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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 (list; graph; refs; listen; history; text; internal format)
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 c-divisor 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 c-divisor 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 c-divisors of 23. But 6=(1)(10) is not a c-divisor of 23.

One can prove a one-to-one correspondence between distinct subsequences for  composition no. n in standard order and c-divisors of n. So, the sequence lists also numbers of c-divisors 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^n-1) = n+1; for n>=2, a(2^n+1) = 4.

For c-equivalent 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 c-divisors of 11: 0,1,2,3,5,11. Thus a(11)=6. Note, that 3=(1)(1) is not a c-divisor of 13=(1)(10)(1) since, although it contains parts of 3=(1)(1), but in non-consecutive order. The c-divisors 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. Adams-Watters, Nov 06 2006

STATUS

approved

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Last modified June 20 17:27 EDT 2019. Contains 324234 sequences. (Running on oeis4.)