A177520 -id:A177520 - cp's OEIS frontend

A242784 Number A(n,k) of permutations of [n] avoiding the consecutive step pattern given by the binary expansion of k, where 1=up and 0=down; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 2, 5, 8, 1, 1, 1, 1, 2, 6, 17, 16, 1, 1, 1, 1, 2, 6, 21, 70, 32, 1, 1, 1, 1, 2, 6, 19, 90, 349, 64, 1, 1, 1, 1, 2, 6, 21, 70, 450, 2017, 128, 1, 1, 1, 1, 2, 6, 23, 90, 331, 2619, 13358, 256, 1, 1

Offset: 0

Alois P. Heinz, May 22 2014

			A(4,5) = 19 because there are 4! = 24 permutations of {1,2,3,4} and only 5 of them do not avoid the consecutive step pattern up, down, up given by the binary expansion of 5 = 101_2: (1,3,2,4), (1,4,2,3), (2,3,1,4), (2,4,1,3), (3,4,1,2).
Square array A(n,k) begins:
  1, 1,   1,     1,     1,     1,     1,     1,     1, ...
  1, 1,   1,     1,     1,     1,     1,     1,     1, ...
  1, 1,   2,     2,     2,     2,     2,     2,     2, ...
  1, 1,   4,     5,     6,     6,     6,     6,     6, ...
  1, 1,   8,    17,    21,    19,    21,    23,    24, ...
  1, 1,  16,    70,    90,    70,    90,   111,   116, ...
  1, 1,  32,   349,   450,   331,   450,   642,   672, ...
  1, 1,  64,  2017,  2619,  1863,  2619,  4326,  4536, ...
  1, 1, 128, 13358, 17334, 11637, 17334, 33333, 34944, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns give: 0, 1: A000012, 2: A011782, 3: A049774, 4, 6: A177479, 5: A177477, 7: A117158, 8, 14: A177518, 9: A177519, 10: A177520, 11, 13: A177521, 12: A177522, 15: A177523, 16, 30: A177524, 17: A177525, 18, 22: A177526, 19, 25: A177527, 20, 26: A177528, 21: A177529, 23, 29: A177530, 24, 28: A177531, 27: A177532, 31: A177533, 32, 62: A177534, 33: A177535, 34, 46: A177536, 35, 49: A177537, 36, 54: A177538, 37, 41: A177539, 38: A177540, 39, 57: A177541, 40, 58: A177542, 42: A177543, 43, 53: A177544, 44, 50: A177545, 45: A177546, 47, 61: A177547, 48, 60: A177548, 51: A177549, 52: A177550, 55, 59: A177551, 56: A177552, 63: A177553, 127: A230051, 255: A230231, 511: A230232, 1023: A230233, 2047: A254523.
Main diagonal gives A242785.
Cf. A242783, A335308.

A:= proc(n, k) option remember; local b, m, r, h;
      if k<2 then return 1 fi;
      m:= iquo(k, 2, 'r'); h:= 2^ilog2(k);
      b:= proc(u, o, t) option remember; `if`(u+o=0, 1,
      `if`(t=m and r=0, 0, add(b(u-j, o+j-1, irem(2*t, h)), j=1..u))+
      `if`(t=m and r=1, 0, add(b(u+j-1, o-j, irem(2*t+1, h)), j=1..o)))
      end; forget(b);
      b(n, 0, 0)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..15);

Clear[A]; A[n_, k_] := A[n, k] = Module[{b, m, r, h}, If[k < 2, Return[1]]; {m, r} = QuotientRemainder[k, 2]; h = 2^Floor[Log[2, k]]; b[u_, o_, t_] := b[u, o, t] = If[u + o == 0, 1, If[t == m && r == 0, 0, Sum[b[u - j, o + j - 1, Mod[2*t, h]], {j, 1, u}]] + If[t == m && r == 1, 0, Sum[b[u + j - 1, o - j, Mod[2*t + 1, h]], {j, 1, o}]]]; b[n, 0, 0]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Sep 22 2014, translated from Maple *)

A230797 Number T(n,k) of permutations of [n] with exactly k (possibly overlapping) occurrences of the consecutive step pattern up, down, up, down; triangle T(n,k), n>=0, 0<=k<=max(0,floor((n-3)/2)), read by rows.

Original entry on oeis.org

1, 1, 2, 6, 24, 104, 16, 528, 192, 3296, 1472, 272, 23168, 12800, 4352, 179712, 132352, 42880, 7936, 1573632, 1366016, 530432, 158720, 15207424, 14781952, 7662336, 1911296, 353792, 158880768, 178102272, 101713920, 31813632, 8491008, 1801996288, 2282645504

Offset: 0

Alois P. Heinz, Oct 30 2013

			T(5,1) = 16: 13254, 14253, 14352, 15243, 15342, 23154, 24153, 24351, 25143, 25341, 34152, 34251, 35142, 35241, 45132, 45231.
T(7,2) = 272: 1325476, 1326475, 1326574, ..., 6735241, 6745132, 6745231.
Triangle T(n,k) begins:
:  0 :       1;
:  1 :       1;
:  2 :       2;
:  3 :       6;
:  4 :      24;
:  5 :     104,      16;
:  6 :     528,     192;
:  7 :    3296,    1472,    272;
:  8 :   23168,   12800,   4352;
:  9 :  179712,  132352,  42880,   7936;
: 10 : 1573632, 1366016, 530432, 158720;

Alois P. Heinz, Rows n = 0..170, flattened

Columns k=0-2 give: A177520, A230832, A264077.
T(2n-1,n-2) gives A000182(n) for n>=3.
Row sums give: A000142.
Cf. A242783, A242784, A295987.

b:= proc(u, o, t) option remember; `if`(u+o=0, 1, expand(
      add(b(u-j, o+j-1, [1, 3, 1, 3][t])*`if`(t=4, x, 1), j=1..u)+
      add(b(u+j-1, o-j, [2, 2, 4, 2][t]), j=1..o)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0, 1)):
seq(T(n), n=0..15);  # Alois P. Heinz, Oct 30 2013

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, Expand[Sum[b[u-j, o+j-1, {1, 3, 1, 3}[[t]]]*If[t == 4, x, 1], {j, 1, u}] + Sum[b[u+j-1, o-j, {2, 2, 4, 2}[[t]]], {j, 1, o}]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 0, 1]]; Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Oct 24 2016, after Alois P. Heinz *)

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A242784 Number A(n,k) of permutations of [n] avoiding the consecutive step pattern given by the binary expansion of k, where 1=up and 0=down; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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