A132264 - OEIS

A132264

Product{0<=k<=floor(log_12(n)), floor(n/12^k)}, n>=1.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232, 236, 300, 305, 310, 315

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OFFSET

1,2

COMMENTS

If n is written in base-12 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 d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

LINKS

Table of n, a(n) for n=1..63.

FORMULA

The following formulas are given for a general parameter p considering the product of terms floor(n/p^k) for 0<=k<=floor(log_p(n)), where p=12 for this sequence.

Recurrence: a(n)=n*a(floor(n/p)); a(n*p^m)=n^m*p^(m(m+1)/2)*a(n).

a(k*p^m)=k^(m+1)*p^(m(m+1)/2), for 0<k<p.

Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.

a(n)<=b(n), where b(n)=n^(1+floor(log_p(n)))/p^((1+floor(log_p(n)))*floor(log_p(n))/2); equality holds for n=k*p^m, 0<k<p, m>=0. b(n) can also be written n^(1+floor(log_p(n)))/p^A000217(floor(log_p(n))).

Also: a(n)<=q^((1-log_p(q))/2)*n^((1+log_p(n))/2)=q^((1-log_p(q))/2)*p^A000217(log_p(n)), equality holds for n=q*p^m, m>=0, where q=floor(sqrt(p)+1/2). Also, equality holds for n=(q+1)*p^m, provided p is a A002378-number (in this case we have p=q*(q+1) and so q^((1-log_p(q))/2)=(q+1)^((1-log_p(q+1))/2)).

a(n)>c*b(n), where c=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... (for p=12 see constant A132265).

Also: a(n)>c*(sqrt(2)/2^log_p(sqrt(2)))*n^((1+log_p(n))/2)=0.612870619...*p^A000217(log_p(n)), (p=12).

lim inf a(n)/b(n)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380198334286365820..., for n-->oo (for p=12 see constant A132265).

lim sup a(n)/b(n)=1, for n-->oo.

lim inf a(n)/n^((1+log_p(n))/2)=(sqrt(2)/2^log_p(sqrt(2)))*product{k>0, 1-1/(2*p^k)}=0.612870619..., for n-->oo, (p=12).

lim sup a(n)/n^((1+log_p(n))/2)=sqrt(q)/q^log_p(sqrt(q))=1.358593737..., for n-->oo, (p=12, q=round(sqrt(p))=3).

lim inf a(n)/a(n+1)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... for n-->oo (for p=12 see constant A132265).

EXAMPLE

a(50)=floor(50/12^0)*floor(50/12^1)=50*4=200.

a(65)=325 since 65=55(base-12) and so a(65)=55*5(base-12)=65*5=325.

CROSSREFS

Cf. A048651, A132019-A132026, A132265, A132267, A000217, A002378.

For the product of terms floor(n/p^k) for p=2 to p=11 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11).

For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Sequence in context: A335282 A370133 A351576 * A214922 A004830 A081330

Adjacent sequences: A132261 A132262 A132263 * A132265 A132266 A132267

KEYWORD

nonn,base

AUTHOR

Hieronymus Fischer, Aug 20 2007

STATUS

approved