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

[jnthn stdhr]

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/
>>>
>>> --
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>>
>>
>>