A287142 Least k such that the number of pairs of consecutive divisors of k equals n.

1, 2, 6, 12, 72, 60, 180, 360, 420, 840, 1260, 3780, 2520, 5040, 13860, 36960, 41580, 27720, 55440, 83160, 166320, 277200, 491400, 471240, 360360, 1113840, 720720, 1081080, 3341520, 2162160, 2827440, 5405400, 4324320, 12972960, 6126120, 16576560, 28274400

Offset: 0

Author

Michel Lagneau, May 20 2017

Keywords

• nonn

Comments

a(n) is even for n > 0.

We observe numbers of the decimal form (abcabc) = 360360, 720720 and numbers of the decimal form (abcabc0) = 1081080, 2162160, 5405400, 4324320, 6126120.

Observation and questions: many terms are products of powers of a contiguous set of the smallest primes. Many early terms of a(n) are in A002182; e.g., a(35) - A002182(44). The smallest exception outside of the empty product a(0) = 1 is a(22) = 491400 = 2^3 * 3^3 * 5^2 * 7 * 13. In other words, many terms have A006530(a(n)) < A053669(a(n)); a(22) is the smallest exception. Other exceptions include {471240, 1113840, 3341520, 2827440, 16576560, 28274400, ...}. A000720(A053669(a(22))) - A000720(A006530(a(22))) = 1, but the first instance of 2 for this function is a(35) = 16576560. This is evident by mapping A054841 across a(n). Are there a finite number of terms of a(n) that are also in A002182? Are there a finite number of terms of a(n) that have A006530(a(n)) < A053669(a(n)); are they becoming less frequent as n increases? - Michael De Vlieger, May 20 2017

In other words, a(n) is the least integer with exactly n divisors that are oblong (A002378). - Bernard Schott, Jul 30 2022

Examples

			a(3) = 12 because the divisors of 12 are {1, 2, 3, 4, 6, 12} with three pairs of consecutive divisors (1, 2), (2, 3) and (3, 4).
		

Crossrefs

Cf. A000005, A002378, A027750, A129308, A130317.

Essentially the same as A088726.

Programs

Maple

with(numtheory):
for n from 0 to 60 do:
ii:=0:
  for k from 1 to 10^8 while(ii=0) do:
    d0:=divisors(k):n0:=nops(d0):c0:=0:
      for i from 1 to n0-1 do:
        if d0[i+1]=d0[i]+1
         then
          c0:=c0+1:
          else
         fi:
       od:
       if c0=n
       then
     ii:=1:printf(“%d %d \n”,n,k):
     else
     fi:
   od:
  od:

Mathematica

Function[s, Function[t, ReplacePart[t, Map[#1 -> #2 & @@ # &, Transpose@{1 + Keys@ s, Values[s][[All, 1]]}]]]@ ConstantArray[0, Max@ Keys@ s]]@ KeySort@ PositionIndex@ Table[DivisorSum[n, 1 &, Divisible[n, # + 1] &], {n, 2 * 10^6}] (* Michael De Vlieger, May 20 2017, Version 10 *)

PARI

isok(n,k) = {dk = divisors(k); ddk = vector(#dk-1, j, dk[j+1] - dk[j]); #select(x->x==1, ddk) == n;}
a(n) = {my(k=1); while (!isok(n, k), k++); k;} \\ Michel Marcus, May 20 2017

Formula

a(n) = 2*A130317(n) for n >= 1. - Bernard Schott, Jul 30 2022