cp's OEIS Frontend

A132272 a(n) = Product_{k>0} (1 + floor(n/10^k)).

%I A132272 #16 Jun 13 2025 10:50:49
%S A132272 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 A132272 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 A132272 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 A132272 a(n) = Product_{k>0} (1 + floor(n/10^k)).
%C A132272 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 A132272 a(n) = A179051(n) for n < 100. [From _Reinhard Zumkeller_, Jun 27 2010]
%H A132272 R. J. Mathar, <a href="/A132272/b132272.txt">Table of n, a(n) for n = 0..500</a>
%F A132272 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 A132272 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 A132272 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 A132272 a(n)=A132271(floor(n/p))=A132271(n)/(1+n).
%F A132272 Asymptotic behavior: a(n)=O(n^((log_p(n)-1)/p)); this follows from the inequalities below.
%F A132272 a(n)<=A067080(n)/(n+1)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.
%F A132272 a(n)>=A067080(n)/((n+1)*product{0<k<=floor(log_p(n)), 1-1/p^k}).
%F A132272 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 A132272 a(n)>n^((1+log_p(n))/2)/(n+1)=p^A000217(log_p(n))/(n+1).
%F A132272 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 A132272 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 A132272 lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.
%F A132272 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 A132272 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 A132272 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 A132272 a(132)=(1+13)*(1+1)(base-10)=14*2=28.
%p A132272 A132272 := proc(n)
%p A132272     a := 1;
%p A132272     for k from 1 do
%p A132272         f := floor(n/10^k) ;
%p A132272         if f <=0 then
%p A132272             return a;
%p A132272         else
%p A132272             a := a*(1+f) ;
%p A132272         end if;
%p A132272     end do:
%p A132272 end proc:
%p A132272 seq(A132272(n),n=1..120) ; # _R. J. Mathar_, Jun 13 2025
%t A132272 Table[Product[1+Floor[n/10^k],{k,n}],{n,0,100}] (* _Harvey P. Dale_, Jul 20 2024 *)
%Y A132272 Cf. A132038, A132270(p=2), A132271, A132325, A132328(p=3), A132271.
%Y A132272 For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.
%K A132272 nonn,base
%O A132272 0,11
%A A132272 _Hieronymus Fischer_, Aug 20 2007