A283751 Least b>1 such that n, when expressed in base b and then interpreted in base b^2, yields a multiple of n.

2, 2, 2, 3, 2, 5, 3, 2, 2, 3, 5, 11, 4, 3, 2, 5, 2, 17, 3, 19, 5, 2, 11, 23, 4, 5, 3, 3, 2, 29, 6, 2, 2, 11, 3, 7, 6, 37, 19, 3, 8, 41, 2, 6, 5, 9, 23, 3, 4, 7, 5, 17, 13, 53, 3, 11, 2, 7, 29, 59, 6, 61, 2, 4, 2, 13, 11, 67, 17, 23, 10, 71, 6, 2, 37, 5, 19, 11

Offset

0,1

Comments

a(n) <= max(2, n) for any n >= 0.

a(n*a(n)) <= a(n) for any n >= 0 (see also A283927).

a(n) = 2 iff n belongs to A062846.

Conjecture: if n is composite, then a(n) < n (see also A283937).

Theorem: If n is composite, then a(n) < n. Proof: If n=ab with 1<a<b, then n_b=a0, which interpreted as ab^2 is a multiple of n; if n=b^2 then n_b=100, which interpreted as b^4 is again a multiple of n. - Michael R Peake, Mar 25 2017

First occurrence of b > 1: 1, 4, 13, 6, 31, 36, 41, 46, 71, 12, 133, 53, 155, 106, 161, 18, 199, 20, 261, ..., . - Robert G. Wilson v, Mar 19 2017

Links

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Example

The number 5, when expressed in base 2 and then interpreted in base 4, yields 17, not a multiple of 5.
The number 5, when expressed in base 3 and then interpreted in base 9, yields 11, not a multiple of 5.
The number 5, when expressed in base 4 and then interpreted in base 16, yields 17, not a multiple of 5.
The number 5, when expressed in base 5 and then interpreted in base 25, yields 25, a multiple of 5.
Hence, a(5)=5.

Mathematica

f[n_] := Block[{b = 2}, While[ Mod[ FromDigits[ IntegerDigits[n, b], b^2], n] > 0, b++]; b]; Array[f, 80, 0] (* Robert G. Wilson v, Mar 19 2017 *)

Prog

(PARI) a(n) = my (b=2); if (n>0, while (fromdigits(digits(n, b), b^2)%n, b++)); return (b)

Crossrefs

Cf. A062846, A155078, A283927, A283937.

Sequence in context: A319478 A341408 A104239 * A132321 A058726 A058614

Adjacent sequences: A283748 A283749 A283750 * A283752 A283753 A283754

Keyword

nonn,base

Author

Rémy Sigrist, Mar 15 2017

Status

approved