. A082405 (first differences).
Sequence in context: A084694 A065945 A076715 * A183252 A183287 A087639
Adjacent sequences: A029546 A029547 A029548 * A029550 A029551 A029552
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KEYWORD
| nonn
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AUTHOR
| Don N. Page (don(AT)phys.ualberta.ca)
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EXTENSIONS
| Additional comments from Christian G. Bower (bowerc(AT)usa.net), Sep 19 2002; Shyam Sunder Gupta (guptass(AT)rediffmail.com), Oct 26 2002; T. D. Noe (noe(AT)sspectra.com), Nov 07 2006; and others.
Edited by N. J. A. Sloane (njas(AT)research.att.com), Apr 18 2007, following suggestions from Andrew Plewe and Tanya Khovanova.
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