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A255905 Expansion of exp( Sum_{n >= 1} R(n,u)*x^n/n ), where R(n,u) denotes the n-th row polynomial of A086646.

%I A255905 #6 Mar 16 2015 01:59:50
%S A255905 1,1,1,3,4,1,23,31,9,1,371,484,128,16,1,10515,13407,3228,360,25,1,
%T A255905 461869,581680,132291,13260,815,36,1,28969177,36241581,7981991,749199,
%U A255905 41167,1603,49,1,2454072147,3058280624,660958100,59706312,3060128,106232,2856,64,1
%N A255905 Expansion of exp( Sum_{n >= 1} R(n,u)*x^n/n ), where R(n,u) denotes the n-th row polynomial of A086646.
%C A255905 Triangle A086646 has the e.g.f. cosh(sqrt(u)*t)/cos(t). The n-th row polynomial of A086646 is given by the formula R(n,u) = Sum_{k = 0..n} binomial(2*n,2*k)*A000364(n-k)*u^k.
%C A255905 It appears that in the expansion of exp( Sum_{n >= 1} R(n,u)*x^n/n ), the coefficient polynomials in u are always integer polynomials. Alternatively expressed, the o.g.f. for A086646 is (apart from its initial element) the logarithmic derivative of the o.g.f. of the present triangle.
%C A255905 The above conjecture can be extensively generalized. The elements of A000364 can be expressed in terms of the Euler polynomial E(n,x) as A000364(n) = (-1)^n*2^(2*n)*E(2*n,1/2). This suggests considering polynomials of the form P(n,u) = Sum_{k = 0..n} binomial(2*n,2*k)*A(n-k)*u^k, where the sequence A(n) is defined in terms of the Euler polynomials. Calculation suggests that in the expansion of exp( Sum_{n >= 1} P(n,u)*x^n/n ), the coefficient polynomials in u are always integer polynomials for the following choices of A(n):
%C A255905 1) A(n) := k^(2*n)*E(2*n,h/k)
%C A255905 2) A(n) := (4*k)^n*E(n,h/(4*k))
%C A255905 3) A(n) := (2*k)^(2*n+1)*E(2*n+1,h/(2*k))
%C A255905 In each case above, h and k are arbitrary integers except that k is nonzero.
%C A255905 The present triangle (up to signs) is simply the case of conjecture 1 with the choices h = 1 and k = 2.
%C A255905 Similar conjectures can be made if, in the above definition of the polynomial P(n,u), the factor equal to binomial(2*n,2*k) is replaced by binomial(m*n,m*k) for some fixed m = 1,2,3,....
%F A255905 O.g.f.: exp( Sum_{n >= 1} R(n,u)*x^n/n ) = exp( (1 + u)*x + (5 + 6*u + u^2)*x^2/2 + (61 + 75*u + 15*u^2 + u^3)*x^3/3 + ... ) = 1 + (1 + u)*x + (3 + 4*u + u^2)*x^2 + (23 + 31*u + 9*u^2 + u^3)*x^3 + ....
%e A255905 The triangle begins
%e A255905 n\k|      0      1      2     3   4   5  6
%e A255905 = = = = = = = = = = = = = = = = = = = = = =
%e A255905 0  |      1
%e A255905 1  |      1      1
%e A255905 2  |      3      4      1
%e A255905 3  |     23     31      9     1
%e A255905 4  |    371    484    128    16   1
%e A255905 5  |  10515  13407   3228   360  25   1
%e A255905 6  | 461869 581680 132291 13260 815  36  1
%p A255905 #A255905
%p A255905 A000364 := n -> (-1)^n*2^(2*n)*euler(2*n, 1/2):
%p A255905 #define row polynomials of A086646
%p A255905 R := proc (n, u) add(binomial(2*n, 2*k)*A000364(n-k)*u^k, k = 0 .. n) end proc:
%p A255905 series(exp(add(R(n, u)*x^n/n, n = 1 .. 9)), x, 9):
%p A255905 seq(seq(coeff(coeftayl(%, x = 0, n), u, k), k = 0 .. n), n = 0 .. 8);
%Y A255905 Cf. A000364, A086646.
%K A255905 nonn,tabl,easy
%O A255905 0,4
%A A255905 _Peter Bala_, Mar 10 2015