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
Keywords
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...
Programs
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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