A348079 Starts of runs of 5 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).

792007675, 2513546971, 2820448771, 3201296272, 4742326672, 4894282924, 5462510272, 5664816448, 6947006272, 7814337424, 8784450448, 9085360624, 10147712524, 10246365547, 11537724975, 11861786572, 11907710548, 12456672496, 13338112048, 13510075471, 13931933948

Offset: 1

Author

Amiram Eldar, Sep 27 2021

Keywords

• nonn

Examples

			792007675 is a term since 792007675 = 2^2 * 31680307, 792007675 + 1 = 792007676 = 2^2 * 198001919, 792007675 + 2 = 792007677 = 3^2 * 88000853, 792007675 + 3 = 792007678 = 2 * 7^2 * 11^2 * 66791 and 792007675 + 4 = 792007679 = 17^2 * 2740511 all have an equal number of even and odd exponents in their prime factorization.
		

Crossrefs

Subsequence of A187039, A348076, A348077 and A348078.

Programs

Mathematica

q[n_] := n == 1 || Count[(e = FactorInteger[n][[;; , 2]]), ?OddQ] == Count[e, ?EvenQ]; v = q /@ Range[5]; seq = {}; Do[v = Append[Drop[v, 1], q[k]]; If[And @@ v, AppendTo[seq, k - 4]], {k, 6, 3*10^9}]; seq

Python

from sympy import factorint
def cond(n):
    evenodd = [0, 0]
    for e in factorint(n).values():
        evenodd[e%2] += 1
    return evenodd[0] == evenodd[1]
def afind(limit, startk=6):
    condvec = [cond(startk+i) for i in range(5)]
    for kp4 in range(startk+4, limit+5):
        condvec = condvec[1:] + [cond(kp4)]
        if all(condvec):
            print(kp4-4, end=", ")
afind(10**9) # Michael S. Branicky, Sep 27 2021