A342803
Primes p whose palindromization A082216(p) is a square or higher power.
Original entry on oeis.org
67, 449, 1367, 10303, 12343, 1003003, 1022141, 1230127, 1234543, 4004009, 121200307, 10022234347, 10201204021, 10203242527, 12100242001, 13310399303, 16151080151, 52281509069, 61584539747, 90608667517, 104190107303, 1020102040201, 1022143262341, 12384043938083
Offset: 1
The prime 449 belongs to sequence because 44944 is a square: 212^2.
The prime 1367 is in the sequence since 1367631 is a cube: 111^3.
The prime 13 is not a term as A082216(13) = 131 and 131 is prime. The prime 10303 is in the sequence since 1030301 is a cube: 101^3. - _Chai Wah Wu_, Aug 26 2021
Cf.
A082216 (smallest palindrome beginning with n).
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Select[Prime@Range@100000,Or@@(GCD@@Last/@FactorInteger@#>1&/@(FromDigits/@(Join[a,Reverse@#]&/@{a=IntegerDigits@#,Most@a})))&] (* Giorgos Kalogeropoulos, Mar 31 2021 *)
Corrected terms and missing terms added by
Chai Wah Wu, Aug 26 2021
A384612
a(n) is the smallest integer k such that k^n is an abelian square; or -1 if no such k exists.
Original entry on oeis.org
11, 836, 11, 207, 624, 818222, 1001, 2776, 100001, 32323107, 100001, 85692627, 10000001, 501249084, 10000001, 27962757, 41695607, 70983559, 72768046, 977688137, 219873071, 112562383, 2338280974, 2435385853, 1231380445, 4557057314, 361499019, 8096434047, 5278552513
Offset: 1
a(1) = 11, since 11^1 = 1|1
a(2) = 836, since 836^2 = 698|896
a(3) = 11, since 11^3 = 13|31
a(4) = 207, since 207^4 = 18360|36801
a(5) = 624, since 624^5 = 9460692|9690624
a(6) = 818222, since 818222^6 = 300072996174564185|100579862765194304
a(7) = 1001, since 1001^7 = 10070210350|35021007001
a(8) = 2776, since 2776^8 = 35265958674713|24535718936576
a(9) = 100001, since 100001^9 = 10000900036000840012600|12600084000360000900001.
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f:= proc(n) local k,d,x,L;
k:= 2;
do
x:= k^n;
d:= ilog10(x)+1;
if d::odd then k:= ceil(10^(d/n)); next fi;
L:= convert(x,base,10);
if sort(L[1..d/2]) = sort(L[d/2+1..d]) then return k fi;
k:= k+1
od;
end proc:
map(f, [$1..30]); # Robert Israel, Jun 05 2025
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from itertools import count
def ok(k, n):
s = str(k**n)
if len(s) % 2 != 0:
return False
mid = len(s) // 2
return sorted(s[:mid]) == sorted(s[mid:])
def a(n):
return next(k for k in count(2) if ok(k, n))
print([a(n) for n in range(1, 10)])
Showing 1-2 of 2 results.
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