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A122649 Difference between the double factorial of the n-th nonnegative odd number and the double factorial of the n-th nonnegative even number. 6

%I

%S 0,1,7,57,561,6555,89055,1381905,24137505,468934515,10033419375,

%T 234484536825,5943863027025,162446292283275,4761954230608575,

%U 149048910271886625,4961463912662882625,175022432901300859875

%N Difference between the double factorial of the n-th nonnegative odd number and the double factorial of the n-th nonnegative even number.

%H Vincenzo Librandi, <a href="/A122649/b122649.txt">Table of n, a(n) for n = 1..410</a>

%F a(n) = (2*n - 1)!! - (2*n - 2)!! = A006882(2*n - 1) - A000165(n - 1).

%F From _Peter Bala_, Jun 22 2016: (Start)

%F a(1) = 0, a(2) = 1 and for n >= 3, a(n) = (4*n - 5)*a(n-1) - (2*n - 4)*(2*n - 3)*a(n-2).

%F E.g.f. assuming an offset of 0: A(x) = 1/(1 - 2*x)^(3/2) - 1/(1 - 2*x) = x + 7*x^2/2! + 57*x^3/3! + ....

%F A( Sum_{n >= 1} n^(n-2)*x^n/n! ) = Sum_{n >= 1} n^(n+1)*x^n/n!.

%F Series reversion (A(x)) = 1/2*Sum_{n >= 1} (-1)^(n+1)*1/(n+1)*

%F binomial(3*n + 1,n)*x^n. Cf. A006013.(End)

%F E.g.f.: -1 + 1/sqrt(1-2*x) + log(1-2*x)/2. - _Ilya Gutkovskiy_, Jun 23 2016

%e a(1) = 0, since 1!! - 0!! = 1 - 1 = 0, where the usual convention 0!! = 1 has been heeded. Note that 1 is the first nonnegative odd and 0 the first nonnegative even number.

%e a(4) = 57, since 7!! - 6!! = 1*3*5*7 - 6*4*2*1 = 105 - 48 = 57.

%p for n from 1 to 24 do: l[n]:=product(2*k-1, k=1..n); od: r[1]:=1; for n from 2 to 24 do: r[n]:=product(2*k, k=1..n-1); od; for k from 1 to 24 do: a[k]:=l[k]-r[k]; od;

%t #[[2]]-#[[1]]&/@Partition[Range[0,40]!!,2] (* _Harvey P. Dale_, Feb 19 2013 *)

%t Rest[Range[0, 100]! CoefficientList[Series[-1 + 1/Sqrt[1 - 2 x] + Log[1 - 2 x]/2, {x, 0, 800}], x]] (* _Vincenzo Librandi_, Jun 24 2016 *)

%Y Cf. A006882, A000165, A006013.

%K easy,nonn

%O 1,3

%A Peter C. Heinig (algorithms(AT)gmx.de), Sep 21 2006

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Last modified January 18 05:09 EST 2020. Contains 330995 sequences. (Running on oeis4.)