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%I #56 Jan 05 2024 15:08:13
%S 4,0,2,0,5,0,3,0,8,0,5,0,13,0,9,0,22,0,16,0,37,0,27,0,60,0,43,0,93,0,
%T 65,0,138,0,94,0,197,0,131,0,272,0,177,0,365,0,233,0,478,0,300,0,613,
%U 0,379,0,772,0,471,0,957,0,577,0,1170,0,698,0,1413,0
%N Number of palindromic squares, k^2, of length n such that k is also palindromic.
%H Patrick De Geest, <a href="http://www.worldofnumbers.com/square.htm">Palindromic Squares</a>
%H M. Kauers and C. Koutschan, <a href="https://arxiv.org/abs/2202.07966">Guessing with Little Data</a>, arXiv:2202.07966 [cs.SC], 2022.
%H Bertrand Teguia Tabuguia and Wolfram Koepf, <a href="https://arxiv.org/abs/2207.01031">FPS In Action: An Easy Way To Find Explicit Formulas For Interlaced Hypergeometric Sequences</a>, arXiv:2207.01031 [cs.SC], 2022.
%H Bertrand Teguia Tabuguia, <a href="https://arxiv.org/abs/2401.00256">Hypergeometric-Type Sequences</a>, arXiv:2401.00256 [cs.SC], 2023.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PalindromicNumber.html">Palindromic Number</a>
%H <a href="/index/Rec#order_16">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,4,0,0,0,-6,0,0,0,4,0,0,0,-1).
%F From _Christoph Koutschan_, Feb 19 2022: (Start)
%F a(2n-1) = A218035(n).
%F a(n) is given by a quasi-polynomial (for a proof, see A218035):
%F a(1) = 4;
%F a(2n) = 0;
%F a(4n+1) = (n^3-3*n^2+11*n+6)/3 (n > 0);
%F a(4n+3) = (n^3+5*n+12)/6 (n >= 0). (End)
%e There are only two palindromic squares of length 3 whose root is also palindromic. 11^2=121 and 22^2=484. Thus, a(3)=2.
%t Table[Length[Select[Range[If[n == 1, 0, Ceiling[Sqrt[10^(n - 1)]]], Floor[Sqrt[10^n]]], # == IntegerReverse[#] && #^2 == IntegerReverse[#^2] &]], {n, 15}]
%Y Cf. A002778, A002779, A034307, A034822, A218035.
%K nonn,base,easy
%O 1,1
%A _Robert Price_, Apr 23 2019
%E a(16)-a(20) from _Robert Price_, Apr 25 2019
%E a(21)-a(70) from _Giovanni Resta_, Apr 28 2019
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