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A040000
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a(0)=1; a(n)=2 for n >= 1.
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165
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1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
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OFFSET
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0,2
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COMMENTS
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Continued fraction expansion of sqrt(2) is 1 + 1/(2 + 1/(2 + 1/(2 + ...))).
Inverse binomial transform of Mersenne numbers A000225(n+1) = 2^(n+1) - 1. - Paul Barry, Feb 28 2003
A Chebyshev transform of 2^n: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)). - Paul Barry, Oct 31 2004
An inverse Catalan transform of A068875 under the mapping g(x)->g(x(1-x)). A068875 can be retrieved using the mapping g(x)->g(xc(x)), where c(x) is the g.f. of A000108. A040000 and A068875 may be described as a Catalan pair. - Paul Barry, Nov 14 2004
Let m=2. We observe that a(n) = Sum_{k=0..floor(n/2)} binomial(m,n-2*k). Then there is a link with A113311 and A115291: it is the same formula with respectively m=3 and m=4. We can generalize this result with the sequence whose g.f. is given by (1+z)^(m-1)/(1-z). - Richard Choulet, Dec 08 2009
With offset 1: number of permutations where |p(i) - p(i+1)| <= 1 for n=1,2,...,n-1. This is the identical permutation and (for n>1) its reversal.
Equals INVERT transform of bar(1, 1, -1, -1, ...).
With offset 1: minimum cardinality of the range of a periodic sequence with (least) period n. Of course the range's maximum cardinality for a purely periodic sequence with (least) period n is n. - Rick L. Shepherd, Dec 08 2014
With offset 1: n*a(1) + (n-1)*a(2) + ... + 2*a(n-1) + a(n) = n^2. - Warren Breslow, Dec 12 2014
With offset 1: decimal expansion of gamma(4) = 11/9 where gamma(n) = Cp(n)/Cv(n) is the n-th Poisson's constant. For the definition of Cp and Cv see A272002. - Natan Arie Consigli, Sep 11 2016
a(n) equals the number of binary sequences of length n where no two consecutive terms differ. Also equals the number of binary sequences of length n where no two consecutive terms are the same. - David Nacin, May 31 2017
a(n) is the period of the continued fractions for sqrt((n+2)/(n+1)) and sqrt((n+1)/(n+2)). - A.H.M. Smeets, Dec 05 2017
Also, number of self-avoiding walks and coordination sequence for the one-dimensional lattice Z. - Sean A. Irvine, Jul 27 2020
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REFERENCES
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A. Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
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LINKS
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FORMULA
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a(n) = 2 - 0^n; a(n) = Sum_{k=0..n} binomial(1, k). - Paul Barry, Oct 16 2004
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*2^(n-2*k)/(n-k). - Paul Barry, Oct 31 2004
Euler transform of length 2 sequence [2, -1]. - Michael Somos, Apr 16 2007
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = (u-v)*(u+v) - 2*v*(u-w). - Michael Somos, Apr 16 2007
a(n) = a(-n) for all n in Z (one possible extension to n<0). - Michael Somos, Apr 16 2007
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EXAMPLE
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sqrt(2) = 1.41421356237309504... = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + ...)))). - Harry J. Smith, Apr 21 2009
G.f. = 1 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 2*x^8 + ...
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MAPLE
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Digits := 100: convert(evalf(sqrt(2)), confrac, 90, 'cvgts'):
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MATHEMATICA
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PROG
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(PARI) allocatemem(932245000); default(realprecision, 21000); x=contfrac(sqrt(2)); for (n=0, 20000, write("b040000.txt", n, " ", x[n+1])); \\ Harry J. Smith, Apr 21 2009
(Haskell)
a040000 0 = 1; a040000 n = 2
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CROSSREFS
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See A003945 etc. for (1+x)/(1-k*x).
Prod_{0<=k<=n} a(k) = A000079(n). (End)
Cf. A000674 (boustrophedon transform).
Other continued fractions for sqrt(a^2+1) = (a, 2a, 2a, 2a....): A040002 (contfrac(sqrt(5)) = (2,4,4,...)), A040006, A040012, A040020, A040030, A040042, A040056, A040072, A040090, A040110 (contfrac(sqrt(122)) = (11,22,22,...)), A040132, A040156, A040182, A040210, A040240, A040272, A040306, A040342, A040380, A040420 (contfrac(sqrt(442)) = (21,42,42,...)), A040462, A040506, A040552, A040600, A040650, A040702, A040756, A040812, A040870, A040930 (contfrac(sqrt(962)) = (31,62,62,...)).
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KEYWORD
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AUTHOR
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STATUS
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approved
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