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
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jul 23 2017
STATUS
approved