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Column k=4 of triangle A039755, Sheffer(exp(x), (exp(2*x) - 1)/2).
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%I #9 Dec 24 2017 09:07:55

%S 1,25,395,5075,58086,618870,6289690,61885450,595122671,5629238615,

%T 52605474285,487197745125,4481780785756,41018845739260,

%U 373968405050180,3399402534376100,30830907772159341,279134548584080805,2523817507756513375,22795663165336810375,205730405672107235426,1855561201430080303250,16727971116048518559870,150747219419372400319950,1358093516662781192486011

%N Column k=4 of triangle A039755, Sheffer(exp(x), (exp(2*x) - 1)/2).

%C For a combinatorial interpretation following from the g.f. and the a(n) = h^{(5)}_n formula below see A039755.

%H Colin Barker, <a href="/A286719/b286719.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (25,-230,950,-1689,945).

%F a(n) = A039755(n+4,4), n >= 0.

%F G.f.: 1/Product_{j=0..4}(1 - (1+2*j)*x).

%F E.g.f.: (d^4/dx^4)(exp(x)*((exp(2*x)-1)/2)^4/4!) = (2187/128)*exp(9*x) - (2401/96)*exp(7*x) + (625/64)*exp(5*x) - (27/32)*exp(3*x) + (1/384)*exp(x).

%F a(n) = h^{(5)}_n, the complete homogeneous symmetric function of degree n of the five symbols 1, 3, 5, 7, 9.

%F From _Colin Barker_, Dec 23 2017: (Start)

%F G.f.: 1 / ((1 - x)*(1 - 3*x)*(1 - 5*x)*(1 - 7*x)*(1 - 9*x)).

%F a(n) = (1 - 4*3^(4+n) + 6*5^(4+n) - 4*7^(4+n) + 9^(4+n)) / 384.

%F a(n) = 25*a(n-1) - 230*a(n-2) + 950*a(n-3) - 1689*a(n-4) + 945*a(n-5) for n>4.

%F (End)

%e a(2) = h^{(5)}_2 = 1^2 + 3^2 + 5^2 + 7^2 + 9^2 + 1^1*(3^1 + 5^1 + 7^1 + 9^1) + 3^1*(5^1 + 7^1 + 9^1) + 5^1*(7^1 + 9^1) + 7^1*9^1 = 165 + 230 = 395. The multichoose(5, 2) = binomial(6, 2) = 15 polytopes are five squares and ten rectangles of total area 165 and 230, respectively.

%o (PARI) Vec(1 / ((1 - x)*(1 - 3*x)*(1 - 5*x)*(1 - 7*x)*(1 - 9*x)) + O(x^40)) \\ _Colin Barker_, Dec 23 2017

%Y Cf. A003462 (k=1), A016209 (k=2), A021424 (k=3), A039755.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, May 26 2017