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Coefficients of a set of infinite sum rational polynomials: p(x,n)=(-1 + x)^(m - 1)*( 1 - (1 + x)/(-1 + x))^(m + 1)*Sum[(k + 1)^(2*m - 1)*((x + 1)/( x - 1))^k, {k, 0, Infinity}]
0

%I #2 Mar 30 2012 17:34:36

%S 1,1,0,-3,4,0,-30,0,30,34,0,-462,0,1050,0,-630,496,0,-10560,0,40320,0,

%T -52920,0,22680,11056,0,-338448,0,1907400,0,-4074840,0,3742200,0,

%U -1247400,349504,0,-14523600,0,112192080,0,-344504160,0,505945440,0

%N Coefficients of a set of infinite sum rational polynomials: p(x,n)=(-1 + x)^(m - 1)*( 1 - (1 + x)/(-1 + x))^(m + 1)*Sum[(k + 1)^(2*m - 1)*((x + 1)/( x - 1))^k, {k, 0, Infinity}]

%C Row sums are:

%C {1, -2, 4, -8, 16, -32, 64, -128, 256, -512,...},

%C The infinite sums is found by doing a modular form substitution on the

%C Eulerian numbers A008292 infinite sum:

%C f((x+1)/(x-1))=(x-1)^(2*n)*f(x).

%C Further substitution of x^2->x removes the alternating zeros.

%F p(x,n)=(-1 + x)^(m - 1)*(1 - (1 + x)/(-1 + x))^(m + 1)*Sum[(k + 1)^(2*m - 1)*((x + 1)/(x - 1))^k, {k, 0, Infinity}]

%e {1},

%e {1, 0, -3},

%e {4, 0, -30, 0, 30},

%e {34, 0, -462, 0, 1050, 0, -630},

%e {496, 0, -10560, 0, 40320, 0, -52920, 0, 22680},

%e {11056, 0, -338448, 0, 1907400, 0, -4074840, 0, 3742200, 0, -1247400},

%e {349504, 0, -14523600, 0, 112192080, 0, -344504160, 0, 505945440, 0, -356756400, 0, 97297200},

%e {14873104, 0, -804913392, 0, 8117600400, 0, -33403209840, 0, 69437768400, 0, -77416138800, 0, 44270226000, 0, -10216206000},

%e {819786496, 0, -55994442240, 0, 712037291520, 0, -3751484405760, 0, 10310701040640, 0, -16108264972800, 0, 14449801766400, 0, -6947020080000, 0, 1389404016000},

%e {56814228736, 0, -4778633088768, 0, 74633142796800, 0, -487475148142080, 0, 1695092146675200, 0, -3469875535526400, 0, 4330639993680000, 0, -3247037185392000, 0, 1346332491504000, 0, -237588086736000}

%t Clear[p, x, n]

%t p[x_, m_] = (-1 + x)^(m - 1)*(1 - (1 + x)/(-1 + x))^(m + 1)*Sum[(k + 1)^(2*m - 1)*((x + 1)/(x - 1))^k, {k, 0, Infinity}]

%t Table[CoefficientList[FullSimplify[ExpandAll[p[x, m]]], x], {m, 1, 10}]

%t Flatten[%]

%Y Cf. A008292

%K sign,uned

%O 1,4

%A _Roger L. Bagula_, Dec 14 2009