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Expansion of x^4/((1-2*x)*(x^2-x+1)*(x-1)^2).
1

%I #9 Mar 09 2024 14:51:30

%S 0,0,0,0,1,5,15,36,78,162,331,671,1353,2718,5448,10908,21829,43673,

%T 87363,174744,349506,699030,1398079,2796179,5592381,11184786,22369596,

%U 44739216,89478457,178956941,357913911,715827852,1431655734,2863311498

%N Expansion of x^4/((1-2*x)*(x^2-x+1)*(x-1)^2).

%C Binomial transform of sequence (0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0). Note: the binomial transform of the sequence (0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0) is A111927; the binomial transform of the sequence (0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0) is A024495 (disregarding first two terms, which are both zero).

%C Floretion Algebra Multiplication Program, FAMP Code: -4ibaseisumseq[ + .5'i + .5'j + .5'k + .5'ij' + .5'jk' + .5'ki' + e], sumtype: Y[8] = (int)Y[6] - (int)Y[7] + Y[8] + sum (internal program code).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,11,-7,2).

%F a(n+2) - a(n+1) + a(n) = A000295(n) = 2^n - n - 1 (Eulerian numbers); a(n) = 1/3*2^n-n+2/3*(1/2+1/2*I*sqrt(3))^n*(-1/4-1/4*I*sqrt(3))+2/3*(1/2-1/2*I*sqrt(3))^n*(-1/4+1/4*I*sqrt(3))

%F a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1, a(n)=5*a(n-1)-10*a(n-2)+ 11*a(n-3)- 7*a(n-4)+2*a(n-5). - _Harvey P. Dale_, Feb 24 2016

%t CoefficientList[Series[x^4/((1-2x)(x^2-x+1)(x-1)^2),{x,0,40}],x] (* or *) LinearRecurrence[{5,-10,11,-7,2},{0,0,0,0,1},40] (* _Harvey P. Dale_, Feb 24 2016 *)

%Y Cf. A000295, A111927, A024495.

%K easy,nonn

%O 0,6

%A _Creighton Dement_, Aug 21 2005