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A240915
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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.
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0
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8, 9, 19, 59, 499, 1889, 17989, 39989, 199999, 4999999, 9899999, 389999999, 9199999999, 9959999999, 99499999999, 899999998999, 64999999999999, 59999999999999, 999999899999999, 9999979999999999, 99999999299999999, 699999989999999999, 5989999999999999999
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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EXAMPLE
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a(6) = 1889 because tau(1889) = 2 and (1+8+8+9)/2 = 13 = prime(6).
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MAPLE
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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:
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MATHEMATICA
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lst={}; Do[k=1; While[Plus@@IntegerDigits[k]/DivisorSigma[0, k]!=Prime[n], k++]; Print[n, " ", k], {n, 1, 10}]
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CROSSREFS
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KEYWORD
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nonn,base,hard
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AUTHOR
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STATUS
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approved
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