A258382 - OEIS

%I #40 Jul 19 2015 11:12:49

%S 144,441,1584,4851,10404,12544,14544,14884,15984,27648,40401,44521,

%T 44541,48841,48951,84672,114444,137984,144144,159984,409739,441441,

%U 444411,489731,489951,937904,1004004,1022121,1024144,1042441,1044484,1050804

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

%C 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.

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

%H Pieter Post and Giovanni Resta, <a href="/A258382/b258382.txt">Table of n, a(n) for n = 1..1124</a> (first 246 terms from Pieter Post, terms < 10^12)

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

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

%t 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 *)

%o (Python)

%o for n in range (1, 10**9):

%o ....y=int(str(n)[::-1])

%o ....ya=int(pow(n*y,1/2))

%o ....if ya==int(str(ya)[::-1]) and n*y==ya**2 and n!=y:

%o ........print (n)

%o (PARI) rev(k) = subst(Polrev(digits(k)), x, 10);

%o isok(n) = {rn = rev(n); if (rn != n, nrn = n*rn; issquare(nrn) && (y=sqrtint(nrn)) && (y == rev(y)););} \\ _Michel Marcus_, May 29 2015

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

%K nonn,base

%O 1,1

%A _Pieter Post_, May 28 2015