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Accumulative sum of the greatest digit of n minus the least digit of n (A037904) <= 10^n.
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%I #12 Jan 28 2020 16:13:18

%S 1,286,4621,56980,640663,6904678,72722233,755339992,7774461355,

%T 79520082490,809705785165,8217213032524,83178920046367,

%U 840306174900622,8475694265094817,85380606857454976,859192675118710099,8638686211723117474,86794540082486097589,871513270875022245748

%N Accumulative sum of the greatest digit of n minus the least digit of n (A037904) <= 10^n.

%C Let b(n) = sum( A037904(k),{k=1..n}), then the lim b(n)/n -> 9. Reason, as the number of digits increases, then the likelihood of the maximum digit -> 9 and the minimum digits -> 0 becomes one.

%H Andrew Howroyd, <a href="/A072817/b072817.txt">Table of n, a(n) for n = 1..200</a>

%F a(n) = 1 + Sum_{w=1..n} Sum_{k=1..9} (10-k)*k*((k+1)^w - 2*k^w + (k-1)^w) - k*((k+1)^(w-1) - k^(w-1)). - _Andrew Howroyd_, Jan 28 2020

%t f[n_] := Block[{d = IntegerDigits[n]}, Max[d] - Min[d]]; s = 0; k = 0; Do[ While[k != 10^n, k++; s = s + f[k]]; Print[s], {n, 1, 8}]

%o (PARI) a(n)={1 + sum(w=1, n, sum(k=1, 9, (10-k)*k*((k+1)^w - 2*k^w + (k-1)^w) - k*((k+1)^(w-1) - k^(w-1))))} \\ _Andrew Howroyd_, Jan 28 2020

%Y Cf. A037904.

%K nonn,base

%O 1,2

%A _Benoit Cloitre_ and _Robert G. Wilson v_, Aug 09 2002

%E a(9)-a(12) from _Donovan Johnson_, Apr 09 2010

%E Terms a(13) and beyond from _Andrew Howroyd_, Jan 28 2020