A319441 Cubes of non-palindromic numbers.

1000, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507, 91125, 97336, 103823, 110592, 117649, 125000, 132651, 140608

Offset: 1

Author

Seiichi Manyama, Sep 19 2018

Keywords

• nonn
• base

Comments

This is not a subsequence of A029742. - Bruno Berselli, Sep 19 2018

Examples

			2201^3 = 10662526601 is a term.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• G. J. Simmons, Palindrome cubes: Problem B-183, Fibonacci Quart. 8 (1970), no. 5, p. 551.

Crossrefs

Cf. A000578, A002113, A029742, A319389, A319440.

Programs

Magma

[n^3: n in [0..65] | Intseq(n) ne Reverse(Intseq(n))]; // Vincenzo Librandi, Sep 19 2018

Mathematica

palQ[n_]:=Module[{idn=IntegerDigits[n]}, idn==Reverse[idn]]; DeleteCases[Range[10, 110], ?palQ]^3 (* _Vincenzo Librandi, Sep 19 2018 *)
Select[Range[100],!PalindromeQ[#]&]^3 (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 13 2019 *)

PARI

is_a029742(n)=my(d=digits(n)); d!=Vecrev(d) \\ after Charles R Greathouse IV in A029742
terms(n) = my(i=0, x=1); while(1, if(i==n, break, if(is_a029742(x), print1(x^3, ", "); i++)); x++)
/* Print initial 40 terms as follows */
terms(40) \\ Felix Fröhlich, Sep 19 2018

Python

def A319441(n):
    def f(x): return n+x//10**((l:=len(s:=str(x)))-(k:=l+1>>1))-(int(s[k-1::-1])>x%10**k)+10**(k-1+(l&1^1))-1
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return m**3 # Chai Wah Wu, Jul 24 2024

Formula

a(n) = A029742(n)^3.