

A124771


Number of distinct subsequences for compositions in standard order.


10



1, 2, 2, 3, 2, 4, 4, 4, 2, 4, 3, 6, 4, 6, 6, 5, 2, 4, 4, 6, 4, 6, 6, 8, 4, 6, 6, 9, 6, 9, 8, 6, 2, 4, 4, 6, 3, 7, 7, 8, 4, 7, 4, 9, 7, 8, 9, 10, 4, 6, 7, 9, 7, 9, 8, 12, 6, 9, 9, 12, 8, 12, 10, 7, 2, 4, 4, 6, 4, 7, 7, 8, 4, 6, 6, 10, 6, 10, 10, 10, 4, 7, 6, 10, 6, 8, 9, 12, 7, 10, 9, 12, 10, 12, 12, 12, 4
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OFFSET

0,2


COMMENTS

The standard order of compositions is given by A066099.
From Vladimir Shevelev, Dec 18 2013: (Start)
Every number in binary is a concatenation of parts of the form 10...0 with k>=0 zeros. For example, 5=(10)(1), 11=(10)(1)(1), 7=(1)(1)(1). We call d>0 a cdivisor of m, if d consists of some consecutive parts of m taking from the left to the right. Note that, to d=0 corresponds an empty set of parts. So it is natural to consider 0 as a cdivisor of every m. For example, 5=(10)(1) is a divisor of 23=(10)(1)(1)(1). Analogously, 1,2,3,7,11,23 are cdivisors of 23. But 6=(1)(10) is not a cdivisor of 23.
One can prove a onetoone correspondence between distinct subsequences for composition no. n in standard order and cdivisors of n. So, the sequence lists also numbers of cdivisors of nonnegative integers.
(End)


LINKS

Alois P. Heinz, Rows n = 0..14, flattened


FORMULA

a(n) = A124770(n) + 1.
From Vladimir Shevelev, Dec 18 2013: (Start)
a(2^n) = 2. Note that in concatenation representations of integers in binary, numbers {2^k}, k>=0, play the role of primes. So the formula is an analog of A000005(prime(n))=2.
a(2^n1) = n+1; for n>=2, a(2^n+1) = 4.
For cequivalent numbers n_1 and n_2 (i.e., differed only by order of parts) we have a(n_1) = a(n_2). For example, a(24)=a(17)=4. If the canonical representation of n is n=(1)^k_1[*](10)^k_2[*](100)^k_3[*]... , where [*] denotes operation of concatenation (cf. A233569), then a(n)<=(k_1+1)*(k_2+1)*...
(End)


EXAMPLE

Composition number 11 is 2,1,1; the subsequences are (empty); 1; 2; 1,1; 2,1; 2,1,1; so a(11) = 6.
The table starts:
1
2
1 2
1 3 3 3
Let n=11=(10)(1)(1). We have the following cdivisors of 11: 0,1,2,3,5,11. Thus a(11)=6. Note, that 3=(1)(1) is not a cdivisor of 13=(1)(10)(1) since, although it contains parts of 3=(1)(1), but in nonconsecutive order. The cdivisors of 13 are 0,1,2,5,6,13. So, a(13)=6.


CROSSREFS

Cf. A066099, A124770, A011782 (row lengths), A114994, A233249, A233312, A000005.
Sequence in context: A049822 A140060 A164341 * A066589 A007897 A180783
Adjacent sequences: A124768 A124769 A124770 * A124772 A124773 A124774


KEYWORD

easy,nonn,tabf,base


AUTHOR

Franklin T. AdamsWatters, Nov 06 2006


STATUS

approved



