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A348079
Starts of runs of 5 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).
0
792007675, 2513546971, 2820448771, 3201296272, 4742326672, 4894282924, 5462510272, 5664816448, 6947006272, 7814337424, 8784450448, 9085360624, 10147712524, 10246365547, 11537724975, 11861786572, 11907710548, 12456672496, 13338112048, 13510075471, 13931933948
OFFSET
1,1
EXAMPLE
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.
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
PROG
(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
CROSSREFS
Subsequence of A187039, A348076, A348077 and A348078.
Sequence in context: A344730 A225389 A166121 * A046186 A166227 A104829
KEYWORD
nonn
AUTHOR
Amiram Eldar, Sep 27 2021
STATUS
approved