|
| |
|
|
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.
|
|
1
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
