cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-14 of 14 results.

A231074 The number of possible ways to arrange the sums x_i + x_j (1 <= i < j <= n) of the items x_1 < x_2 <...< x_n in nondecreasing order.

Original entry on oeis.org

1, 1, 1, 1, 2, 12, 244
Offset: 0

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Author

Vladimir Letsko, Nov 03 2013

Keywords

Comments

For n<=5, a(n) = A003121(n), but for n > 5, a(n) < A003121(n).
Every possible ordering of sums provides a way of placing 1,2,...,n(n+1)/2 in a triangular array such that both rows and columns are increasing, but the reverse is generally not true.

Examples

			Let a < b < c < d. There are two possible ways to arrange the sums in nondecreasing order:
1) a+b <= a+c <= a+d <= b+c <= b+d <= c+d, (for instance, a = 1, b = 3, c = 4, d = 5);
2) a+b <= a+c <= b+c <= a+d <= b+d <= c+d, (for instance, a = 1, b = 2, c = 3, d = 5).
Hence a(4) = 2.
		

Crossrefs

Extensions

Term a(0)=1 prepended by Max Alekseyev, Feb 23 2014

A231085 The number of possible ways to arrange the sums x_i + x_j (1 <= i < j <= n) of the items x_1 < x_2 <...< x_n in increasing order provided that all sums are different.

Original entry on oeis.org

1, 1, 1, 1, 2, 12, 168, 4680
Offset: 0

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Author

Vladimir Letsko, Nov 03 2013

Keywords

Comments

For n<=5, a(n) = A231074(n), but for n>5, a(n) < A231074(n). For instance, let n = 6 and a < b < c < d < e < f. Then the arrangement a+b <= a+c <= a+d <= a+e <= b+c <= b+d <= a+f <= b+e <= b+f <= c+d <= c+e <= c+f <= d+e <= d+f <= e+f is possible (e.g., for a = 1, b = 5, c = 9, d = 12, e=13, f = 16), while the same arrangement with "<" instead of "<=" is not possible.

Examples

			Let a < b < c < d. There are two possible ways to arrange the sums in increasing order:
1) a+b < a+c < a+d < b+c < b+d < c+d, (for instance, a = 1, b = 3, c = 4, d = 5);
2) a+b < a+c < b+c < a+d < b+d < c+d, (for instance, a = 1, b = 2, c = 3, d = 5).
Hence a(4) = 2.
		

Crossrefs

Extensions

a(7) from Anton Nikonov, Feb 07 2014
Edited and a(0)=1 prepended by Max Alekseyev, Feb 19 2014
a(7) corrected by Max Alekseyev, Apr 02 2024

A131811 Number of symbolic sequences on n symbols that can be realized by the arrangement of the real roots of some polynomial of degree n and its derivatives.

Original entry on oeis.org

1, 1, 2, 10, 116
Offset: 1

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Author

Peter Bala, Jul 18 2007

Keywords

Comments

A symbolic sequence on n symbols 0, 1, ..., n - 1 is a sequence of length n*(n+1)/2 consisting of n occurrences of 0, (n - 1) occurrences of 1, ..., one occurrence of n-1 and satisfying the condition that between any two consecutive occurrences of the symbol i it has exactly one occurrence of the symbol i+1. For example, the two symbolic sequences on 3 symbols are 012010 and 010210.
The number of symbolic sequences on n symbols is listed in A003121. Anderson shows that out of the possible 12 symbolic sequences on 4 symbols only 10 can be realized as the arrangement of the 10 roots of some real polynomial of degree 4 and its derivatives.
Kostov shows that out of the possible 286 symbolic sequences on 5 symbols only 116 can be realized as the arrangement of the 15 roots of some real polynomial of degree 5 and its derivatives. V. I. Arnold has asked if the limit as n -> infinity of a(n)/A003121(n) is zero.

