A334916 a(n) is the smallest number > 1 whose base n digits yield the original number when added and multiplied left to right; or 0 if no such number exists.

0, 0, 0, 6, 12, 160, 324, 405, 12, 8385, 36, 189, 784, 32, 1656, 20, 721, 25215, 80, 45, 559, 2585, 5525, 323844, 30, 160, 60, 90, 150, 1071, 11650, 1038448, 6275, 2669, 77, 42, 2224, 324224, 1817, 2016, 252, 7425, 1593074855, 96, 5450, 192, 345906, 23541, 56

Offset: 1

Author

Matej Veselovac, May 16 2020

Keywords

• nonn
• base
• hard

Comments

These numbers have been called "baseless in base n".

a(n) is divisible by its last base n digit.

The number 8385 = ((((8)8+3)3+8)8+5)5 is known to be the unique baseless number in base 10. Are there number bases n, other than 6 and 10, that have a unique example?

If the term a(107) is not zero, then it is at least a(107) > 107^6 > 1.5*10^12. Is it true that a(n)>0 for all n>3?

Examples

			Every number can be written as A = (...((((a)N+b)N+c)N+d)...) where a,b,c,d,... are digits of number A in base N. If we take that expression and replace the "multiplications by base N" with "multiplications by digits a,b,c,d,..." and also multiply it with the last digit to use up all digits, we get some number A*. If it holds A = A*, then we say number A is a baseless number.
For example, the decimal number base has only one baseless number:
.
a(10) = 8385 = ((((8)*10+3)*10+8)*10+5) = ((((8)*8+3)*3+8)*8+5)*5.
.
There are at most finitely many baseless numbers for every fixed number base. For example, the number base 4 has exactly three baseless numbers:
.
6 = ((1)*4+2)        = ((1)*1+2)*2       = 12_4;
27 = (((1)*4+2)*4+3) = (((1)*1+2)*2+3)*3 = 123_4;
46 = (((2)*4+3)*4+2) = (((2)*2+3)*3+2)*2 = 232_4;
.
The smallest of them is 6, hence a(4)=6.
		

Links

• Math StackExchange user "Vepir" (Matej Veselovac), Terms a(n) < 10^10 for n < 500, and including the 11th record a(73) ~ 2*10^10.
• Math StackExchange, Does every number base have at least one "baseless number"?

Crossrefs

Cf. A000290 (perfect squares), A334917 (indices of records).

Programs

PARI

\\ for n>=4
isok(k,n) = {my(d=digits(k, n), s=0); for (i=1, #d, s = (s+d[i])*d[i];); s == k;}
a(n) = {my(k=2); while (!isok(k, n), k++); k;} \\ Michel Marcus, Jun 18 2020

Formula

If n is a perfect square, then a(n) = n + sqrt(n). Otherwise, a(n) > 2n.