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%I #6 Mar 31 2012 13:21:03
%S 1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,4,4,4,4,
%T 4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,
%U 7,7,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10
%N Product{k>0, 1+floor(n/10^k)}.
%C If n is written in base-10 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product (1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).
%C a(n) = A179051(n) for n < 100. [From _Reinhard Zumkeller_, Jun 27 2010]
%F The following formulas are given for a general parameter p considering the product of terms 1+floor(n/p^k) for 0<k<=floor(log_p(n)), where p=10 for this sequence.
%F Recurrence: a(n)=(1+floor(n/p))*a(floor(n/p)); a(pn)=(1+n)*a(n); a(n*p^m)=product{0<=k<m, 1+n*p^k}*a(n).
%F a(k*p^m-j)=k^m*p^(m(m-1)/2), for 0<k<p, 0<j<p, m>=1. a(p^m)=p^(m(m-1)/2)*product{0<=k<m, 1+1/p^k}
%F a(n)=A132271(floor(n/p)=A132271(n)/(1+n).
%F Asymptotic behavior: a(n)=O(n^((log_p(n)-1)/p)); this follows from the inequalities below.
%F a(n)<=A067080(n)/(n+1)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.
%F a(n)>=A067080(n)/((n+1)*product{0<k<=floor(log_p(n)), 1-1/p^k}).
%F a(n)<c*n^((1+log_p(n))/2)/(n+1)=c*2^A000217(log_p(n))/(n+1), where c=product{k>0, 1+1/p^k}=2.2244691382741012... (for p=10 see constant A132325).
%F a(n)>n^((1+log_p(n))/2)/(n+1)=p^A000217(log_p(n))/(n+1).
%F lim sup n*a(n)/A067080(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
%F lim inf n*a(n)/A067080(n)=1/product{k>0, 1-1/p^k}=1/0.8900100999989990000001000..., for n-->oo (for p=10 s. constant A132038).
%F lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.
%F lim sup a(n)/n^((1+log_p(n))/2)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
%F lim inf a(n+1)/a(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012... for n-->oo (for p=10 see constant A132325).
%e a(121)=(1+floor(121/10^1))*(1+floor(121/10^2))=13*2=26; a(132)=28 since 132=132(base-10) and so
%e a(132)=(1+13)*(1+1)(base-10)=14*2=28.
%Y Cf. A132038, A132270(p=2), A132271, A132325, A132328(p=3), A132271.
%Y For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.
%K nonn,base
%O 0,11
%A _Hieronymus Fischer_, Aug 20 2007
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