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.

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A381612 Irregular triangle read by rows: T(n,k) for n-1 <= k <= A328378(n) is the number of permutations of length n having a sum of differences equal to k.

Original entry on oeis.org

2, 2, 4, 2, 4, 12, 4, 2, 2, 4, 14, 32, 18, 28, 14, 8, 2, 4, 16, 36, 92, 68, 128, 92, 122, 72, 64, 16, 8, 2, 4, 18, 40, 112, 240, 256, 448, 438, 668, 502, 696, 480, 496, 264, 240, 88, 48, 2, 4, 20, 44, 134, 288, 696, 776, 1566, 1620, 2788, 2524, 3914, 3192, 4544, 3376, 4056, 2720, 2960, 1776, 1712, 816, 576, 144, 72
Offset: 2

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Author

Norman Whitehead, Mar 01 2025

Keywords

Comments

The first number in a row, T(n,n-1)=2, corresponding to the permutations having the numbers in order and reverse order.

Examples

			Triangle begins:
  n:  row n
---------------
  2:  2;
  3:  2, 4;
  4:  2, 4, 12,  4,   2;
  5:  2, 4, 14, 32,  18,  28,  14,   8;
  6:  2, 4, 16, 36,  92,  68, 128,  92, 122,  72,  64,  16,   8;
  7:  2, 4, 18, 40, 112, 240, 256, 448, 438, 668, 502, 696, 480, 496, 264, 240, 88, 48;
  ...
First item in each row corresponds to sum at n-1, second item to sum at n, etc.
Row 4:
    |1, 2, 3, 4|=3, |1, 3, 2, 4|=5, |3, 1, 2, 4|=5, |1, 4, 2, 3|=6,
    |4, 3, 2, 1|=3, |1, 3, 4, 2|=5, |3, 2, 1, 4|=5, |2, 3, 1, 4|=6,
    |1, 2, 4, 3|=4, |1, 4, 3, 2|=5, |3, 4, 1, 2|=5, |3, 2, 4, 1|=6,
    |2, 1, 3, 4|=4, |2, 1, 4, 3|=5, |4, 1, 2, 3|=5, |4, 1, 3, 2|=6,
    |3, 4, 2, 1|=4, |2, 3, 4, 1|=5, |4, 2, 1, 3|=5, |2, 4, 1, 3|=7,
    |4, 3, 1, 2|=4, |2, 4, 3, 1|=5, |4, 2, 3, 1|=5, |3, 1, 4, 2|=7,
Thus, for example, T(4,5)=12 because there are 12 permutations with a sum of differences equal to 5.
		

Crossrefs

Cf. A000982 (number of possible sums for n).
Cf. A328378 (maximal sums for n).

Programs

  • Python
    from itertools import permutations
    def count_l1_sums(n):
        nm1     = n-1
        a000982 = (((nm1*nm1)+1)//2) # from OEIS
        tally   = a000982*[0]    # number of different totals
        perms   = permutations([i for i in range(n)]) # length is factorial(n)
        # Compute each permutation's first difference
        for perm in perms:
            l1_norm = 0
            for i in range(nm1):
                diff = perm[i+1]-perm[i]
                if diff>0: l1_norm += diff
                else:      l1_norm -= diff
            tally[l1_norm-nm1] += 1 # diff==(n-1) goes into first position
        return tally
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