A085315 -id:A085315 - cp's OEIS frontend

A069494 Numbers n such that (reversal(n))^3 = reversal(n^3). Ignore leading 0's.

0, 1, 2, 7, 10, 11, 20, 70, 100, 101, 110, 111, 200, 700, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 2000, 7000, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 11000, 11001, 11010, 11011, 11100, 20000, 70000, 100000, 100001, 100010, 100011, 100100, 100101

Offset: 1

Joseph L. Pe, Apr 15 2002

For an arithmetical function f, call the arguments n such that f(reverse(n)) = reverse(f(n)) the "palinpoints" of f. This sequence is the sequence of palinpoints of f(n) = n^3.

			Let f(n) = n^3. Then f(1011) = 1033364331, f(1101) = 1334633301, so f(reverse(1011)) = reverse(f(1011)). Therefore 1011 belongs to the sequence.

Cf. A000578, A004086, A085315.

r:= n-> (s-> parse(cat(seq(s[-i], i=1..length(s)))))(""||n):
q:= n-> is(r(n^3)=r(n)^3):
select(q, [$0..200000])[];  # Alois P. Heinz, Oct 22 2021

rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; f[n_] := n^3; Select[Range[10^5], f[rev[ # ]] == rev[f[ # ]] &]

More terms from Alois P. Heinz, Oct 22 2021

A107340 Number of solutions to rev(x^3)=rev(x)^3 below 10^n.

Original entry on oeis.org

3, 4, 6, 9, 14, 25, 40, 62, 89, 129, 185, 271

Offset: 1

Martin Renner, May 22 2005

Cf. A085315.

A107341 Number of solutions to rev(x^3)=rev(x)^3 with n digits.

Original entry on oeis.org

3, 1, 2, 3, 5, 11, 15, 22, 27, 40, 56, 86

Offset: 1

Martin Renner, May 22 2005

Cf. A085315.

cp's OEIS Frontend

A069494 Numbers n such that (reversal(n))^3 = reversal(n^3). Ignore leading 0's.

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A107340 Number of solutions to rev(x^3)=rev(x)^3 below 10^n.

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A107341 Number of solutions to rev(x^3)=rev(x)^3 with n digits.

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