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 A034910 One quarter of octo-factorial numbers. 8
 1, 12, 240, 6720, 241920, 10644480, 553512960, 33210777600, 2258332876800, 171633298636800, 14417197085491200, 1326382131865190400, 132638213186519040000, 14324927024144056320000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 LINKS FORMULA 4*a(n) = (8*n-4)(!^8) := product(8*j-4, j=1..n) = 4^n*A001147(n) = 2^n*(2*n)!/n!, A001147(n) = (2*n-1)!!; E.g.f. (-1+(1-8*x)^(-1/2))/4. a(n) = A090802(2n-1, n). - Ross La Haye, Oct 18 2005 a(n) = (2*n)!/(n)!*2^(n-2). - Zerinvary Lajos, Sep 25 2006 G.f.: x/(1-12x/(1-8x/(1-20x/(1-16x/(1-28x/(1-24x/(1-36x/(1-32x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2011 From Peter Bala, Feb 01 2015: (Start) 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..inf} (-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) From Peter Bala, Feb 03 2015: (Start) 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 EXAMPLE G.f. = x + 12*x^2 + 240*x^3 + 6720*x^4 + 241920*x^5 + 10644480*x^6 + ... MAPLE [seq((2*n)!/(n)!*2^(n-2), n=1..14)]; # Zerinvary Lajos, Sep 25 2006 MATHEMATICA s=1; lst={s}; Do[s+=n*s; AppendTo[lst, s], {n, 11, 5!, 8}]; lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *) 0[ n_] := Pochhammer[ 1/2, n] 8^n / 4; (* Michael Somos, Feb 04 2015 *) PROG (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 */ CROSSREFS Cf. A001147, A045755, A034908, A034909, A034911, A034912, A203516. Sequence in context: A009150 A009080 A002166 * A091745 A012344 A282789 Adjacent sequences:  A034907 A034908 A034909 * A034911 A034912 A034913 KEYWORD easy,nonn AUTHOR STATUS approved

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Last modified February 26 18:29 EST 2020. Contains 332293 sequences. (Running on oeis4.)