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A348078 Starts of runs of 4 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039). 3
906596, 1141550, 1243275, 12133673, 13852924, 19293209, 20738672, 22997761, 23542001, 26587348, 30731822, 31237450, 39987773, 41419024, 43627148, 54040975, 54652148, 56487148, 70289225, 75855625, 77449300, 79677772, 80665072, 82126448, 91420721, 93883850, 95162849 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
LINKS
EXAMPLE
906596 is a term since 906596 = 2^2 * 226649, 906596 + 1 = 906597 = 3^2 * 100733, 906596 + 2 = 906598 = 2 * 7^2 * 11 * 29^2 and 906596 + 3 = 906599 = 71 * 113^2 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[4]; seq = {}; Do[v = Append[Drop[v, 1], q[k]]; If[And @@ v, AppendTo[seq, k - 3]], {k, 5, 2*10^7}]; 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=5):
condvec = [cond(startk+i) for i in range(4)]
for kp3 in range(startk+3, limit+4):
condvec = condvec[1:] + [cond(kp3)]
if all(condvec):
print(kp3-3, end=", ")
afind(125*10**4) # Michael S. Branicky, Sep 27 2021
CROSSREFS
Subsequence of A187039, A348076 and A348077.
Sequence in context: A251520 A250542 A251799 * A348101 A179735 A015333
KEYWORD
nonn
AUTHOR
Amiram Eldar, Sep 27 2021
STATUS
approved

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Last modified April 15 02:01 EDT 2024. Contains 371667 sequences. (Running on oeis4.)