A342031 Starts of runs of 5 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

1, 16, 241, 2644, 4372, 1431124, 12502348, 112753348, 750031648, 2844282247, 5882272324, 6741230497, 8004453748, 87346072024, 130489991521, 218551872247, 245127093748, 460925878624, 804065433748, 1176638279524, 2210511903748, 2404792968748, 2483167488748, 3121595927521

Offset: 1

Author

Amiram Eldar, Feb 25 2021

Keywords

• nonn

Comments

Bernardo Recamán Santos (2015) showed that there is no run of more than 23 consecutive numbers, since numbers of the form 36*k - 6 and 36*k + 6 do not have distinct exponents. Pace Nielsen and Adam P. Goucher showed that there can be only finitely many runs of 23 consecutive numbers (see MathOverflow link).

Aktaş and Ram Murty (2017) gave an explicit upper bound to such a run of 23 numbers. They found the first 5 terms of this sequence (and stated that there are a few more known up to 7*10^8), and said that we may conjecture (based on numerical evidence) that there are no 6 consecutive numbers.

Examples

			16 is a term since 16 = 2^4, 17, 18 = 2*3^2, 19 and 20 = 2^2*5 all have distinct exponents in their prime factorization.
		

Links

• Martin Ehrenstein, Table of n, a(n) for n = 1..43
• 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.

Crossrefs

Subsequence of A130091, A342028, A342029 and A342030.

Programs

Mathematica

q[n_] := Length[(e = FactorInteger[n][[;; , 2]])] == Length[Union[e]]; v = q /@ Range[5]; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 5]]; v = Join[Rest[v], {q[k]}], {k, 6, 1.3*10^6}]; seq

Extensions

a(15) and beyond from Martin Ehrenstein, Mar 08 2021