A255905 - OEIS

%I #6 Mar 16 2015 01:59:50

%S 1,1,1,3,4,1,23,31,9,1,371,484,128,16,1,10515,13407,3228,360,25,1,

%T 461869,581680,132291,13260,815,36,1,28969177,36241581,7981991,749199,

%U 41167,1603,49,1,2454072147,3058280624,660958100,59706312,3060128,106232,2856,64,1

%N 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 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 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 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 1) A(n) := k^(2*n)*E(2*n,h/k)

%C 2) A(n) := (4*k)^n*E(n,h/(4*k))

%C 3) A(n) := (2*k)^(2*n+1)*E(2*n+1,h/(2*k))

%C In each case above, h and k are arbitrary integers except that k is nonzero.

%C The present triangle (up to signs) is simply the case of conjecture 1 with the choices h = 1 and k = 2.

%C 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 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 The triangle begins

%e n\k|      0      1      2     3   4   5  6

%e = = = = = = = = = = = = = = = = = = = = = =

%e 0  |      1

%e 1  |      1      1

%e 2  |      3      4      1

%e 3  |     23     31      9     1

%e 4  |    371    484    128    16   1

%e 5  |  10515  13407   3228   360  25   1

%e 6  | 461869 581680 132291 13260 815  36  1

%p #A255905

%p A000364 := n -> (-1)^n*2^(2*n)*euler(2*n, 1/2):

%p #define row polynomials of A086646

%p R := proc (n, u) add(binomial(2*n, 2*k)*A000364(n-k)*u^k, k = 0 .. n) end proc:

%p series(exp(add(R(n, u)*x^n/n, n = 1 .. 9)), x, 9):

%p seq(seq(coeff(coeftayl(%, x = 0, n), u, k), k = 0 .. n), n = 0 .. 8);

%Y Cf. A000364, A086646.

%K nonn,tabl,easy

%O 0,4

%A _Peter Bala_, Mar 10 2015