A292788 For n > 1, a(n) = least positive k, not a power of n, such that the digital sum of k in base n equals the digital sum of k^3 in base n.

56953, 13, 2, 3, 20, 2, 6, 3, 8, 5, 1110, 3, 65, 8, 4, 7, 86, 9, 2374, 4, 8, 12, 114, 3, 99, 12, 135, 15, 3567, 4, 185, 15, 11, 16, 6, 19, 73, 20, 12, 5, 81, 6, 85, 23, 19, 24, 93, 7, 97, 24, 18, 27, 796, 28, 44, 7, 19, 28, 413, 4, 365, 32, 8, 31, 26, 21, 200

Offset: 2

Author

Rémy Sigrist, Sep 23 2017

Keywords

• nonn
• base
• look

Comments

The term a(10) = 8 belongs to A070276.

For any n > 1, a(n^2) <= n.

Is this sequence defined for any n > 1 ?

Apparently, a(k) < k for any odd k > 3.

Among the first 99 999 terms, the digital sum of a(n) in base n is > n for n = 2, 12, 20, 30.

The scatterplot of the sequence shows beams on the upper part, which correspond to clusters of close points for which a(n) = k*n + (n-k-e) for some k > 0 and e in { 0, 2 }.

See also A292787 for a similar sequence involving squares instead of cubes.

The least positive k, not a power of 2, such that the hamming weight of k equals the hamming weight of k^4 is 34225258495.

Examples

			For n = 3:
- let d_3 denote the digital sum in base 3 (d_3 = A053735),
- 1 is a power of 3,
- d_3(2) = 2 and d_3(2^3) = 4,
- 3 is a power of 3,
- d_3(4) = 2 and d_3(4^3) = 4,
- d_3(5) = 3 and d_3(5^3) = 7,
- d_3(6) = 2 and d_3(6^3) = 4,
- d_3(7) = 3 and d_3(7^3) = 5,
- d_3(8) = 4 and d_3(8^3) = 8,
- 9 is a power of 3,
- d_3(10) = 2 and d_3(10^3) = 4,
- d_3(11) = 3 and d_3(11^3) = 9,
- d_3(12) = 2 and d_3(12^3) = 4,
- d_3(13) = 3 and d_3(13^3) = 3,
- hence a(3) = 13.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 2..10000
• Rémy Sigrist, Colorized scatterplot of the sequence for n=2..100000

Crossrefs

Cf. A053735, A070276, A292787.

Programs

Mathematica

With[{kk = 10^5}, Table[SelectFirst[Complement[Range[2, kk], n^Range@ Floor@ Log[n, kk]], Total@ IntegerDigits[#, n] == Total@ IntegerDigits[#^3, n] &] /. k_ /; MissingQ@ k -> -1, {n, 2, 68}]] (* Michael De Vlieger, Sep 24 2017 *)

PARI

a(n) = my (p=1); for (k=1, oo, if (k==p, p*=n, if (sumdigits(k,n) == sumdigits(k^3,n), return (k))))