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

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, 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, 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

Offset: 0

Author

Hieronymus Fischer, Aug 20 2007

Keywords

• nonn
• base

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))*(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)).

a(n) = A179051(n) for n < 100. [From Reinhard Zumkeller, Jun 27 2010]

Examples

			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
a(132)=(1+13)*(1+1)(base-10)=14*2=28.
		

Links

• R. J. Mathar, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A132038, A132270(p=2), A132271, A132325, A132328(p=3), A132271.

For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Programs

Maple

A132272 := proc(n)
    a := 1;
    for k from 1 do
        f := floor(n/10^k) ;
        if f <=0 then
            return a;
        else
            a := a*(1+f) ;
        end if;
    end do:
end proc:
seq(A132272(n),n=1..120) ; # R. J. Mathar, Jun 13 2025

Mathematica

Table[Product[1+Floor[n/10^k],{k,n}],{n,0,100}] (* Harvey P. Dale, Jul 20 2024 *)

Formula

The following formulas are given for a general parameter p considering the product of terms 1+floor(n/p^k) for 0

Recurrence: a(n)=(1+floor(n/p))*a(floor(n/p)); a(pn)=(1+n)*a(n); a(n*p^m)=product{0<=k

a(k*p^m-j)=k^m*p^(m(m-1)/2), for 0=1. a(p^m)=p^(m(m-1)/2)*product{0<=k

a(n)=A132271(floor(n/p))=A132271(n)/(1+n).

Asymptotic behavior: a(n)=O(n^((log_p(n)-1)/p)); this follows from the inequalities below.

a(n)<=A067080(n)/(n+1)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.

a(n)>=A067080(n)/((n+1)*product{0

a(n)A000217(log_p(n))/(n+1), 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)/(n+1)=p^A000217(log_p(n))/(n+1).

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).

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).

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).

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