A225562 -id:A225562 - cp's OEIS frontend

A226326 a(n) = smallest k such that prime(n) is the n-th largest divisor of k.

2, 6, 20, 42, 154, 156, 306, 836, 552, 1044, 1488, 2960, 2460, 3870, 7050, 12084, 8496, 10248, 14070, 12780, 18396, 31284, 50796, 38448, 55872, 82416, 37080, 51360, 65400, 88140, 146304, 169776, 123300, 133440, 150192, 181200, 131880, 176040, 260520, 326970

Offset: 1

Irina Gerasimova, Jun 04 2013

nonn

			a(5) = 165 because the divisors of 165 are (165, 55, 33, 15, 11, 5, 3, 1) and prime(5) = 11 is the 5th divisor of 165.

Charles R Greathouse IV, Table of n, a(n) for n = 1..200

Cf. A225562.

with(numtheory):
a:= proc(n) local k, p; p:= ithprime(n);
      for k from p by p while tau(k)`)[n]<>p do od; k
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Jun 04 2013

a[n_] := Module[{k, p}, p = Prime[n]; For[k = p, DivisorSigma[0, k] < n || Reverse[Divisors[k]][[n]] != p, k = k + p]; k];
Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Mar 25 2017, after Alois P. Heinz *)

a(n)=my(p=prime(n),k,d);while(k+=p, d=divisors(k); if(#d>=n && d[#d-n+1]==p, return(k))) \\ Charles R Greathouse IV, Jun 04 2013

a(5) corrected, a(6)-a(40) from Charles R Greathouse IV, Jun 04 2013

A256605 Least k such that n+1 is the n-th divisor of k.

Original entry on oeis.org

3, 4, 20, 12, 84, 120, 360, 360, 3960, 2520, 32760, 27720, 27720, 55440, 942480, 720720, 13693680, 12252240, 12252240, 12252240, 281801520, 232792560, 1163962800, 1163962800, 3491888400, 3491888400, 101264763600, 80313433200, 2489716429200, 4658179125600

Offset: 2

Michel Lagneau, Apr 04 2015

The case n = 1 is not possible because the number 2 is never the first divisor of k (1 is the first divisor).

			a(6) = 84 because the divisors of 84 are {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84} and 7 is the 6th divisor of 84.

Cf. A003418, A007917, A225562.

See A221647 for other sequences giving the smallest number whose n-th divisor satisfies some condition.

with(numtheory):for n from 2 to 31 do:ii:=0:for  k from 1 to 10^9 while(ii=0) do:x:=divisors(k):n1:=nops(x):if n<=n1 and x[n]=n+1 then ii:=1: printf ( "%d %d \n",n,k):else fi:od:od:

nn=20;t=Table[0,{nn}];found=1;n=2;While[found

a(n) = {k = 1; ok = 0; while (!ok, d = divisors(k); if ((#d >= n) && (d[n] == n+1), ok = 1, k++);); k;} \\ Michel Marcus, Apr 04 2015

a(n) = A003418(n+1)/A007917(n). - Peter Munn, May 14 2025

cp's OEIS Frontend

A226326 a(n) = smallest k such that prime(n) is the n-th largest divisor of k.

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