0,2

If n is written in base 2 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)).

From Gary W. Adamson, Aug 25 2016: (Start)

Given the following production matrix M =

1, 0, 0, 0, 0, ...

2, 0, 0, 0, 0, ...

0, 3, 0, 0, 0, ...

0, 4, 0, 0, 0, ...

0, 0, 5, 0, 0, ...

0, 0, 6, 0, 0, ...

0, 0, 0, 7, 0, ...

...

the sequence is the left-shifted vector as lim_{n->infinity} M^n. (End)

Robert Israel, Table of n, a(n) for n = 0..10000

Recurrence: a(n)=(1+n)*a(floor(n/2)); a(2n)=(1+2n)*a(n); a(n*2^m) = (Product_{k=1..m} (1 + n*2^k))*a(n).

a(2^m-1) = 2^(m*(m+1)/2), a(2^m) = 2^(m*(m+1)/2)*Product_{k=0..m} (1 + 1/2^k), m>=1.

a(n) = A132270(2n) = (1+n)*A132270(n).

Asymptotic behavior: a(n) = O(n^((1+log_2(n))/2)); this follows from the inequalities below.

a(n) <= A098844(n)*Product_{k=0..floor(log_2(n))} (1 + 1/2^k).

a(n) >= A098844(n)/Product_{k=1..floor(log_2(n))} (1 - 1/2^k).

a(n) < c*n^((1+log_2(n))/2) = c*2^A000217(log_2(n)), where c = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627... (see constant A081845).

a(n) > n^((1+log_2(n))/2) = 2^A000217(log_2(n)),

lim sup a(n)/A098844(n) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627..., for n->oo (see constant A081845).

lim inf a(n)/A098844(n) = 1/Product_{k>=1} (1 - 1/2^k) = 1/0.288788095086602421..., for n->oo (see constant A048651).

lim inf a(n)/n^((1+log_2(n))/2) = 1, for n->oo.

lim sup a(n)/n^((1+log_2(n))/2) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627..., for n->oo (see constant A081845).

lim inf a(n+1)/a(n) = Product_{k>=0} (1 + 1/2^k) = 4.7684620580627... for n->oo (see constant A081845).

G.f. g(x) satisfies g(x) = (1+2x)*g(x^2) + 2*x^2*(1+x)*g'(x^2). - Robert Israel, Aug 26 2016

a(10) = (1 + floor(10/2^0))*(1 + floor(10/2^1))*(1 + floor(10/2^2))*(1 + floor(10/2^3)) = 11*6*3*2 = 396;

a(17) = 4860 since 17 = 10001_2 and so a(17) = (1+10001_2)*(1+1000_2)*(1+100_2)*(1+10_2)*(1+1) = 18*9*5*3*2 = 4860.

f:= proc(n) option remember; (1+n)*procname(floor(n/2)) end proc:

f(0):= 1:

map(f, [$0..100]); # Robert Israel, Aug 26 2016

Table[Product[1 + Floor[2 n/2^k], {k, 2 n}], {n, 0, 42}] (* or *)

Table[Function[w, Times @@ Map[1 + FromDigits[PadRight[w, #], 2] &, Range@ Length@ w]]@ IntegerDigits[n, 2], {n, 0, 42}] (* Michael De Vlieger, Aug 26 2016 *)

(Magma) [1] cat [n le 1 select 2 else (1+n)*Self(Floor(n/2)): n in [1..50]]; // Vincenzo Librandi, Aug 26 2016

Cf. A048651, A081845, A132270, A132271(p=10), A132272, A132327(p=3), A132328.

For formulas regarding a general parameter p (i.e., terms 1+floor(n/p^k)) see A132271.

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

Sequence in context: A290249 A321471 A216762 * A053287 A086323 A336618

Adjacent sequences: A132266 A132267 A132268 * A132270 A132271 A132272

Hieronymus Fischer, Aug 20 2007

approved