A381596 a(n) = number of real zeros (counted with multiplicity) of the polynomial P(n,z) = Sum_{i=1..n} A001223(i)*z^(i-1) where A001223(i) = differences between consecutive primes.

0, 1, 0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1

Offset: 1

Author

Michel Lagneau, Mar 01 2025

Keywords

• nonn

Examples

			a(4) = 1 because P(4,z) = Sum_{i=1..4} A001223(i)*z^(i-1) = 1 + 2*z + 2*z^2 + 4*z^3 = (2*z + 1)*(2*z^2 + 1) = 0 for z = -1/2.
a(5) = 2 because P(5,z) = Sum_{i=1..5} A001223(i)*z^(i-1) = 1 + 2*z + 2*z^2 + 4*z^3 + 2*z^4 = 0 for z = -1.6499348..., -0.5606729...
		

Crossrefs

Cf. A000040, A001223, A381604.

Programs

Maple

with(numtheory):
for n from 1 to 100 do :
  P:=add((ithprime(i+1)-ithprime(i))*x^(i-1),i=1..n):
   y:=fsolve(P,x,real):
   z:=evalf({%}):k:=nops(z):
   printf(`%d, `,k):
od:

PARI

a(n) = my(v=primes(n+1)); #polrootsreal(sum(i=1, n, (v[i+1]-v[i])*z^(i-1))); \\ Michel Marcus, Mar 01 2025