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.

2, 4, 30, 16, 140, 64, 264, 144, 336, 525, 144, 800, 1200, 576, 1600, 2016, 1440, 1296, 2160, 2304, 7980, 6440, 3360, 8360, 4080, 3960, 2772, 16100, 9108, 10608, 7392, 12320, 14688, 37240, 21780, 18200, 45760, 20160, 9240, 24624, 14364, 8400, 22176, 23760

Offset

2,1

Comments

The corresponding pairs of primes are (3, 3), (7, 7), (11, 61), (31, 31), (29, 307), (127, 127), (47, 673), (61, 379), (73, 919), ...

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).

The squares of the terms of the sequence are 4, 16, 64, 144, 576, 1296, 1600, 2304, ...

Links

Chai Wah Wu, Table of n, a(n) for n = 2..1000

Example

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.

Mathematica

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}]

Crossrefs

Cf. A000043, A000668, A019279, A289776, A290126.

Sequence in context: A216026 A241589 A289776 * A232173 A067195 A080230

Adjacent sequences: A290166 A290167 A290168 * A290170 A290171 A290172

Keyword

nonn

Author

Michel Lagneau, Jul 23 2017

Status

approved