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A254620
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a(n) = 9^n*(2*n + 1)!/n!.
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2
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1, 54, 4860, 612360, 99202320, 19642059360, 4596241890240, 1240985310364800, 379741504971628800, 129871594700297049600, 49091462796712284748800, 20323865597838885886003200, 9145739519027498648701440000, 4444829406247364343268899840000
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OFFSET
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0,2
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LINKS
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FORMULA
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E.g.f.: 1/(1 - 36*x)^(3/2) = 1 + 54*x + 4860*x^2/2! + 612360*x^3/3! + ....
Recurrence equation: a(n) = 18*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = (40*n + 16)*a(n-1) - 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*9^k) beginning [1, 56, 5052, 636672, 103142544, 20422253952, 4778808090048, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion log(2) = 2/3*Sum {k >= 0} 1/((2*k + 1)*9^k) = 2/3*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))).
Alternative 2nd order recurrence equation: a(n) = (32*n + 20)*a(n-1) + 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*9^k) beginning [1, 52, 4692, 591072, 95755344, 18959527872, 4436530187328, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion arctan(1/3) = 1/3*Sum {k >= 0} (-1)^k/((2*k + 1)*9^k) = 1/3*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ... ))))). Cf. A254381 and A254619.
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MAPLE
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seq(9^n*(2*n + 1)!/n!, n = 0..14);
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MATHEMATICA
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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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STATUS
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approved
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