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A080242 Table of coefficients of polynomials P(n,x) defined by the relation P(n,x) = (1+x)*P(n-1,x) + (-x)^(n+1). 5

%I #21 Feb 24 2019 14:27:49

%S 1,1,1,1,1,2,2,1,3,4,2,1,1,4,7,6,3,1,5,11,13,9,3,1,1,6,16,24,22,12,4,

%T 1,7,22,40,46,34,16,4,1,1,8,29,62,86,80,50,20,5,1,9,37,91,148,166,130,

%U 70,25,5,1,1,10,46,128,239,314,296,200,95

%N Table of coefficients of polynomials P(n,x) defined by the relation P(n,x) = (1+x)*P(n-1,x) + (-x)^(n+1).

%C Values generate solutions to the recurrence a(n) = a(n-1) + k(k+1)* a(n-2), a(0)=1, a(1) = k(k+1)+1. Values and sequences associated with this table are included in A072024.

%H G. C. Greubel, <a href="/A080242/b080242.txt">Rows n = 0..100 of coefficients, flattened</a>

%F Rows are generated by P(n,x) = ((x+1)^(n+2) - (-x)^(n+2))/(2*x+1).

%F The polynomials P(n,-x), n > 0, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re x = 1/2 in the complex plane.

%F O.g.f.: (1+x*t+x^2*t)/((1+x*t)(1-t-x*t)) = 1 + (1+x+x^2)*t + (1+2x+2x^2)*t^2 + ... . - _Peter Bala_, Oct 24 2007

%F T(n,k) = if(k<=2*floor((n+1)/2), Sum_{j=0..floor((n+1)/2)} binomial(n-2j,k-2j), 0). - _Paul Barry_, Apr 08 2011 (This formula produces the odd numbered rows correctly, but not the even. - _G. C. Greubel_, Feb 22 2019)

%e Rows are {1}, {1,1,1}, {1,2,2}, {1,3,4,2,1}, {1,4,7,6,3}, ... This is the same as table A035317 with an extra 1 at the end of every second row.

%e Triangle begins

%e 1;

%e 1, 1, 1;

%e 1, 2, 2;

%e 1, 3, 4, 2, 1;

%e 1, 4, 7, 6, 3;

%e 1, 5, 11, 13, 9, 3, 1;

%e 1, 6, 16, 24, 22, 12, 4;

%e 1, 7, 22, 40, 46, 34, 16, 4, 1;

%e 1, 8, 29, 62, 86, 80, 50, 20, 5;

%t Table[CoefficientList[Series[((1+x)^(n+2) -(-1)^n*x^(n+2))/(1+2*x), {x, 0, n+2}], x], {n, 0, 10}]//Flatten (* _G. C. Greubel_, Feb 18 2019 *)

%Y Similar to the triangles A059259, A035317, A108561, A112555. Cf. A059260.

%Y Cf. A001045 (row sums).

%K easy,nonn,tabf

%O 0,6

%A _Paul Barry_, Feb 12 2003

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