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A318529 a(n) begins the first run of at least n consecutive numbers with same number of exponential divisors.

%I A318529 #29 Feb 16 2025 08:33:56
%S A318529 1,1,1,242,3624,22020,671346,8870024,49250144,463239475,1407472722,
%T A318529 82462576220,82462576220,5907695879319
%N A318529 a(n) begins the first run of at least n consecutive numbers with same number of exponential divisors.
%C A318529 From _David A. Corneth_, Aug 28 2018: (Start)
%C A318529 For 4 <= n <= 10, a(n) has two exponential divisors. Most numbers have 1 or 2 exponential divisors.
%C A318529 For n > 3, a(n) isn't squarefree. (End)
%C A318529 For n >= 2^(k+1), A049419(a(n)) must be divisible by A051548(k), because for 1 <= j <= k at least one of a(n),...,a(n)+n-1 has 2-adic order j. - _Robert Israel_, Sep 07 2018
%H A318529 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/e-Divisor.html">e-Divisor</a>
%e A318529 a(4) = 242 since the number of exponential divisors of 242, 243, 244, and 245 is 2, and this is the first run of 4 consecutive numbers.
%t A318529 edivnum[1] = 1; edivnum [p_?PrimeQ] = 1; edivnum [p_?PrimeQ, e_] := DivisorSigma[ 0, e ]; edivnum [n_] := Times @@ (edivnum [#[[1]], #[[2]]] & ) /@ FactorInteger[ n ]; Seq[n_,q_] := Map[edivnum, Range[n, n + q - 1]]; findConsec[q_, nmin_, nmax_] := Module[{}, s = Seq[1, q]; n = q + 1; found = False; Do[ If[ CountDistinct[s] == 1, found = True; Break[] ]; s = Rest[AppendTo[s, edivnum[n] ]]; n++, {k, nmin, nmax}]; If[found, n - q, 0]]; seq = {1}; nmax = 100000000; Do[n1 = Last[seq]; s1 = findConsec[m, n1, nmax]; If[s1 == 0, Break[]]; AppendTo[ seq, s1 ], {m, 2, 7}]; seq (* after _Jean-François Alcover_ in A049419 *)
%Y A318529 Cf. A005117, A006558, A025487, A045983, A049419, A051548, A304463, A318166.
%K A318529 nonn,more
%O A318529 1,4
%A A318529 _Amiram Eldar_, Aug 28 2018
%E A318529 a(11)-a(13) from _Giovanni Resta_, Aug 28 2018
%E A318529 a(14) from _Giovanni Resta_, Sep 07 2018