

A353025


Terms of A352991 which are perfect powers.


2



1, 13527684, 34857216, 65318724, 73256481, 81432576, 139854276, 152843769, 157326849, 215384976, 245893761, 254817369, 326597184, 361874529, 375468129, 382945761, 385297641, 412739856, 523814769, 529874361, 537219684, 549386721, 587432169, 589324176, 597362481, 615387249, 627953481, 653927184
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OFFSET

1,2


COMMENTS

It appears that all terms are terms of A062503.
We note that a(n)=A352329(n) up to a(36)=A352329(36)=923187456, while the mentioned match does not hold starting from a(37)=14102987536 (since A352329(37)=1234608769).
There are no perfect powers among terms t which are permutations of 123_...(m  1)_m for m == {2, 3, 5, 6} (mod 9). This is since 10 == 1 (mod 9) and also (1 + 0) == 1 (mod 9), so digit position has no effect. Hence, t == A134804(m) (mod 9). Now, if m is such that A134804(m) = {3, 6}, there is a lone factor of 3, which is not a perfect power (indeed).
Therefore, all terms are necessarily congruent modulo 9 to 0 or 1 (see Marco Ripà link).
All terms up to 10^34 are squares (in particular, there are 67 squares with no more than 17 digits).  Aldo Roberto Pessolano, May 12 2022


LINKS



FORMULA

Digit sum of a(n) is always congruent to 0 or 1 modulo 9.
a(n) = m^2, where the integer m := m(n) is not a perfect power itself (conjectured).


EXAMPLE

75910168324 is a term since 75910168324 = 275518^2.


MATHEMATICA

z = 1; Do[r = Range[k];
n = ToExpression[StringJoin[ToString[#] & /@ r]];
If[And[Mod[n, 9] != 3, Mod[n, 9] != 6], d = DigitCount[n];
s = IntegerPart[Sqrt[10^(IntegerLength[n]  1)]];
f = IntegerPart[Sqrt[10^(IntegerLength[n])]];
Do[y = x^2;
If[DigitCount[y] == d, c = True;
Do[If[Not[StringContainsQ[ToString[y], ToString[i]]],
c = False], {i, 10, k}]; If[c, Print[z, " ", y]; z++]], {x, s,


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



