OFFSET
0,2
COMMENTS
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)d(0))*(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)).
FORMULA
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.
Recurrence: a(n)=(1+n)*a(floor(n/p)); a(pn)=(1+pn)*a(n); a(n*p^m)=product{1<=k<=m, 1+n*p^k}*a(n).
a(k*p^m-j)=(k*p^m-j+1)*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}, m>=1.
Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
a(n)<=A067080(n)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.
a(n)>=A067080(n)/product{1<=k<=floor(log_p(n)), 1-1/p^k}.
a(n)<c*n^((1+log_p(n))/2)=c*p^A000217(log_p(n)), where c=product{k>=0, 1+1/p^k}=2.2244691382741012... (for p=10 see constant A132325).
a(n)>n^((1+log_p(n))/2)=p^A000217(log_p(n)).
lim sup a(n)/A067080(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
lim inf a(n)/A067080(n)=1/product{k>0, 1-1/p^k}=1/0.8900100999989990000001000..., for n-->oo (for p=10 see constant A132038).
lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.
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).
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).
EXAMPLE
a(12)=(1+floor(12/10^0))*(1+floor(12/10^1))=13*2=26; a(21)=63 since 21=21(base-10) and so
a(21)=(1+21)*(1+2)(base-10)=22*3=66.
MATHEMATICA
f[n_] := Block[{k = 0, p = 1}, While[a = Floor[n/10^k]; a > 0, p *= 1 + a; k++]; p]; Array[f, 61, 0] (* Robert G. Wilson v, May 10 2011 *)
Table[Product[1+Floor[n/10^k], {k, 0, n}], {n, 0, 60}] (* Harvey P. Dale, May 14 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Aug 20 2007
STATUS
approved