

A269435


T(n,k)=Number of lengthn 0..k arrays with no repeated value greater than the previous repeated value.


9



2, 3, 4, 4, 9, 8, 5, 16, 27, 15, 6, 25, 64, 78, 28, 7, 36, 125, 250, 222, 51, 8, 49, 216, 615, 964, 622, 92, 9, 64, 343, 1281, 2995, 3674, 1722, 164, 10, 81, 512, 2380, 7536, 14455, 13868, 4719, 290, 11, 100, 729, 4068, 16408, 44021, 69235, 51917, 12821, 509, 12, 121
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OFFSET

1,1


COMMENTS

Table starts
...2.....3......4.......5........6.........7.........8..........9.........10
...4.....9.....16......25.......36........49........64.........81........100
...8....27.....64.....125......216.......343.......512........729.......1000
..15....78....250.....615.....1281......2380......4068.......6525.......9955
..28...222....964....2995.....7536.....16408.....32152......58149......98740
..51...622...3674...14455....44021....112476....252932.....516189.....976135
..92..1722..13868...69235...255576....767172...1981512....4566213....9621220
.164..4719..51917..329430..1475871...5209554..15465934...40265487...94574110
.290.12821.192980.1558430..8482276..35236110.120310016..354051015..927338710
.509.34575.712868.7334806.48543777.237479970.933059856.3105016479.9072298237
From Robert Israel, May 30 2019: (Start)
For each of the A000110 partitions pi of the set {1,...,n}, let A_pi(n,k) be the number of lengthn 0..k arrays v, such that v(i)=v(j) if and only if i and j are in the same part, and with no repeated value greater than the previous repeated value. There are restrictions on the values in the parts: if two parts a and b each have cardinality >= 2 and a_2 < b_2 (where the parts are indexed in increasing order), then v(b_i) < v(a_i). Thus if there are m partitions with cardinality >= 2, the values on those m parts are decreasing (listing these parts in order of their second entries). So for a partition with j parts of which m have cardinality >= 2, we have A_pi(n,k) = (k+1)*k*...*(k+2j)/m!, which is a polynomial in k of degree j. The partition of largest cardinality is the partition into singletons, which has m=0. The result is that for each n, T(n,k) is a monic polynomial of degree n. To verify the "empirical" formula for a row, only n terms in that row need to be computed. (End)


LINKS

R. H. Hardin, Table of n, a(n) for n = 1..9999


FORMULA

Empirical for column k:
k=1: a(n) = 2*a(n1) +a(n2) 2*a(n3) a(n4)
k=2: a(n) = 6*a(n1) 9*a(n2) 4*a(n3) +9*a(n4) +6*a(n5) +a(n6)
k=3: [order 8]
k=4: [order 10]
k=5: [order 12]
k=6: [order 14]
k=7: [order 16]
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = n^2 + 2*n + 1
n=3: a(n) = n^3 + 3*n^2 + 3*n + 1
n=4: a(n) = n^4 + 4*n^3 + (11/2)*n^2 + (7/2)*n + 1
n=5: a(n) = n^5 + 5*n^4 + (17/2)*n^3 + 8*n^2 + (9/2)*n + 1
n=6: a(n) = n^6 + 6*n^5 + 12*n^4 + (44/3)*n^3 + (23/2)*n^2 + (29/6)*n + 1
n=7: a(n) = n^7 + 7*n^6 + 16*n^5 + (71/3)*n^4 + (139/6)*n^3 + (43/3)*n^2 + (35/6)*n + 1


EXAMPLE

Some solutions for n=6 k=4
..1. .2. .3. .3. .2. .2. .2. .4. .3. .2. .3. .4. .3. .2. .1. .2
..4. .3. .3. .4. .0. .0. .4. .3. .1. .0. .2. .3. .4. .0. .3. .4
..3. .0. .1. .3. .3. .0. .3. .4. .3. .4. .3. .3. .3. .3. .4. .0
..3. .4. .3. .2. .1. .0. .3. .2. .1. .3. .4. .3. .2. .0. .0. .0
..3. .1. .4. .2. .2. .2. .2. .4. .3. .0. .3. .3. .3. .4. .4. .2
..2. .4. .3. .1. .3. .3. .0. .4. .2. .1. .2. .0. .0. .3. .4. .3


CROSSREFS

Column 1 is A029907(n+1).
Column 2 is A268013.
Column 3 is A267975.
Diagonal is A268104.
Row 1 is A000027(n+1).
Row 2 is A000290(n+1).
Row 3 is A000578(n+1).
Sequence in context: A269494 A269776 A269619 * A269656 A223949 A224133
Adjacent sequences: A269432 A269433 A269434 * A269436 A269437 A269438


KEYWORD

nonn,tabl


AUTHOR

R. H. Hardin, Feb 26 2016


STATUS

approved



