A348077
Starts of runs of 3 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).
Original entry on oeis.org
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
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.
-
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
-
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
A348078
Starts of runs of 4 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).
Original entry on oeis.org
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
Offset: 1
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.
-
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
-
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
A356415
a(n) is the least start of exactly n consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039), or -1 if no such run of consecutive numbers exists.
Original entry on oeis.org
1, 44, 603, 906596, 792007675
Offset: 1
a(2) = 44 since 44 = 2^2 * 11 and 45 = 3^2 * 5 both have one even and one odd exponent in their prime factorization, 43 and 46 have no even exponent, and 44 is the least number with this property.
-
f[p_, e_] := (-1)^e; q[1] = True; q[n_] := Plus @@ f @@@ FactorInteger[n] == 0; seq[len_, nmax_] := Module[{s = Table[0, {len}], v = {1}, n = 2, c = 0, m}, While[c <= len && n <= nmax, If[q[n], v = Join[v, {n}], m = Length[v]; v = {}; If[0 <= m <= len && s[[m]] == 0, c++; s[[m]] = n - m]]; n++]; s]; seq[4, 10^6]
A348079
Starts of runs of 5 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).
Original entry on oeis.org
792007675, 2513546971, 2820448771, 3201296272, 4742326672, 4894282924, 5462510272, 5664816448, 6947006272, 7814337424, 8784450448, 9085360624, 10147712524, 10246365547, 11537724975, 11861786572, 11907710548, 12456672496, 13338112048, 13510075471, 13931933948
Offset: 1
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.
-
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
-
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
Showing 1-4 of 4 results.
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