A061226 -id:A061226 - cp's OEIS frontend

A319603 a(n) = n^3 + reversal of digits of n^3.

0, 2, 16, 99, 110, 646, 828, 686, 727, 1656, 1001, 2662, 9999, 10109, 7216, 9108, 11000, 8107, 8217, 16445, 8008, 10890, 95249, 88288, 56655, 68276, 85147, 58374, 47864, 122731, 27072, 49583, 119491, 109890, 79697, 100699, 112320, 86258, 82717, 150714, 64046, 81907, 162135, 150104

Offset: 0

Seiichi Manyama, Sep 24 2018

Seiichi Manyama, Table of n, a(n) for n = 0..10000

n^b + reversal of digits of n^b: A056964 (b=1), A061226 (b=2), this sequence (b=3).

Cf. A000578, A004165, A072384 (subsequence of primes).

{a(n) = n^3+fromdigits(Vecrev(digits(n^3)))}

a(n) = A000578(n) + A004165(n).

A113800 Numbers k such that k^2 plus the reverse of k^2 gives a perfect power.

Original entry on oeis.org

2, 231, 1010, 102010, 451429, 1000100, 9426681, 1000001000, 8803095102, 10002000100, 56017891104, 1000000010000, 4811618419542

Offset: 1

Giovanni Resta, Jan 22 2006

Sequence is infinite since it contains all the numbers of the form (1+100^k)*10^k and (1+100^k)^2*10^k, for k >= 1. - Giovanni Resta, Sep 24 2013

			451429^2 = 203788142041 and 203788142041 + 140241887302 = 7007^3,
9426681^2 = 88862314675761 and 88862314675761 + 16757641326888 = 10277157^2.

Cf. A000290, A002942, A061226.

ppQ[n_] := n != 1 && GCD @@ (Transpose[FactorInteger[n]][[2]]) > 1; rev[n_] := FromDigits@Reverse@IntegerDigits@n ; lst = {}; Do[If[ppQ[n^2 + rev[n^2]], AppendTo[lst, n]], {n, 10^6}]; lst

a(8)-a(13) from Giovanni Resta, Sep 24 2013

cp's OEIS Frontend

A319603 a(n) = n^3 + reversal of digits of n^3.

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A113800 Numbers k such that k^2 plus the reverse of k^2 gives a perfect power.

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