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 A289776 Least k such that the sum of the first n divisors of k is a prime number. 5
 2, 4, 30, 16, 84, 36, 60, 144, 144, 144, 144, 210, 324, 360, 630, 756, 756, 576, 660, 840, 840, 2040, 900, 900, 2304, 1980, 1980, 1980, 4320, 5184, 3300, 4620, 5460, 7056, 3960, 4680, 2520, 3600, 3600, 3600, 10080, 8100, 3600, 6300, 9900, 7920, 11088, 14400 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,1 COMMENTS The corresponding primes are 3, 7, 11, 31, 23, 37, 43, 61, 79, 103, 139, 191, 523, 167, 263, 347, 431, 787, 641, ... The squares in the sequence are 4, 16, 36, 144, 324, 576, 900, 2304, 3600, 5184, 7056, 8100, 14400, ... LINKS Chai Wah Wu, Table of n, a(n) for n = 2..1000 EXAMPLE a(4)=30 because the sum of the first 4 divisors of 30 is 1 + 2 + 3 + 5 = 11, which is prime, and there is no integer below 30 with this property. MAPLE with(numtheory):nn:=10^6: for n from 2 to 50 do: ii:=0:    for k from 2 to nn while(ii=0) do:      x:=divisors(k):n0:=nops(x):        for l from 1 to n0 while(ii=0) do:         p:=sum('x[i]', 'i'=1..l):         if type(p, prime)=true and l=n          then          ii:=1:printf (`%d %d \n`, n, k):          else fi:         od:       od:   od: MATHEMATICA Table[k = 1; While[Nand[Length@ # >= n, PrimeQ@ Total@ Take[PadRight[#, n], n]] &@ Divisors@ k, k++]; k, {n, 2, 49}] (* Michael De Vlieger, Jul 12 2017 *) PROG (PARI) a(n) = k=1; while((d=divisors(k)) && ((#d

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Last modified January 17 18:14 EST 2020. Contains 330987 sequences. (Running on oeis4.)