For how many integers having between 1 and 10 digits (base 10) are all of their digits when read from left to right monotonically increasing? In other words, every digit is less than or equal to all of those to its right. For example, 244467889 is one of them, and 0 is another, but there are more.
- found my write-up on the mapping -
I have finally succeeded to define the mapping of all n-digit increasing numbers into a binary word of n+9 bits.
Lemma:
Every binary string of length n+9 bits (n ones, 9 zeroes) represents an n-digit decimal number written in “all leading zeroes included” notation i.e. all numbers from 0 to 10^n-1.
Take all 2 digit increasing numbers:
00000000011…..…99
00000000101……89
00000000110……88
00000001001……79
00000001010……78
00000001100……77
……………………
00010010000……35
……………………
00011000000……33
……………………
10000001000…….07
……………………
11000000000……..00
The first 1 - just write the digit equal to the number of zeroes preceding it, the second 1 transforms into the digit equal to the first augmented by the number of zeroes between the ones.
There are COMB(11,2) =55 increasing numbers below 100 and COMB(19,10) =92378 increasing numbers below 10^10.
One can prove that the sum for comb(8+m,8) , (m from 0 to m) equals comb(9+m,m).
Edited on January 7, 2016, 1:34 am
My mapping
