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A124771
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Number of distinct subsequences for compositions in standard order.
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39
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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
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0,2
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COMMENTS
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The standard order of compositions is given by A066099.
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)
These are contiguous subsequences, or restrictions to a subinterval. The case for all subsequences is A334299. - Gus Wiseman, Jun 02 2020
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LINKS
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FORMULA
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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)
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EXAMPLE
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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.
The c-divisors of n are given in column n below:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 2 1 4 1 1 1 8 1 2 1 1 1 1 1 16 1 2
3 2 2 3 4 10 2 4 2 2 3 8 4
5 6 7 9 3 12 5 3 7 17 18
5 6 6 15
11 13 14
(End)
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MATHEMATICA
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stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
Table[Length[Union[ReplaceList[stc[n], {___, s___, ___}:>{s}]]], {n, 0, 100}] (* Gus Wiseman, Jun 01 2020 *)
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CROSSREFS
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Not allowing empty subsequences gives A124770.
The case for not just contiguous subsequences is A334299.
Positions of first appearances are A335279.
Compositions where every subinterval has a different sum are A333222.
Cf. A000120, A003022, A029931, A070939, A108917, A124767, A325680, A325770, A333224, A334967, A334968.
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KEYWORD
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easy,nonn,tabf,base
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AUTHOR
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STATUS
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approved
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