A348077 Starts of runs of 3 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).

603, 1250, 1323, 2523, 4203, 4923, 4948, 7442, 10467, 12591, 18027, 20402, 21123, 23823, 31507, 31850, 36162, 40327, 54475, 54511, 55323, 58923, 63747, 64386, 71523, 73204, 79011, 83151, 85291, 88047, 97675, 103923, 104211, 118323, 120787, 122571, 124891, 126927

Offset: 1

Author

Amiram Eldar, Sep 27 2021

Keywords

• nonn

Examples

			603 is a term since 603 = 3^2 * 67, 603 + 1 = 604 = 2^2 * 151 and 603 + 2 = 605 = 5 * 11^2 all have one even and one odd exponent in their prime factorization.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Subsequence of A187039 and A348076.

Programs

Mathematica

q[n_] := n == 1 || Count[(e = FactorInteger[n][[;; , 2]]), ?OddQ] == Count[e, ?EvenQ]; v = q /@ Range[3]; seq = {}; Do[v = Append[Drop[v, 1], q[k]]; If[And @@ v, AppendTo[seq, k - 2]], {k, 4, 130000}]; seq

Python

from sympy import factorint
def aupto(limit):
    alst, condvec = [], [False, False, False]
    for kp2 in range(4, limit+3):
        evenodd = [0, 0]
        for e in factorint(kp2).values():
            evenodd[e%2] += 1
        condvec = condvec[1:] + [evenodd[0] == evenodd[1]]
        if all(condvec):
            alst.append(kp2-2)
    return alst
print(aupto(126927)) # Michael S. Branicky, Sep 27 2021