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%I #40 Jun 24 2022 05:22:08
%S 698896,637832238736,4099923883299904,6916103777337773016196,
%T 40460195511188111559106404,4872133543202112023453312784,
%U 9658137819052882509187318569,46501623417708833880771432610564,1635977102407987117897042017795361,163296619873968186681869378916692361
%N Palindromic squares with an even number of digits.
%D 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]
%H M. F. Hasler, <a href="/A027829/b027829.txt">Table of n, a(n) for n = 1..14</a>
%H K. S. Brown, <a href="http://www.mathpages.com/home/kmath359.htm">On General Palindromic Numbers</a>
%H Patrick De Geest, <a href="http://www.worldofnumbers.com/square.htm">Palindromic Squares</a>
%H Patrick De Geest, <a href="http://www.worldofnumbers.com/subsquar.htm">Subsets of Palindromic Squares</a>
%H Michael Keith, <a href="/A002778/a002778_1.pdf">Classification and enumeration of palindromic squares</a>, J. Rec. Math., 22 (No. 2, 1990), 124-132. [Annotated scanned copy]
%F a(n) = A016113(n)^2. - _M. F. Hasler_, Jun 08 2014
%e 836^2 = 698896, which is palindromic, so 698896 is in the sequence.
%e 1001^2 = 1002001, which is palindromic, but it has an odd number of digits, so it's not in the sequence.
%t Select[Range[1000000]^2, PalindromeQ[#] && OddQ[Floor[Log[10, #]]] &] (* _Alonso del Arte_, Oct 11 2019 *)
%o (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
%o (Scala) def isPalindromic(n: BigInt): Boolean = n.toString == n.toString.reverse
%o val squares = ((1: BigInt) to (1000000: BigInt)).map(n => n * n)
%o squares.filter(n => isPalindromic(n) && n.toString.length % 2 == 0) // _Alonso del Arte_, Oct 07 2019
%o (Python)
%o from math import isqrt
%o from itertools import count, islice
%o def A027829_gen(): # generator of terms
%o 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))))
%o A027829_list = list(islice(A027829_gen(),3)) # _Chai Wah Wu_, Jun 23 2022
%Y Cf. A002113, A002778, A002779, A016113.
%K nonn,base
%O 1,1
%A Keith Devlin, via Boon Leong (boon_leong(AT)hotmail.com)
%E Two new terms were recently found by Bennett from UK (communication from _Patrick De Geest_, Dec. 1999 or before)
%E Edited by _M. F. Hasler_, Jun 08 2014
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