A258382 Non-palindromic numbers n such that the square root of n multiplied by the reversal of n is a palindrome.

144, 441, 1584, 4851, 10404, 12544, 14544, 14884, 15984, 27648, 40401, 44521, 44541, 48841, 48951, 84672, 114444, 137984, 144144, 159984, 409739, 441441, 444411, 489731, 489951, 937904, 1004004, 1022121, 1024144, 1042441, 1044484, 1050804

Offset

1,1

Comments

This sequence is infinite, because it contains several infinite subsequences such as: sqrt(1584*4851)=2772, sqrt(15984*48951)=27972, sqrt(159..984*489...951)=279...972.

It appears that the first (or last) digit is never 5, 6 or 7.

Links

Pieter Post and Giovanni Resta, Table of n, a(n) for n = 1..1124 (first 246 terms from Pieter Post, terms < 10^12)

Formula

Numbers n such that sqrt(n*reversal(n)) is a palindrome, where n is not a palindrome.

Example

27648 is in the sequence because sqrt(27648*84672)=48384.

Mathematica

palQ[n_] := Block[{d = IntegerDigits@ n}, And[IntegerQ@ n, d == Reverse@ d]]; Select[Range@ 100000, And[! palQ@ #, palQ[Sqrt[# FromDigits@ Reverse@ IntegerDigits@ #]]] &] (* Michael De Vlieger, May 28 2015 *)

Prog

(Python)
for n in range (1, 10**9):
....y=int(str(n)[::-1])
....ya=int(pow(n*y, 1/2))
....if ya==int(str(ya)[::-1]) and n*y==ya**2 and n!=y:
........print (n)
(PARI) rev(k) = subst(Polrev(digits(k)), x, 10);
isok(n) = {rn = rev(n); if (rn != n, nrn = n*rn; issquare(nrn) && (y=sqrtint(nrn)) && (y == rev(y)); ); } \\ Michel Marcus, May 29 2015

Crossrefs

Cf. A002113, A002778, A059744, A004086, A117281.

Sequence in context: A178972 A250787 A188246 * A151820 A322905 A281240

Adjacent sequences: A258379 A258380 A258381 * A258383 A258384 A258385

Keyword

nonn,base

Author

Pieter Post, May 28 2015

Status

approved