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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. 0
8, 9, 19, 59, 499, 1889, 17989, 39989, 199999, 4999999, 9899999, 389999999, 9199999999, 9959999999, 99499999999, 899999998999, 64999999999999, 59999999999999, 999999899999999, 9999979999999999, 99999999299999999, 699999989999999999, 5989999999999999999 (list; graph; refs; listen; history; text; internal format)
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

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Last modified September 17 06:00 EDT 2019. Contains 327119 sequences. (Running on oeis4.)