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A124647
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a(n) = (2n + 1)*3^n.
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10
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1, 9, 45, 189, 729, 2673, 9477, 32805, 111537, 373977, 1240029, 4074381, 13286025, 43046721, 138706101, 444816117, 1420541793, 4519905705, 14334558093, 45328197213, 142958160441, 449795187729, 1412147682405, 4424729404869, 13839047287569, 43211719081593, 134718888901437
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OFFSET
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0,2
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COMMENTS
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1 - 1/9 + 1/45 - 1/189 + ... = Pi/(2*sqrt(3)) = A093766. [Jolley eq 271].
If X_1,X_2,...,X_n are 3-blocks of a (4n+1)-set X then, for n>=1, a(n) is the number of (n+1)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 23 2007
Sum_{k>=0} 1/a(k) = log(2+sqrt(3))*sqrt(3)/2 = 1.1405189944... - Jaume Oliver Lafont, Nov 30 2009
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REFERENCES
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L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 50
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LINKS
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FORMULA
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a(n) = 6*a(n-1) - 9*a(n-2) for n > 1; a(0) = 1, a(1) = 9. - Klaus Brockhaus, Sep 23 2009
a(n) = Sum_{i=1..2*3^n-1} gcd(i,2*3^n) = A018804(2*3^n) -2*3^n. This is an application of the multiplicative property of the gcd sum-function A018804. So we get: 2*3^0 * phi(3^n) + ... + 2*3^(n-1) * phi(3^1) + 2*3^n * phi(3^0)+3^0 * phi(2*3^n) + ... + 3^n * phi(2*3^0) - gcd(2*3^n,2*3^n) = a(n), where phi=A000010 is Euler's totient. A general formula is Sum_{i=1..2*p^n-1} gcd(i,2*p^n) = n*3*p^n * n - 3*n*p^(n-1) + p^n, for p an odd prime. This sequence correspondes to p=3. - Jeffrey R. Goodwin, Nov 10 2011
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EXAMPLE
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a(3) = 189 = 7*(3^3).
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MATHEMATICA
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PROG
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(Sage) [3^n*(2*n+1) for n in (0..30)] # G. C. Greubel, May 01 2021
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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