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A024395 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 2 mod 3.

%I A024395 #39 Jul 09 2019 17:25:43
%S A024395 1,7,66,806,12164,219108,4591600,109795600,2951028000,88084714400,
%T A024395 2891353030400,103521905491200,4015191638617600,167714507921497600,
%U A024395 7506196028811110400,358368551285791692800,18180562447078051328000
%N A024395 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 2 mod 3.
%C A024395 Comment by _R. J. Mathar_, Oct 01 2016 (Start):
%C A024395 The k-th elementary symmetric functions of the integers 2+j*3, j=0..n-1, form a triangle T(n,k), 0<=k<=n, n>=0:
%C A024395 1
%C A024395 1 2
%C A024395 1 7 10
%C A024395 1 15 66 80
%C A024395 1 26 231 806 880
%C A024395 1 40 595 4040 12164 12320
%C A024395 1 57 1275 14155 80844 219108 209440
%C A024395 1 77 2415 39655 363944 1835988 4591600 4188800
%C A024395 1 100 4186 95200 1276009 10206700 46819324 109795600 96342400
%C A024395 This here is the first subdiagonal. The diagonal seems to be A008544. The first columns are A000012, A005449, A024391, A024392. (End)
%H A024395 Vincenzo Librandi, <a href="/A024395/b024395.txt">Table of n, a(n) for n = 0..200</a>
%F A024395 E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(2/3). - _Vladeta Jovovic_, Sep 26 2003
%F A024395 For n >= 1, a(n-1) = 3^(n-1)*n!*sum(binomial(k-1/3,k)/(n-k), k = 0..n-1). - _Milan Janjic_, Dec 14 2008, corrected by _Peter Bala_, Oct 08 2013
%F A024395 a(n) ~ (n+1)! * 3^n * (log(n) + gamma - Pi*sqrt(3)/6 + 3*log(3)/2) / (n^(1/3)*GAMMA(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - _Vaclav Kotesovec_, Oct 07 2013
%F A024395 a(n+1) = (6*n+7) * a(n) - (3*n+2)^2 * a(n-1). - _Gheorghe Coserea_, Aug 30 2015
%F A024395 a(n) = A225470(n+1, 1), n >= 0. - _Wolfdieter Lang_, May 29 2017
%e A024395 From _Gheorghe Coserea_, Dec 24 2015: (Start)
%e A024395 For n=1 we have a(1) = 2*5*(1/2 + 1/5) = 7.
%e A024395 For n=2 we have a(2) = 2*5*8*(1/2 + 1/5 + 1/8) = 66.
%e A024395 For n=3 we have a(3) = 2*5*8*11*(1/2 + 1/5 + 1/8 + 1/11) = 806.
%e A024395 (End)
%t A024395 Table[ (-1)^(n+1)*Sum[(-3)^(n - k) k (-1)^(n - k) StirlingS1[n+1, k + 1], {k, 0, n}], {n, 1, 30}]
%t A024395 Join[{1},Table[Module[{c=NestList[3+#&,2,n+1]},Times@@c*Total[1/c]],{n,0,20}]] (* _Harvey P. Dale_, Jul 09 2019 *)
%o A024395 (PARI)
%o A024395 n = 16; a = vector(n); a[1] = 7; a[2] = 66;
%o A024395 for (k=2, n-1, a[k+1] = (6*k+7) * a[k] - (3*k+2)^2 * a[k-1]);
%o A024395 print(concat(1,a))  \\ _Gheorghe Coserea_, Aug 30 2015
%Y A024395 Cf. A024216, A225470 (second column).
%K A024395 nonn,easy
%O A024395 0,2
%A A024395 _Clark Kimberling_
%E A024395 Formula (see Mathematica line), correction and more terms from Victor Adamchik (adamchik(AT)cs.cmu.edu), Jul 21 2001