[seqfan] n consecutive perfect powers sum to a perfect power

[jnthn stdhr]

Tim,

  Would you please add a comment about a(59924) ?

  And, unless you care to do the tasks below for yourself, would you email
me your list of solutions so that I can double-check my terms in the DATA
section? I will then update the section with any new terms from your list
and credit you.  I could also create a b-file from your list.

  I would run your code, but I use Qpython on my phone and that uses python
version 3.6, so it doesn't understand the  := operator.

Thank you.


Jack,

  Would you please add a comment regarding a(9).

Thank you.


On Wednesday, December 13, 2023, jnthn stdhr <jstdhr at gmail.com> wrote:

> Since there are no objections and a bit of interest, this sequence is now
> https://oeis.org/A368161.
>
> Cheers
>
> On Wednesday, December 13, 2023, Jack Brennen <jfb at brennen.net> wrote:
>
>> I've been looking for a solution for 9 consecutive powers.
>>
>> If one exists, it's greater than 6.8 * 10^19.
>>
>> Note that any such solution must include an odd power (cube, 5th power,
>> etc.).
>>
>> This is because the sum of 9 consecutive squares is of the form 9k^2 +
>> 60, which can't be a perfect power of course, because it's divisible by 3
>> with multiplicity of 1.
>>
>>
>>
>> On 12/13/2023 8:21 PM, Tim Peters wrote:
>>
>>> I'll attach a Python program that works to cut off sums when it becomes
>>> futile to pursue them. I assume your algorithm isn't doing that, because
>>> this only takes about 2 seconds to finish 10^10 (running under PyPy).
>>>
>>> At 10^14 it took noticeable (very noticeable ;-) ) time, but still didn't
>>> find a solution for sums of 9 or 10 consecutive powers. It found 744
>>> solutions in all, the longest being a sum of 59924 consecutive powers,
>>> starting at 37711881, summing to 92121066512784 (the square of 9597972).
>>>
>>> The code:
>>>
>>> from math import isqrt
>>>
>>> TOP = 10**10
>>> SHOW_DELETES = False
>>>
>>> ppset = {1}
>>> for i in range(2, isqrt(TOP) + 1):
>>>     power = i
>>>     while (power := power * i) <= TOP:
>>>         ppset.add(power)
>>> pp = sorted(ppset)
>>> npp = len(pp)
>>> MAXPP = pp[-1]
>>> print(f"len(pp)={npp:_} through {TOP=:_}; {MAXPP=:_}")
>>>
>>> sums = pp[:]
>>> nterms = 0
>>> delta = 1
>>> while sums:
>>> ##    # Crucial invariant, but expensive to check.
>>> ##    assert all(sums[i] == sum(pp[i : i + delta])
>>> ##               for i in range(len(sums)))
>>>      winner = -1
>>>      for i, s in enumerate(sums):
>>>          if s in ppset:
>>>              winner = s
>>>              assert sum(pp[i : i + delta]) == s
>>>              break
>>>      if winner > 0:
>>>          nterms += 1
>>>          print(nterms, delta, winner,
>>>                f"starting at pp[{i}] = {pp[i]}, {len(sums)=:_}")
>>>      if len(sums) + delta > npp:
>>>          sums.pop()
>>>          assert len(sums) + delta == npp
>>>      for i in range(len(sums)):
>>>          sums[i] += pp[i + delta]
>>>          if sums[i] > MAXPP:
>>>              if SHOW_DELETES:
>>>                  print("    deleting", len(sums) - i,
>>>                        "of", len(sums),
>>>                        "at delta", delta)
>>>              del sums[i:]
>>>              break
>>> ##    assert sums == sorted(set(sums))
>>>      delta += 1
>>>
>>> On Wed, Dec 13, 2023 at 12:44 AM jnthn stdhr <jstdhr at gmail.com> wrote:
>>>
>>> Howdy, all.
>>>>
>>>>    What is the least perfect power m (in A001597) that is the sum of n
>>>> consecutive perfect powers.
>>>>
>>>>    The sequence isn't in the database, and begins 1, 25, 441, 100, 169,
>>>> 289,
>>>> 121, 2395417249, -1, -1, 676, 232324, -1, -1, -1, 64866916, 3721,
>>>> 3622354596, 279936, ..., with -1 representing no solution found up to
>>>> ~10^10.
>>>>
>>>>    For the first few terms, we have:
>>>>
>>>> {1}=1, {9+16}=25, {128+144+169}=441, {16+25+27+32}=100, etc.
>>>>
>>>>    Should I add this? If so, up to a(8) only, or include the -1s?
>>>>
>>>>   When I first searched for solutions, the maximum value of the set of
>>>> perfect powers was ~10^8, and both a(8) and a(18) came out -1.  But
>>>> when I
>>>> increased the max to 10^10 solutions for those two terms were found.  At
>>>> 10^8 I was able to get to 100+ terms in a somewhat reasonable time, with
>>>> solutions becoming more and more sparse.  At 10^10, things get very
>>>> bogged
>>>> down, but more solutions are found along the way.  Also, the erratic
>>>> nature
>>>> of the terms seems to persist.
>>>>
>>>>    Can a solution always be found if the set of perfect powers is large
>>>> enough?
>>>>
>>>> -Jonathan
>>>>
>>>> --
>>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>>
>>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>>
>>>