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.

%I A382108 #7 Mar 25 2025 14:03:04
%S A382108 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,
%T A382108 9,10,11,14,11,14,11,12,13,12,13,12,15,12,7,18,19,16,11,14,11,14,11,
%U A382108 18,11,18,15,18,19,22,7,16,21,20,17,22,15,18,21,20,25,20
%N 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.
%e A382108 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.
%e A382108 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:
%e A382108 x1 = -1
%e A382108 x2 = +1
%e A382108 x2 = -.6666666667 - .7453559925*i
%e A382108 x3 = -.6666666667 + .7453559925*i
%e A382108 x5 = -.500000000 - .8660254038*i
%e A382108 x6 = -.500000000 + .8660254038*i
%p A382108 A382108:=proc(n) local m,y,it:
%p A382108 y:=[fsolve(add((binomial(n+1,k+1)-binomial(n+1,k))*x^k,k=0..n),x,complex)]:it:=0:
%p A382108  for m from 1 to nops(y) do:
%p A382108     if ((Re(y[m]))^2+(Im(y[m]))^2)=1
%p A382108      then it:=it+1:
%p A382108      else
%p A382108     fi:
%p A382108    od:
%p A382108     A382108(n):=it:end proc:seq(A382108(n),n=1..80);
%Y A382108 Cf. A007318, A214292, A382019 (on and inside the circle).
%K A382108 nonn
%O A382108 0,3
%A A382108 _Michel Lagneau_, Mar 15 2025