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A034910
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One quarter of octo-factorial numbers.
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14
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1, 12, 240, 6720, 241920, 10644480, 553512960, 33210777600, 2258332876800, 171633298636800, 14417197085491200, 1326382131865190400, 132638213186519040000, 14324927024144056320000, 1661691534800710533120000, 206049750315288106106880000
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OFFSET
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1,2
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COMMENTS
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A034910 occurs in connection with the Vandermonde permanent of (1,3,5,7,9,...); see the Mathematica section of A203516. - Clark Kimberling, Jan 03 2012
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LINKS
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FORMULA
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4*a(n) = (8*n-4)(!^8) = Product_{j=1..n} (8*j-4) = 4^n*A001147(n) = 2^n*(2*n)!/n!, A001147(n) = (2*n-1)!!.
E.g.f. (-1+(1-8*x)^(-1/2))/4.
G.f.: x/(1-12*x/(1-8*x/(1-20*x/(1-16*x/(1-28*x/(1-24*x/(1-36*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2011
Recurrence equation: a(n) = (7*n - 3)*a(n-1) + 4*(n - 1)*(2*n - 3)*a(n-2).
The sequence b(n) := a(n)* Sum_{k = 0..n-1} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) beginning [1, 11, 222, 6210, 223584, ...] satisfies the same recurrence. This leads to the finite continued fraction expansion b(n)/a(n) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/(7*n - 3) )))) for n >= 3.
Letting n tend to infinity gives the continued fraction expansion Sum_{k>=0} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (4/3)*log(2) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/((7*n - 3) + ... ))))). (End)
This sequence satisfies several other second order recurrence equations leading to some continued fraction expansions.
1) a(n) = (9*n + 4)*a(n-1) - 4*n*(2*n - 1)*a(n-2).
This recurrence is also satisfied by the (integer) sequence c(n) := a(n)*Sum_{k = 0..n} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (8/sqrt(7))*arctan(sqrt(7)/7) = (8/sqrt(7))*A195699 = 1 + 1/(12 - 24/(22 - 60/(31 - ... - 4*n*(2*n - 1)/((9*n + 4) - ... )))).
2) a(n) = (12*n + 2)*a(n-1) - 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence d(n) := a(n)*Sum_{k = 0..n} 1/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( (2*k + 1)*2^k ) = (1/sqrt(2))*log(3 + 2*sqrt(2)) = 1 + 2/(12 - 8*3^2/(26 - 8*5^2/(38 - ... - 8*(2*n - 1)^2/((12*n + 2) - ... )))). Cf. A002391.
3) a(n) = (4*n + 6)*a(n-1) + 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence e(n) := a(n)*Sum_{k = 0..n} (-1)^k/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} (-1)^k/( (2*k + 1)*2^k ) = (1/sqrt(2))*arctan(sqrt(2)/2) = 1 - 2/(12 + 8*3^2/(14 + 8*5^2/(18 + ... + 8*(2*n - 1)^2/((4*n + 6) + ... )))). Cf. A073000. (End)
a(n) = (-1)^n / (16*a(-n)) for all n in Z. - Michael Somos, Feb 04 2015
Sum_{n>=1} 1/a(n) = e^(1/8)*sqrt(2*Pi)*erf(1/(2*sqrt(2))), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = e^(-1/8)*sqrt(2*Pi)*erfi(1/(2*sqrt(2))), where erfi is the imaginary error function. (End)
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EXAMPLE
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G.f. = x + 12*x^2 + 240*x^3 + 6720*x^4 + 241920*x^5 + 10644480*x^6 + ...
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MAPLE
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MATHEMATICA
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a[ n_] := Pochhammer[ 1/2, n] 8^n / 4; (* Michael Somos, Feb 04 2015 *)
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PROG
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(PARI) {a(n) = if( n==1, 1, n>1, a(n-1) * (8*n - 4), a(n+1) / (8*n + 4))}; /* Michael Somos, Feb 04 2015 */
(Magma) [n le 2 select 12^(n-1) else (7*n-3)*Self(n-1) +4*(n-1)*(2*n-3)*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 20 2022
(SageMath) [2^(3*n-2)*rising_factorial(1/2, n) for n in range(1, 40)] # G. C. Greubel, Oct 20 2022
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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