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A008542
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Sextuple factorial numbers: Product_{k=0..n-1} (6*k+1).
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39
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1, 1, 7, 91, 1729, 43225, 1339975, 49579075, 2131900225, 104463111025, 5745471106375, 350473737488875, 23481740411754625, 1714167050058087625, 135419196954588922375, 11510631741140058401875, 1047467488443745314570625, 101604346379043295513350625
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OFFSET
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0,3
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COMMENTS
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a(n), n>=1, enumerates increasing heptic (7-ary) trees with n vertices. - Wolfdieter Lang, Sep 14 2007; see a D. Callan comment on A007559 (number of increasing quarterny trees).
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LINKS
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FORMULA
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E.g.f.: (1-6*x)^(-1/6).
a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(1/6)^-1*n^(-1/3)*6^n*e^-n*n^n*{1 + 1/72*n^-1 - ...}. - Joe Keane (jgk(AT)jgk.org), Nov 24 2001
G.f.: 1+x/(1-7x/(1-6x/(1-13x/(1-12x/(1-19x/(1-18x/(1-25x/(1-24x/(1-... (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-5)^n*Sum_{k=0..n} (6/5)^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
G.f.: 1/Q(0) where Q(k) = 1 - x*(6*k+1)/(1 - x*(6*k+6)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
D-finite with recurrence: a(n) +(-6*n+5)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
Sum_{n>=0} 1/a(n) = 1 + (e/6^5)^(1/6)*(Gamma(1/6) - Gamma(1/6, 1/6)). - Amiram Eldar, Dec 18 2022
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MAPLE
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a := n -> mul(6*k+1, k=0..n-1);
G(x):=(1-6*x)^(-1/6): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1], x) od: x:=0: seq(f[n], n=0..15); # Zerinvary Lajos, Apr 03 2009
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MATHEMATICA
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Table[6^n*Pochhammer[1/6, n], {n, 0, 20}] (* G. C. Greubel, Aug 17 2019 *)
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PROG
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(Magma) [1] cat [(&*[(6*k+1): k in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Aug 17 2019
(Sage) [product((6*k+1) for k in (0..n-1)) for n in (0..20)] # G. C. Greubel, Aug 17 2019
(GAP) List([0..20], n-> Product([0..n-1], k-> (6*k+1) )); # G. C. Greubel, Aug 17 2019
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CROSSREFS
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Cf. k-fold factorials: A000142, A001147 (and A000165, A006882), A007559 (and A032031, A008544, A007661), A007696 (and A001813, A008545, A047053, A007662), A008548 (and A052562, A047055, A085157), A045754 (and A084947, A114799), A045755.
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KEYWORD
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nonn
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AUTHOR
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Joe Keane (jgk(AT)jgk.org)
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STATUS
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approved
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