Let us call "pretty" a number in the sequence which is $\ge 100$ and is in the form $x=\overbrace{aa...aa}^{\times p}\overbrace{(a-1)(a-1)...(a-1)(a-1)}^{\times q}\overbrace{(a-2)(a-2)...(a-2)(a-2)}^{\times r}$ where $p,q$ are positive integers, $r$ is a nonnegative integer and $a$ is some integer in $\{1,2,...,9\}$ (with $r=0$ if $a=1$). (not all the numbers in the sequence are pretty) Let us write such a pretty number $x$ as $(a,p,q,r)$ (this number contains $p+q+r$ decimal digits) 1.1) successor of a pretty number SuccessorsSuccessor of pretty number $(a,p,q,r)$ is : If $r\ge 1$ : pretty number $(a,p+1,q,r-1)$ If $r=0$ and $a\le 8$ : pretty number $(a+1,1,p,q-1)$ If $r=0$ and $a=9$ and $q=1$ : pretty number $(1,1,p+1,0)$ and we increased the number of digits If $r=0$ and $a=9$ and $q\ge 2$, successor is not a pretty number : $\overbrace{99...99}^{\times p+1}\overbrace{88...88}^{\times q-2}0$ next successor is not a pretty number : $\overbrace{99...99}^{\times p+1}\overbrace{88...88}^{\times q-2}2$ next successor is not a pretty number : $\overbrace{99...99}^{\times p+1}\overbrace{88...88}^{\times q-2}4$ next successor is not a pretty number : $\overbrace{99...99}^{\times p+1}\overbrace{88...88}^{\times q-2}6$ next successor is pretty number $(9,p+1,q-1,0)$ 1.2) We need $4(p+q)$ steps to go from $(1,p,q,0)$ to $(9,p,q,0)$ ProofFrom $(1,p,q,0)$, we need one step to reach $(2,1,p,q-1)$ Then $q-1$ steps to reach $(2,q,p,0)$ One step to reach $(3,1,q,p-1)$ Then $p-1$ steps to reach $(3,p,q,0)$ One step to reach $(4,1,p,q-1)$ Then $q-1$ steps to reach $(4,q,p,0)$ One step to reach $(5,1,q,p-1)$ Then $p-1$ steps to reach $(5,p,q,0)$ One step to reach $(6,1,p,q-1)$ Then $q-1$ steps to reach $(6,q,p,0)$ One step to reach $(7,1,q,p-1)$ Then $p-1$ steps to reach $(7,p,q,0)$ One step to reach $(8,1,p,q-1)$ Then $q-1$ steps to reach $(8,q,p,0)$ One step to reach $(9,1,q,p-1)$ Then $p-1$ steps to reach $(9,p,q,0)$ And so $4(p+q)$ steps to go from $(1,p,q,0)$ to $(9,p,q,0)$ Q.E.D. 1.3) We need $4p+5$ steps to go from $(1,p,1,0)$ to $(1,1,p+1,0)$ ProofWe need $4(p+1)$ step to go from $(1,p,1,0)$ to $(9,p,1,0)$ And successor is $(1,1,p+1,0)$ Q.E.D. 1.4) We need $9q$ steps to go from $(1,1,q,0)$ to $(1,1,q+1,0)$ where $q\ge 2$ ProofWe need $4(1+q)$ steps to go from $(1,1,q,0)$ to $(9,1,q,0)$ Then we need $5$ steps to reach (thru four nonpretty numbers $\overbrace{99...99}^{\times 2}\overbrace{88...88}^{\times q-2}0$ to $\overbrace{99...99}^{\times 2}\overbrace{88...88}^{\times q-2}6$) pretty number $(9,2,q-1,0)$ And so $5(q-1)$ steps to go from $(9,1,q,0)$ to $(9,q,1,0)$ And one step to reach $(1,1,q+1,0)$ Q.E.D.