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Search: id:A066770
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| A066770 |
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5^n sin(2n arctan(1/2)) or numerator of tan(2n arctan(1/2)). |
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+0 7
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| 4, 24, 44, -336, -3116, -10296, 16124, 354144, 1721764, 1476984, -34182196, -242017776, -597551756, 2465133864, 29729597084, 116749235904, -42744511676, -3175197967656, -17982575014036, -28515500892816, 278471369994004, 2383715742284424, 7340510203856444
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Four times the numerators of g.f. 1/(1-6x/5+x^2). - Ralf Stephan, Jun 12 2003
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REFERENCES
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J. M. Borwein and R. Girgensohn, Addition theorems and binary expansions, Canadian J. Math. 47 (1995) 262-273.
E. Eckert, The group of primitive Pythagorean triangles, Mathematics Magazine 57 (1984) 22-27.
S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 430-433.
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LINKS
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S. R. Finch, Plouffe's Constant
S. Plouffe, The Computation of Certain Numbers Using a Ruler and Compass, J. Integer Seqs. Vol. 1 (1998), #98.1.3.
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FORMULA
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A(n)=5^n sin(2n arctan(1/2)). A recursive formula for T(n) = tan(2n arctan(1/2)) is T(n+1)=(4/3+T(n))/(1-4/3*T(n)). Unsigned A(n) is the absolute value of numerator of T(n)
a(n) is the imaginary part of (2+I)^(2n) = sum(k=0, n, 2^(2*n-2*k-1)*(-1)^k*binomial(2*n, 2*k+1) ) - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 03 2002
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MAPLE
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a[1] := 4/3; for n from 1 to 40 do a[n+1] := (4/3+a[n])/(1-4/3*a[n]):od: seq(abs(numer(a[n])), n=1..40); # a[n]=tan(2n arctan(1/2))
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PROGRAM
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(PARI) a(n)=if(n<0, 0, imag((2+I)^(2*n))
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CROSSREFS
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Cf. A066771 5^n cos(2n arctan(1/2)), A000351 powers of 5 and also hypotenuse of right triangle with legs given by A066770 and A066771.
Note that A066770, A066771 and A0000351 are primitive Pythagorean triples with hypotenuse 5^n. The offset of A0000351 is zero, but the offset is 1 for A066770, A066771.
Adjacent sequences: A066767 A066768 A066769 this_sequence A066771 A066772 A066773
Sequence in context: A120622 A031117 A139245 this_sequence A080380 A039935 A090821
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KEYWORD
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sign,easy,frac
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AUTHOR
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Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002
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