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A290169 a(n) = least k such that both the sum of the smallest n divisors of k and the sum of its greatest n divisors are prime numbers.

%I A290169 #10 Aug 08 2017 21:38:03
%S A290169 2,4,30,16,140,64,264,144,336,525,144,800,1200,576,1600,2016,1440,
%T A290169 1296,2160,2304,7980,6440,3360,8360,4080,3960,2772,16100,9108,10608,
%U A290169 7392,12320,14688,37240,21780,18200,45760,20160,9240,24624,14364,8400,22176,23760
%N A290169 a(n) = least k such that both the sum of the smallest n divisors of k and the sum of its greatest n divisors are prime numbers.
%C A290169 The corresponding pairs of primes are (3, 3), (7, 7), (11, 61), (31, 31), (29, 307), (127, 127), (47, 673), (61, 379), (73, 919), ...
%C A290169 The sequence contains a subsequence of numbers having the property that the sum of the first n divisors is equal to the sum of the last n divisors; for instance, for a(n) = 2, 4, 16 and 64 with n = 2, 3, 5 and 7. Is it possible to conjecture that this subsequence contains all the superperfect numbers (A019279)? The answer is no: for instance, A019279(5) = 4096 = 2^12 => the sum of the 13 terms 1 + 2 + 4 + 8 + ... + 4096 = 8191 is a Mersenne prime, but a(13) = 800 instead 4096 > 800, and we obtain the corresponding pair of primes (293, 1933) instead (8191, 8191).
%C A290169 The squares of the terms of the sequence are 4, 16, 64, 144, 576, 1296, 1600, 2304, ...
%H A290169 Chai Wah Wu, <a href="/A290169/b290169.txt">Table of n, a(n) for n = 2..1000</a>
%e A290169 a(4)=30 because both the sum of the first 4 divisors of 30 (1 + 2 + 3 + 5 = 11) and the sum of its last 4 divisors (30 + 15 + 10 + 6 = 61) are prime numbers.
%t A290169 Table[k=1;While[Nand[Length@#>=n,PrimeQ[Total@Take[PadRight[#,n],n]]]||Nand[Length@#>=n,PrimeQ[Total@Take[PadLeft[#,n],n]]]&@Divisors@k,k++];k,{n,2,10}]
%Y A290169 Cf. A000043, A000668, A019279, A289776, A290126.
%K A290169 nonn
%O A290169 2,1
%A A290169 _Michel Lagneau_, Jul 23 2017