A240915 a(n) is the smallest number k such that digsum(k)/tau(k) = prime(n) where tau(k) is the number of divisors of k and digsum(k) is the sum of the digits of k.

8, 9, 19, 59, 499, 1889, 17989, 39989, 199999, 4999999, 9899999, 389999999, 9199999999, 9959999999, 99499999999, 899999998999, 64999999999999, 59999999999999, 999999899999999, 9999979999999999, 99999999299999999, 699999989999999999, 5989999999999999999

Offset

1,1

Comments

Observation: digsum(k) = tau(k)*prime(n) is minimum if tau(k) = 2 => k prime.

So, a(n) is prime if n > 2 and contains a majority of digits "9". For n > 3, digsum(a(n)) = A100484(n) = 10, 14, 22, 26, 34, 38, 46, 58, 62, ... (even semiprimes).

Links

Table of n, a(n) for n=1..23.

Example

a(6) = 1889 because tau(1889) = 2 and (1+8+8+9)/2 = 13 = prime(6).

Maple

with(numtheory):for n from 1 to 18 do: p:=ithprime(n):ii:=0:for k from 1 to 10^8 while(ii=0)do:x:=convert(k, base, 10):n1:=nops(x):s:=sum('x[j]', 'j'=1..n1):s:=s/tau(k):if s=p then printf ( "%d %d \n", n, k):ii:=1:else fi:od:od:

Mathematica

lst={}; Do[k=1; While[Plus@@IntegerDigits[k]/DivisorSigma[0, k]!=Prime[n], k++]; Print[n, " ", k], {n, 1, 10}]

Crossrefs

Cf. A000005, A000040, A007953, A100484.

Sequence in context: A022313 A120311 A061414 * A281225 A071869 A326444

Adjacent sequences: A240912 A240913 A240914 * A240916 A240917 A240918

Keyword

nonn,base,hard

Author

Michel Lagneau, Apr 14 2014

Status

approved