

A225603


Palindromic primes whose square is also a palindrome.


2



2, 3, 11, 101, 100111001, 110111011, 111010111, 1100011100011, 1100101010011, 1101010101011, 100110101011001, 101000010000101, 101011000110101, 101110000011101, 10000010101000001, 10011010001011001, 10100110001100101, 10110010001001101, 10111000000011101
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OFFSET

1,1


COMMENTS

Palindromes in A161721. Conjecture: a(n) for n >=3 consists only of the digits 0,1.  Chai Wah Wu, Jan 06 2015


LINKS



EXAMPLE

101 is a member since it is a palindromic prime such that 101^2=10201 is a palindrome.


MATHEMATICA

palQ[n_]:=FromDigits[Reverse[IntegerDigits[n]]]==n; t={}; Do[If[palQ[p=Prime[n]] && palQ[p^2], AppendTo[t, p]], {n, 10^7}]; t


PROG

(Python)
from __future__ import division
from sympy import isprime
def paloddgenrange(t, l, b=10): # generator of oddlength palindromes in base b of 2*t <=length <= 2*l
....if t == 0:
........yield 0
....else:
........for x in range(t+1, l+1):
............n = b**(x1)
............n2 = n*b
............for y in range(n, n2):
................k, m = y//b, 0
................while k >= b:
....................k, r = divmod(k, b)
....................m = b*m + r
................yield y*n + b*m + k
for i in paloddgenrange(1, 10):
....s = str(i*i)
....if s == s[::1] and isprime(i):


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



