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

1, 1, 1, 242, 3624, 22020, 671346, 8870024, 49250144, 463239475, 1407472722, 82462576220, 82462576220, 5907695879319

Offset

1,4

Comments

From David A. Corneth, Aug 28 2018: (Start)

For 4 <= n <= 10, a(n) has two exponential divisors. Most numbers have 1 or 2 exponential divisors.

For n > 3, a(n) isn't squarefree. (End)

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

Links

Table of n, a(n) for n=1..14.

Eric Weisstein's World of Mathematics, e-Divisor

Example

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.

Mathematica

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 *)

Crossrefs

Cf. A005117, A006558, A025487, A045983, A049419, A051548, A304463, A318166.

Sequence in context: A354563 A354565 A165935 * A006601 A283723 A035748

Adjacent sequences: A318526 A318527 A318528 * A318530 A318531 A318532

Keyword

nonn,more

Author

Amiram Eldar, Aug 28 2018

Extensions

a(11)-a(13) from Giovanni Resta, Aug 28 2018

a(14) from Giovanni Resta, Sep 07 2018

Status

approved