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
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"?
Programs
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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.
Comments