A382019 -id:A382019 - cp's OEIS frontend

A382108 Number of zeros (counted with multiplicity) on the unit circle of the polynomial P(n,z) = Sum_{k=0..n} T(n,k)*z^k where T(n,k) = A214292(n,k) is the first differences of rows in Pascal's triangle.

0, 1, 2, 3, 4, 5, 6, 3, 4, 3, 6, 5, 6, 5, 6, 7, 8, 9, 10, 3, 8, 7, 10, 9, 10, 7, 10, 11, 8, 11, 12, 9, 10, 11, 14, 11, 14, 11, 12, 13, 12, 13, 12, 15, 12, 7, 18, 19, 16, 11, 14, 11, 14, 11, 18, 11, 18, 15, 18, 19, 22, 7, 16, 21, 20, 17, 22, 15, 18, 21, 20, 25, 20

Offset: 0

Michel Lagneau, Mar 15 2025

nonn

			a(4)=4 because P(4,z)= 4 + 5*z -5*z^3 -4*z^4  with 4 roots z1, z2, z2, z4 on the unit circle : z1 = -1, z2 = +1, z3 = -.625000 -.7806247*i, z4 = -.625000 +.7806247*i.
a(6)=6 because P(6,z)= 6 + 14*z +14*z^2 -14*z^4-14*z^5-6z^6  with 6 roots on the unit circle:
x1 = -1
x2 = +1
x2 = -.6666666667 - .7453559925*i
x3 = -.6666666667 + .7453559925*i
x5 = -.500000000 - .8660254038*i
x6 = -.500000000 + .8660254038*i

Cf. A007318, A214292, A382019 (on and inside the circle).

A382108:=proc(n) local m,y,it:
y:=[fsolve(add((binomial(n+1,k+1)-binomial(n+1,k))*x^k,k=0..n),x,complex)]:it:=0:
 for m from 1 to nops(y) do:
    if ((Re(y[m]))^2+(Im(y[m]))^2)=1
     then it:=it+1:
     else
    fi:
   od:
    A382108(n):=it:end proc:seq(A382108(n),n=1..80);

cp's OEIS Frontend

A382108 Number of zeros (counted with multiplicity) on the unit circle of the polynomial P(n,z) = Sum_{k=0..n} T(n,k)*z^k where T(n,k) = A214292(n,k) is the first differences of rows in Pascal's triangle.

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