querying multidimensional sequences in OEIS

[Jonathan Post]

I think the example givem without A number, is:

 A007188 <http://www.research.att.com/%7Enjas/sequences/A007188>
 Multiplicative encoding of Pascal triangle: Product p(i+1)^C(n,i).
(Formerly M1722)
 +0
2
2, 6, 90, 47250, 66852843750, 2806877704512541816406250,
1216935896582703898519354781702537118597533386230468750
There might be other Godel-coded sequences, and the like, besides Conway's
brilliant "Fractran", above and beyond those with "decimal Godelization" --
but usually these quickly lead to numbers too large for OEIS.

-- Jonathan Vos Post

On 11/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> <Me>
> >Off hand, I can't think of any interesting functions of finite
> sequences of
> >positive integers that depend on the order of the sequence.
>
> Of course, I no sooner wrote that than I started thinking of such
> functions.
> I'm not sure any of these are tremendously interesting, but they're at
> least
> somewhat so.
>
> In the following, b(i), i = 1..k is the finite sequence.  I'm starting
> all these
> sequences with the empty sequence, but I searched for them without
> that first term.  All are shown here in A066099 order.
>
> Alternating sum
> Sum_i (-1)^{i+1) b(i)
> 0,1,2,0,3,1,-1,1,4,2,0,2,-2,0,2,0
>
> Binomial sum
> Sum_i C(k-1,i-1) b(i)
> 0,1,2,2,3,3,3,4,4,4,4,5,4,6,5,8
>
> Inverse binomial sum
> Sum_i (-1)^{i+1) C(k-1,i-1) b(i)
> 0,1,2,0,3,1,-1,0,4,2,0,1,-2,-2,1,0
>
> Weighted sum
> sum_i i b(i)
> 0,1,2,3,3,4,5,6,4,5,6,7,7,8,9,10
> This is A029931
>
> Zero-based weighted sum
> sum_i (i-1) b(i)
> 0,0,0,1,0,1,2,3,0,1,2,3,3,4,5,6
>
> Sum of products of consecutive elements
> sum_{i=1}^{k-1} b(i) b(i+1)
> 0,0,0,1,0,2,2,2,0,3,4,3,3,4,3,3
>
> Number of rises
> sum_{b(i)>b(i-1)} 1
> 0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0
>
> Number of falls
> sum_{b(i)<b(i-1)} 1
> 0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0
>
> Number of unchanged
> sum_{b(i)=b(i-1)} 1
> 0,0,0,1,0,0,0,2,0,0,1,1,0,0,1,3
>
> Number of non-rises
> sum_{b(i)<=b(i-1)} 1
> 0,0,0,1,0,0,1,2,0,0,1,1,1,1,2,3
>
> Number of non-falls
> sum_{b(i)>=b(i-1)} 1
> 0,0,0,1,0,1,0,2,0,1,1,2,0,1,1,3
>
> Number of monotonically increasing runs
> 1 + number of falls, but 0 for empty sequence
> 0,1,1,1,1,2,1,1,1,2,1,2,1,2,1,1
>
> Number of monotonically decreasing runs
> 1 + number of rises, but 0 for empty sequence
> 0,1,1,1,1,1,2,1,1,1,1,1,2,2,2,1
>
> Number of runs of equal terms
> 1 + number of unchanged, but 0 for empty sequence
> 0,1,1,2,1,1,1,3,1,1,2,2,1,1,2,4
>
> Number of distinct non-empty subsequences
> 0,1,1,2,1,3,3,3,1,3,2,5,3,5,5,4
> E.g., f(1,1) = 2, for the sequences [1], and [1,1].
>
> Number of distinct subsequences
> 1 + number of distinct non-empty subsequences
> 1,2,2,3,2,4,4,4,2,4,3,6,4,6,6,5
>
> Number of set partitions
> List parts of set partition by their smallest element, and count the
> part sizes.
> E.g., 1|2,4|3 (also known as {1,2,3,2}) would count for the sequence
> [1,2,1].
> 1,1,1,1,1,2,1,1,1,3,3,3,1,2,1,1
>
> Number of permutations
> List permutations in cycle form, sorted by the smallest element in the
> cycle,
> and count the cycle lengths.
> Number of set partitions * Product_i (b(i)-1)!
> 1,1,1,1,2,2,1,1,6,6,3,3,2,2,1,1
>
> Number of partially ordered sets (unlabelled) by rank
> The rank of an element in a poset is the length of the longest chain of
> which it
> is the largest element.
> 1,1,1,1,1,2,1,1,1,3,4,3,1,2,1,1
>
> Number of partially ordered sets (labelled) by rank
> 1,1,1,2,1,9,3,6,1,28,54,60,4,36,12,24
>
> Number of partially ordered sets (naturally labelled) by rank
> Naturally labelled means labels are consistent with the ordering.
> 1,1,1,1,1,4,1,1,1,11,13,8,1,4,1,1
>
> Number of forests (unlabelled, unordered) with b(i) nodes at height i.
> 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,2,2,1,1,1,1,2,1,1,1,1,1
>
> ... and, of course, various other forest options.
>
> If there is agreement that this is the right order for these sequences,
> I'll try
> to find time to submit them.
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