A342030 Starts of runs of 4 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).
1, 2, 16, 17, 47, 96, 241, 242, 575, 1249, 2644, 2645, 4049, 4372, 4373, 4799, 9124, 12248, 33749, 72250, 120049, 130436, 281249, 303748, 1431124, 1431125, 1531250, 2101247, 3693761, 4085656, 4910975, 12502348, 12502349, 14268481, 22997761, 25486324, 26693549
Offset: 1
Keywords
Examples
2 is a term since 2, 3, 4 = 2^2, and 5 all have a single exponent in their prime factorization. 3 is not a term since in the run {3, 4, 5, 6} the fourth member 6 = 2*3 has two equal exponents (1) in its prime factorization.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..196 (terms below 10^11)
- Kevser Aktaş and M. Ram Murty, On the number of special numbers, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; alternative link.
- Bernardo Recamán Santos, Consecutive numbers with mutually distinct exponents in their canonical prime factorization, MathOverflow, Mar 30 2015.
Programs
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Mathematica
q[n_] := Length[(e = FactorInteger[n][[;; , 2]])] == Length[Union[e]]; v = q /@ Range[4]; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 4]]; v = Join[Rest[v], {q[k]}], {k, 5, 10^5}]; seq