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%I #53 Sep 08 2022 08:44:48
%S 1,5,39,418,5714,95064,1864456,42124592,1077459120,30777463360,
%T 971142388160,33547112941440,1259204418129280,51032742579123200,
%U 2220990565060377600,103308619261574809600,5114702794181847910400
%N a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 1 mod 3.
%C Comment by _R. J. Mathar_, Oct 01 2016: (Start)
%C The k-th elementary symmetric functions of the integers 1+j*3, j=0..n-1, form a triangle T(n,k), 0 <= k <= n, n >= 0:
%C 1
%C 1 1
%C 1 5 4
%C 1 12 39 28
%C 1 22 159 418 280
%C 1 35 445 2485 5714 3640
%C 1 51 1005 9605 45474 95064 58240
%C 1 70 1974 28700 227969 959070 1864456 1106560
%C 1 92 3514 72128 859369 5974388 22963996 42124592 24344320
%C This here is the first subdiagonal. The diagonal seems to be A007559. The first columns are A000012, A000326, A024212, A024213, A024214. (End)
%H Vincenzo Librandi, <a href="/A024216/b024216.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(1/3). - _Vladeta Jovovic_, Sep 26 2003
%F For n >= 1, a(n-1) = 3^(n-1)*n!*Sum_{k=0..n-1} binomial(k-2/3, k)/(n-k). - _Milan Janjic_, Dec 14 2008, corrected by _Peter Bala_, Oct 08 2013
%F a(n) ~ (n+1)! * GAMMA(2/3) * 3^(n+3/2) * (log(n) + gamma + Pi*sqrt(3)/6 + 3*log(3)/2) / (6*Pi*n^(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - _Vaclav Kotesovec_, Oct 07 2013
%F a(n+1) = (6*n+5) * a(n) - (3*n+1)^2 * a(n-1). - _Gheorghe Coserea_, Aug 29 2015
%F E.g.f.: (3 - log(1-3*x))/(3*(1-3*x)^(4/3)). - _Robert Israel_, Aug 30 2015
%F a(n) = A286718(n+1, 1), n >= 0.
%F Boas-Buck type recurrence: a(0) = 1 and for n >= 1: a(n) = ((n+1)!/n) * Sum_{p=1..n} 3^(n-p)*(1 + 3*beta(n-p))*a(p-1)/p!, with beta(k) = A002208(k+1) / A002209(k+1). Proof from a(n) = A286718(n+1, 1). - _Wolfdieter Lang_, Aug 09 2017
%e From _Gheorghe Coserea_, Dec 24 2015: (Start)
%e For n = 1 we have a(1) = 1*4*(1/1 + 1/4) = 5.
%e For n = 2 we have a(2) = 1*4*7*(1/1 + 1/4 + 1/7) = 39.
%e For n = 3 we have a(3) = 1*4*7*10*(1/1 + 1/4 + 1/7 + 1/10) = 418.
%e (End)
%p f:= gfun:-rectoproc({-(3*n+1)^2*a(n-1)+(6*n+5)*a(n)-a(n+1), a(0) = 1, a(1) = 5, a(2) = 39}, a(n), remember):
%p map(f, [$0..30]); # _Robert Israel_, Aug 30 2015
%t Rest[CoefficientList[Series[-(1/3)*Log[1-3*x]/(1-3*x)^(1/3), {x, 0, 20}], x]* Range[0, 20]!] (* _Vaclav Kotesovec_, Oct 07 2013 *)
%o (PARI)
%o n = 33; a = vector(n); a[1] = 5; a[2] = 39;
%o for (k = 2, n-1, a[k+1] = (6*k+5) * a[k] - (3*k+1)^2 * a[k-1]);
%o print(concat(1,a)); \\ _Gheorghe Coserea_, Aug 29 2015
%o (Magma) I:=[5,39]; [1] cat [n le 2 select I[n] else (6*n-1) * Self(n-1) - (3*n-2)^2 * Self(n-2) : n in [1..30]]; // _Vincenzo Librandi_, Aug 30 2015
%Y Cf. A024395, A024382, A286718 (first column).
%K nonn,easy
%O 0,2
%A _Clark Kimberling_
%E More terms from _Vladeta Jovovic_, Sep 26 2003
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