Big numbers, programming contest

[T. D. Noe]

At 9:33 PM -0500 7/6/05, Ed Pegg Jr wrote:
>Don Reble wrote:
>> Ed Pegg Jr wrote:
>>
>>>http://www.research.att.com/projects/OEIS?Anum=A023199
>>
>>
>>>{3!, 5!, 11#3!2, 17#5!2, 29#7!4, 53#7!12, 89#11!2,
>>>157#17#8!6, ???, 487#*29#*10!/2}
>>
>>
>> The missing term is 271#23#10! .
>
>I verify this
>
>>
>> That last term (487#29#10!/2) is wrong; it has sigma/n = 10.995-.
>> It should be just 487#29#10! .
>
>I verify this, too.  My original code had a rounding error.
>
>>
>> The next term is 857#37#11!42 .
>
>I verify this, too.  Beats the number I had.
>
>For the next term I have 1487#59#15!/6   -- but it may be beatable.
>
>Ed Pegg Jr

Let c(p) be the smallest colossally-abundant number having the prime factor
p.  See A073751 for info about computing these numbers.

Then the terms of this sequence can be expressed

 a(2) = c(3)
 a(3) = c(5)     * 2
 a(4) = c(11)    / 2
 a(5) = c(17)    / 3
 a(6) = c(29)    * 14
 a(7) = c(53)
 a(8) = c(89)    * 4
 a(9) = c(157)   * 34
a(10) = c(271)   * 23
a(11) = c(487)   / 2
a(12) = c(857)   / 2
a(13) = c(1487)  * 212
a(14) = c(2621)  * 710
a(15) = c(4561)  * 506
a(16) = c(8011)  / 2
a(17) = c(13999) * 1630

The terms through a(12) match the results posted above.  I don't agree with
your a(13).  My 1487#53#15!2 is smaller.

Tony