A256437 Nonnegative integers i such that i^2 + reverse(i)^2 is a palindrome.

0, 1, 2, 10, 11, 12, 20, 21, 30, 100, 101, 102, 110, 111, 120, 200, 201, 210, 220, 300, 310, 1000, 1001, 1002, 1010, 1011, 1012, 1020, 1021, 1022, 1100, 1101, 1102, 1110, 1111, 1120, 1200, 1201, 1210, 1300, 2000, 2001, 2010, 2011, 2100, 2101, 2110, 2200, 2201

Offset

1,3

Comments

This sequence generates A256398.

Links

Bui Quang Tuan, Table of n, a(n) for n = 1..554

Example

12 is in the sequence because 12^2 + 21^2 = 585, a palindrome.

Mathematica

palQ[n_] := Reverse@ IntegerDigits@ n == IntegerDigits@ n; Select[
Range@2210, palQ[#^2 + FromDigits[Reverse[IntegerDigits@ #]]^2] &] (* Michael De Vlieger, Mar 29 2015 *)

Prog

(PARI) rev(n)=r=""; d=digits(n); for(i=1, #d, r=concat(Str(d[i]), r)); eval(r);
ispal(n) = d = digits(n); Vecrev(d) == d;
isok(n) = ispal(n^2+rev(n)^2) \\ Michel Marcus, Apr 01 2015
(Python)
A256437_list = [i for i in range(10**6) if str(i**2 + int(str(i)[::-1])**2) == str(i**2 + int(str(i)[::-1])**2)[::-1]] # Chai Wah Wu, Apr 09 2015

Crossrefs

Cf. A002113, A004086, A256398.

Sequence in context: A007089 A159952 A136820 * A136810 A136821 A381579

Adjacent sequences: A256434 A256435 A256436 * A256438 A256439 A256440

Keyword

nonn,base

Author

Bui Quang Tuan, Mar 29 2015

Status

approved