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Number of n-digit numbers x such that rev(x^2) = rev(x)^2 and x does not contain any zero digits, where rev(x) is the digit reversal of x.
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%I #20 Feb 20 2021 00:32:05

%S 3,6,9,11,10,7,7,1,1

%N Number of n-digit numbers x such that rev(x^2) = rev(x)^2 and x does not contain any zero digits, where rev(x) is the digit reversal of x.

%C The number of solutions of rev(x^2) = rev(x)^2 increases but the solutions with a 0 don't. Any number with more than 9 digits can't be a solution, due to the development of x^2.

%e The 7 solutions with 7 digits are 1111111, 1111112, 1111121, 1111211, 1121111, 1211111, 2111111.

%o (PARI) isok(k) = my(d=digits(k)); vecmin(d) && (fromdigits(Vecrev(digits(k^2))) == fromdigits(Vecrev(d))^2);

%o a(n) = sum(k=10^(n-1), 10^n-1, isok(k)); \\ _Michel Marcus_, Jan 16 2021

%Y Cf. A098701 (number of solutions) and A085305 (the solutions), where digit 0 is not forbidden.

%Y Cf. A004086 (digit reversal), A052382 (zeroless numbers).

%K nonn,base,fini,full

%O 1,1

%A _Sébastien Dumortier_, Jan 10 2021