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

1,1

Links

Table of n, a(n) for n=1..70.

Patrick De Geest, Palindromic Squares

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).

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)

Example

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.

Mathematica

Table[Length[Select[Range[If[n == 1, 0, Ceiling[Sqrt[10^(n - 1)]]], Floor[Sqrt[10^n]]], # == IntegerReverse[#] && #^2 == IntegerReverse[#^2] &]], {n, 15}]

Crossrefs

Cf. A002778, A002779, A034307, A034822, A218035.

Sequence in context: A319037 A067565 A243154 * A226782 A010635 A272198

Adjacent sequences: A307714 A307715 A307716 * A307718 A307719 A307720

Keyword

nonn,base,easy

Author

Robert Price, Apr 23 2019

Extensions

a(16)-a(20) from Robert Price, Apr 25 2019

a(21)-a(70) from Giovanni Resta, Apr 28 2019

Status

approved