A307717 Number of palindromic squares, k^2, of length n such that k is also palindromic.
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, 65, 0, 138, 0, 94, 0, 197, 0, 131, 0, 272, 0, 177, 0, 365, 0, 233, 0, 478, 0, 300, 0, 613, 0, 379, 0, 772, 0, 471, 0, 957, 0, 577, 0, 1170, 0, 698, 0, 1413, 0
Offset: 1
Examples
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.
Links
- Patrick De Geest, Palindromic Squares in bases 2 to 17
- M. Kauers and C. Koutschan, Guessing with Little Data, arXiv:2202.07966 [cs.SC], 2022.
- Bertrand Teguia Tabuguia and Wolfram Koepf, FPS In Action: An Easy Way To Find Explicit Formulas For Interlaced Hypergeometric Sequences, arXiv:2207.01031 [cs.SC], 2022.
- Bertrand Teguia Tabuguia, Hypergeometric-Type Sequences, arXiv:2401.00256 [cs.SC], 2023.
- Eric Weisstein's World of Mathematics, Palindromic Number
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,4,0,0,0,-6,0,0,0,4,0,0,0,-1).
Programs
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Mathematica
Table[Length[Select[Range[If[n == 1, 0, Ceiling[Sqrt[10^(n - 1)]]], Floor[Sqrt[10^n]]], # == IntegerReverse[#] && #^2 == IntegerReverse[#^2] &]], {n, 15}]
Formula
From Christoph Koutschan, Feb 19 2022: (Start)
a(2n-1) = A218035(n).
a(n) is given by a quasi-polynomial (for a proof, see A218035):
a(1) = 4;
a(2n) = 0;
a(4n+1) = (n^3-3*n^2+11*n+6)/3 (n > 0);
a(4n+3) = (n^3+5*n+12)/6 (n >= 0). (End)
Extensions
a(16)-a(20) from Robert Price, Apr 25 2019
a(21)-a(70) from Giovanni Resta, Apr 28 2019