A147630 a(1) = 1; for n>1, a(n) = Product_{k = 1..n-1} (9k - 3).
1, 6, 90, 2160, 71280, 2993760, 152681760, 9160905600, 632102486400, 49303993939200, 4289447472710400, 411786957380198400, 43237630524920832000, 4929089879840974848000, 606278055220439906304000, 80028703289098067632128000, 11284047163762827536130048000
Offset: 1
Crossrefs
Programs
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Magma
[Round(9^n*Gamma(n+6/9)/Gamma(6/9)): n in [0..20]]; // Vincenzo Librandi, Feb 21 2015
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Mathematica
s=1;lst={s};Do[s+=n*s;AppendTo[lst,s],{n,5,2*5!,9}];lst Table[Product[9k-3,{k,1,n-1}],{n,20}] (* Harvey P. Dale, Sep 01 2016 *) -
Maxima
a(n):=n!*sum(binomial(k,n-k-1)*3^k*(-1)^(n-k-1)*binomial(n+k-1,n-1),k,1,n-1)/n; /* Vladimir Kruchinin, Apr 01 2011 */
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PARI
a(n) = n--; prod(k=1, n, 9*k-3); \\ Michel Marcus, Feb 28 2015
Formula
a(n+1) = Sum_{k, 0<=k<=n}A132393(n,k)*6^k*9^(n-k). - Philippe Deléham, Nov 09 2008
a(n) = n!*(Sum_{k=1..n-1} binomial(k,n-k-1)*3^k*(-1)^(n-k-1)*binomial(n+k-1,n-1))/n for n>1, also a(n) = n!*A097188(n-1). - Vladimir Kruchinin, Apr 01 2011
a(n) = (-3)^n*sum_{k=0..n} 3^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = round(9^n * Gamma(n+6/9) / Gamma(6/9)). - Vincenzo Librandi, Feb 21 2015
Sum_{n>=1} 1/a(n) = 1 + (e/9^3)^(1/9)*(Gamma(2/3) - Gamma(2/3, 1/9)). - Amiram Eldar, Dec 21 2022
Extensions
New name from Peter Bala, Feb 20 2015
More terms from Michel Marcus, Feb 28 2015
Comments