A027829 Palindromic squares with an even number of digits.

698896, 637832238736, 4099923883299904, 6916103777337773016196, 40460195511188111559106404, 4872133543202112023453312784, 9658137819052882509187318569, 46501623417708833880771432610564, 1635977102407987117897042017795361, 163296619873968186681869378916692361

Offset: 1

Author

Keith Devlin, via Boon Leong (boon_leong(AT)hotmail.com)

Keywords

• nonn
• base

Examples

			836^2 = 698896, which is palindromic, so 698896 is in the sequence.
1001^2 = 1002001, which is palindromic, but it has an odd number of digits, so it's not in the sequence.
		

References

• Charles Ashbacher, More on palindromic squares, J. Rec. Math. 22, no. 2 (1990), 133-135. [A scan of the first page of this article is included with the last page of the Keith (1990) scan]

Links

• M. F. Hasler, Table of n, a(n) for n = 1..14
• K. S. Brown, On General Palindromic Numbers
• Patrick De Geest, Sporadic palindromic squares of even length
• Michael Keith, Classification and enumeration of palindromic squares, J. Rec. Math., 22 (No. 2, 1990), 124-132. [Annotated scanned copy]

Crossrefs

Cf. A002113, A002778, A002779, A016113.

Programs

Mathematica

Select[Range[1000000]^2, PalindromeQ[#] && OddQ[Floor[Log[10, #]]] &] (* Alonso del Arte, Oct 11 2019 *)

PARI

is_A027829(n)={issquare(n)&&Vecrev(n=digits(n))==n&&!bittest(#n, 0)} \\ This is faster than first checking for even length if applied to numbers known to have an even number of digits, as should be the case for a systematic search. For this, one should only consider squares, i.e., rather use is_A016113.  - M. F. Hasler, Jun 08 2014

Python

from math import isqrt
from itertools import count, islice
def A027829_gen(): # generator of terms
    return filter(lambda n: (s:=str(n))[:(t:=(len(s)+1)//2)]==s[:-t-1:-1], map(lambda n: n**2, (d for l in count(2,2) for d in range(isqrt(10**(l-1))+1,isqrt(10**l)+1))))
A027829_list = list(islice(A027829_gen(),3)) # Chai Wah Wu, Jun 23 2022

Scala

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 == 0) // Alonso del Arte, Oct 07 2019

Formula

a(n) = A016113(n)^2. - M. F. Hasler, Jun 08 2014

Extensions

Two new terms were recently found by Bennett from UK (communication from Patrick De Geest, Dec 1999 or before)

Edited by M. F. Hasler, Jun 08 2014