A065378
Primes p such that p^2 is a palindromic square.
Original entry on oeis.org
2, 3, 11, 101, 307, 30001253, 100111001, 110111011, 111010111, 111091111, 1011099011101, 1100011100011, 1100101010011, 1101010101011, 100110101011001, 100110990111001, 101000010000101, 101011000110101, 101110000011101
Offset: 1
E.g. a(6) = 900075181570009 = p^2 with p = 30001253 and prime.
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t={}; Do[p=Prime[n]; If[FromDigits[Reverse[IntegerDigits[p^2]]]==p^2,AppendTo[t,p]],{n,10^7}]; t (* Jayanta Basu, May 11 2013 *)
A028817
Palindromic squares with an odd number of digits.
Original entry on oeis.org
1, 4, 9, 121, 484, 676, 10201, 12321, 14641, 40804, 44944, 69696, 94249, 1002001, 1234321, 4008004, 5221225, 6948496, 100020001, 102030201, 104060401, 121242121, 123454321, 125686521, 400080004, 404090404, 522808225, 617323716, 942060249, 10000200001
Offset: 1
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id[n_] := IntegerDigits[n]; palQ[n_] := FromDigits[Reverse[id[n]]] == n; t = {}; Do[If[palQ[x = n^2] && OddQ[Length[id[x]]], AppendTo[t, x]],{n, 101000}]; t (* Jayanta Basu, May 13 2013 *)
Select[Range[100000]^2, PalindromeQ[#] && EvenQ[Floor[Log[10, #]]] &] (* Alonso del Arte, Oct 11 2019 *)
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def isPalindromic(n: BigInt): Boolean = n.toString == n.toString.reverse
val squares = ((1: BigInt) to (1000000: BigInt)).map(n => n * n)
squares.filter(n => isPalindromic(n) && n.toString.length % 2 == 1) // Alonso del Arte, Oct 07 2019
A065379
Palindromic squares with a prime root.
Original entry on oeis.org
4, 9, 121, 10201, 94249, 900075181570009, 12124434743442121, 12323244744232321, 12341234943214321, 1022321210249420121232201, 1210024420147410244200121, 1210222232227222322220121
Offset: 1
a(6) = 900075181570009 = p^2 with p = 30001253, a prime.
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