|
%I #40 Apr 28 2023 08:16:07
%S 0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,
%T 3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,
%U 6,6,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,11,11,11
%N a(n) = Sum_{k>0} floor(n/10^k).
%C The old definition of this sequence was "Highest power of 10 dividing n!", but that is wrong (see A027868). For example, the highest power of 10 dividing 5!=120 is 1; however, a(5)=0. - _Hieronymus Fischer_, Jun 18 2007
%C Highest power of 10 dividing the quotient of multifactorials Product_{k>=1} M(10^k, 10^k*floor(n/10^k)) /( Product_{k>=1} M(10^(k-1), 10^(k-1) * floor(n/10^k)) ) where M(r,s) is the r-multifactorial (r>=1) of s which is defined by M(r,s) = s*M(r,s-r) for s > 0, M(r,s) = 1 for s <= 0. This is because that quotient of multifactorials evaluates to the product 10^floor(n/10)*10^floor(n/100)*... - _Hieronymus Fischer_, Jun 14 2007
%C Partial sums of A122840. - _Hieronymus Fischer_, Jun 06 2012
%H Hieronymus Fischer, <a href="/A054899/b054899.txt">Table of n, a(n) for n = 0..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Multifactorial.html">Multifactorial</a>
%F a(n) = floor(n/10) + floor(n/100) + floor(n/1000) + ...
%F a(n) = (n - A007953(n))/9.
%F From _Hieronymus Fischer_, Jun 14 2007, Jun 25 2007, and Aug 13 2007: (Start)
%F a(n) = Sum_{k>0} floor(n/10^k).
%F a(n) = Sum_{k=10..n} Sum_{j|k, j>=10} ( floor(log_10(j)) -floor(log_10(j-1)) ).
%F G.f.: g(x) = ( Sum_{k>0} x^(10^k)/(1-x^(10^k)) )/(1-x).
%F G.f. expressed in terms of Lambert series:
%F g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k)=1, if k>1 is a power of 10, else b(k)=0.
%F G.f.: g(x) = ( Sum_{k>0} c(k)*x^k )/(1-x), where c(k) = Sum_{j>1, j|k} (floor(log_10(j)) - floor(log_10(j-1)) ).
%F a(n) = Sum_{k=0..floor(log_10(n))} ds_10(floor(n/10^k))*10^k - n where ds_10(x) = digital sum of x in base 10.
%F a(n) = Sum_{k=0..floor(log_10(n))} A007953(floor(n/10^k))*10^k - n.
%F Recurrence:
%F a(n) = floor(n/10) + a(floor(n/10)).
%F a(10*n) = n + a(n).
%F a(n*10^m) = n*(10^m-1)/9 + a(n).
%F a(k*10^m) = k*(10^m-1)/9, for 0 <= k < 10, m >= 0.
%F Asymptotic behavior:
%F a(n) = n/9 + O(log(n)),
%F a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
%F a(n) <= (n - 1)/9; equality holds for powers of 10.
%F a(n) >= n/9 - 1 - floor(log_10(n)); equality holds for n=10^m-1, m>0.
%F lim inf (n/9 - a(n)) = 1/9, for n --> oo.
%F lim sup (n/9 - log_10(n) - a(n)) = 0, for n --> oo.
%F lim sup (a(n+1) - a(n) - log_10(n)) = 0, for n --> oo. (End)
%e a(11) = 1
%e a(111) = 12.
%e a(1111) = 123.
%e a(11111) = 1234.
%e a(111111) = 12345.
%e a(1111111) = 123456.
%e a(11111111) = 1234567.
%e a(111111111) = 12345678.
%e a(1111111111) = 123456789.
%t Table[t=0; p=10; While[s=Floor[n/p]; t=t+s; s>0, p*=10]; t, {n,0,100}]
%o (PARI) a(n)=my(s);while(n\=10,s+=n);s \\ _Charles R Greathouse IV_, Jul 19 2011
%o (Magma)
%o m:=10;
%o function a(n) // a = A054899, m = 10
%o if n eq 0 then return 0;
%o else return a(Floor(n/m)) + Floor(n/m);
%o end if; end function;
%o [a(n): n in [0..103]]; // _G. C. Greubel_, Apr 28 2023
%o (SageMath)
%o m=10 # a = A054899
%o def a(n): return 0 if (n==0) else a(n//m) + (n//m)
%o [a(n) for n in range(104)] # _G. C. Greubel_, Apr 28 2023
%Y Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
%Y Different from the highest power of 10 dividing n! (see A027868 for reference).
%Y Cf. A007953, A027868, A067080, A098844, A132027, A122840.
%K nonn
%O 0,21
%A _Henry Bottomley_, May 23 2000
%E An incorrect g.f. was deleted by _N. J. A. Sloane_, Sep 13 2009
%E Examples added by _Hieronymus Fischer_, Jun 06 2012
|
