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A348077
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Starts of runs of 3 consecutive numbers that have an equal number of even and odd exponents in their prime factorization (A187039).
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4
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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
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OFFSET
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1,1
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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