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%I #16 Aug 01 2013 18:16:08
%S 1,4,12,36,116,344,1104,3280,10456,31152,98804,295988,935876,2811540,
%T 8870324,26695724,84060148,253376840,796635360,2404558304,7549431884,
%U 22820942416,71541295984,216562743948,677938097756,2054922521644
%N a(n) = sum of absolute values of coefficients in (1-x-x^2+x^3)^n.
%C Conjecture: limit a(n)^(1/n) = 16*sqrt(3)/9 = 3.079201..., which is substantiated by the observation that the sums of the coefficients squared in (1-x-x^2+x^3)^n equals binomial(4n,n) (cf. A005810).
%e The triangle A227964 of coefficients in (1+x-x^2-x^3)^n begins:
%e n=0: [1];
%e n=1: [1, -1, -1, 1];
%e n=2: [1, -2, -1, 4, -1, -2, 1];
%e n=3: [1, -3, 0, 8, -6, -6, 8, 0, -3, 1];
%e n=4: [1, -4, 2, 12, -17, -8, 28, -8, -17, 12, 2, -4, 1];
%e n=5: [1, -5, 5, 15, -35, -1, 65, -45, -45, 65, -1, -35, 15, 5, -5, 1];
%e n=6: [1, -6, 9, 16, -60, 24, 116, -144, -66, 220, -66, -144, 116, 24, -60, 16, 9, -6, 1]; ...
%e This sequence gives the sums of the absolute values of the coefficients for n>=0.
%t Table[Total[Abs[CoefficientList[Expand[(1-x-x^2+x^3)^n],x]]],{n,0,30}] (* _Harvey P. Dale_, Mar 03 2013 *)
%o (PARI) {a(n)=sum(k=0,3*n,abs(polcoeff((1-x-x^2+x^3)^n,k)))}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A192210, A084611, A084613, A084615, A084775, A084776, A084777, A084778, A084779, A084780.
%Y Cf. A227964 (triangle).
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jun 25 2011
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