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.

Showing 1-2 of 2 results.

A120909 Triangle read by rows: T(n,k) is the number of ternary words of length n having k runs (i.e., subwords of maximal length) of identical letters (1 <= k <= n).

Original entry on oeis.org

3, 3, 6, 3, 12, 12, 3, 18, 36, 24, 3, 24, 72, 96, 48, 3, 30, 120, 240, 240, 96, 3, 36, 180, 480, 720, 576, 192, 3, 42, 252, 840, 1680, 2016, 1344, 384, 3, 48, 336, 1344, 3360, 5376, 5376, 3072, 768, 3, 54, 432, 2016, 6048, 12096, 16128, 13824, 6912, 1536, 3, 60
Offset: 1

Views

Author

Emeric Deutsch, Jul 15 2006

Keywords

Comments

Row sums are the powers of 3 (A000244).

Examples

			T(3,2)=12 because we have 001,002,011,022,100,110,112,122,200,211,220 and 221.
Triangle starts:
  3;
  3,  6;
  3, 12, 12;
  3, 18, 36, 24;
  3, 24, 72, 96, 48;
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->3*2^(k-1)*binomial(n-1,k-1): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
  • Mathematica
    nn=15;f[list_]:=Select[list,#>0&];a=y x/(1-x) +1;b=a^2/(1-(a-1)^2 );Drop[Map[f,CoefficientList[Series[b a/(1-(a-1)(b-1)),{x,0,nn}],{x,y}]],1]//Grid  (* Geoffrey Critzer, Nov 20 2012 *)

Formula

T(n,k) = 3*2^(k-1)*binomial(n-1,k-1).
G(t,z) = 3*t*z/(1-z-2*t*z).
T(n,k) = 3*A013609(n-1,k-1).
T(n,k) = A120910(n,n-k).
Sum_{k>=1} k*T(n,k) = 3*A081038(n-1).

A381349 Triangle read by rows: T(n,k) is the number of distinct tuples E each corresponding to some k-ary word W = (w_1, ..., w_n), where E is a tuple (e_1, ..., e_{n-1}) with e_i being the number of pairs of equal letters (w_j,w_k) in W such that j + i = k.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 6, 9, 10, 1, 10, 22, 26, 27, 1, 20, 54, 73, 78, 79, 1, 36, 163, 249, 269, 275, 276, 1, 72, 447, 791, 915, 942, 949, 950, 1, 135, 1350, 3136, 3776, 3899, 3934, 3942, 3943, 1, 272, 4088, 11315, 14849, 15650, 15811, 15855, 15864, 15865
Offset: 1

Views

Author

John Tyler Rascoe, Feb 21 2025

Keywords

Comments

Two different words of the same length can have the same corresponding E tuple.

Examples

			Triangle begins:
    k=1   2    3    4    5    6    7
  n=1 1;
  n=2 1,  2;
  n=3 1,  3,   4;
  n=4 1,  6,   9,  10;
  n=5 1, 10,  22,  26,  27;
  n=6 1, 20,  54,  73,  78,  79;
  n=7 1, 36, 163, 249, 269, 275, 276;
  ...
Considering the pairs of equal letters in W = (1,3,3,1) there is 1 pair with no letters between them, 0 pairs with a single letter between them, and 1 pair seperated by two letters, giving E = (1,0,1).
T(4,2) = 6 counts the following values of E each listed with a corresponding word:
     E        W
  (3,2,1) (1,1,1,1)
  (2,1,0) (1,1,1,2)
  (2,0,0) (1,1,2,2)
  (1,0,1) (1,2,2,1)
  (1,1,1) (2,1,2,2)
  (0,2,0) (2,1,2,1)
		

Crossrefs

Cf. A000312, (empirical column k=2) A006606, A120910, A226873.

Programs

  • Python
    from itertools import combinations_with_replacement, permutations
    def pairs(m): return tuple([sum(1 for j in range(len(m)-i) if m[j] == m[j+i]) for i in range(1,len(m))])
    def A381349(n,k):
        S = set()
        for y in combinations_with_replacement(range(1,k+1),n-1):
            S.update(z for z in permutations((1,)+y))
        return len({pairs(i) for i in S})
    
  • Python
    # see links for a different algorithm
    from sympy.utilities.iterables import multiset_permutations
    from itertools import combinations, combinations_with_replacement
    def E(w):
        return tuple(sum(1 for j, k in combinations(range(len(w)), 2) if w[j] == w[k] and j+i == k) for i in range(len(w)))
    def row(n): # generator of row n
        v, S = 0, set()
        for k in range(1, n+1):
            for c in combinations_with_replacement(range(1, k+1), n-k):
                if len(c) > 0 and c[0] == 2: break
                S.update(E(e) for e in multiset_permutations(tuple(range(1, k+1))+c))
            yield len(S)
    print([e for n in range(1, 9) for e in row(n)]) # Michael S. Branicky, Feb 26 2025

Formula

T(n,k) = T(n,n) for k > n.
From Michael S. Branicky, Feb 25 2025: (Start)
T(n, 2) = A006606(n).
T(n, n-1) = T(n, n-2) + n-1, arising from E's of permutations of 1, ..., n-2, n-1, n-1.
T(n, n) = T(n, n-1) + 1, arising from E(1, ... , n) = (0, ..., 0). (End)

Extensions

a(37) and beyond from Michael S. Branicky, Feb 25 2025
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