Examples

			a(3)=2. Let p be a real polynomial of degree 3 such that the 3 zeros x_1 < x_2 < x_3 of p are real and distinct. Let y_1 and y_2 denote the zeros of p' and let z_1 denote the zero of p''. Then by Rolle's theorem, x_1 < y_1 < x_2 < y_2 < x_3 and y_1 < z_1 < y_2. If z_1 does not coincide with x_2 (the generic case) then there are two possible arrangements for the 6 roots of p and its derivatives: either ( x_1 < y_1 < z_1 < x_2 < y_2 < x_3 ) or ( x_1 < y_1 < x_2 < z_1 < y_2 < x_3 ). These arrangements may be encoded in a symbolic sequence on 3 symbols namely either 012010 or 010210 (replace the x's by 0's, the y's by 1's and the z's by 2's). Both these arrangements are actually realized for some degree 3 polynomial. For example, the six roots of p = x*(x-1)*(x-3/2) and its derivatives are arranged symbolically as 012010 and the six roots of p = x*(x-1)*(x-3) and its derivatives are arranged symbolically as 010210.
		

Crossrefs

Cf. A003121.

A347608 Number of interlacing triangles of size n.

Original entry on oeis.org

1, 2, 20, 1744, 2002568, 42263042752, 21686691099024768, 344069541824691045987328, 226788686879114461294165127878656
Offset: 1

Views

Author

James B. Sidoli, Sep 08 2021

Keywords

Comments

An interlacing triangle of size n is a triangular array of the numbers 1, 2, ..., (n+1)*n/2 such that if T(i,j) denotes the j-th number in the i-th row then either T(i-1,j+1) < T(i,j) < T(i-1,j) or T(i-1,j) < T(i,j) < T(i-1,j+1) for 1 < i <= n and 1 <= j <= n-i+1.
Generalizes A003121 for the case when rows are not strictly increasing. See comment from Mar 25 2012 and comment from Dec 02 2014.

Examples

			For n = 2, a(2) = 2. The interlacing triangles are given below:
    2             2
  1   3   and   3   1.
		

Crossrefs

Cf. A003121.

Programs

  • Sage
    def interlacing(n):
        C_2=[]
        part=[j for j in range(n-1,-1,-1)]
        box=[]
        big_box=[]
        pos=0
        d=0
        C_2_star=[]
        for g in Words([0,1],n*(n-1)/2).list():
            C_2.append(list(g))
        for h in C_2:
            relations=[]
            pos=0
            big_box=[]
            for j in range(len(part)-1):
                for k in list(h)[pos:pos+part[j]]:
                    box.append(k)
                big_box.append(box)
                box=[]
                pos=pos+part[j]
            x=0
            for k in range(1,len(big_box)):
                for r in range(len(big_box[k])):
                    if big_box[k][r]==1 and big_box[k-1][r]==0 and big_box[k-1][r+1]==0 or big_box[k][r]==0 and big_box[k-1][r]==1 and big_box[k-1][r+1]==1:
                        continue
                    else:
                        x=x+1
            if x==(n-1)*(n-2)/2:
                q=q+1
                C_2_star.append(big_box)
        position=range(n*(n+1)/2)
        for tri in C_2_star:
            P=[]
            relations=[]
            counter=0
            collect=[]
            for j in range(len(tri)):
                for r in range(len(tri[j])):
                    if tri[j][r]==0:
                        relations.append([position[counter],position[counter+n-j]])
                        relations.append([position[counter+n-j],position[counter+1]])
                    if tri[j][r]==1:
                        relations.append([position[counter+n-j],position[counter]])
                        relations.append([position[counter+1],position[counter+n-j]])
                    counter=counter+1
                counter=counter+1
            P=Poset([range(n*(n+1)/2),relations])
            d=d+P.linear_extensions().cardinality()
        return d

Extensions

a(7)-a(9) from Dylan Nelson, May 09 2022
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