A011973 -id:A011973 - cp's OEIS frontend

A180178 Triangle read by rows: T(n,k) is the number of compositions of n without 3's and having k parts; 1 <= k <= n.

1, 1, 1, 0, 2, 1, 1, 1, 3, 1, 1, 2, 3, 4, 1, 1, 4, 4, 6, 5, 1, 1, 4, 9, 8, 10, 6, 1, 1, 5, 12, 17, 15, 15, 7, 1, 1, 6, 15, 28, 30, 26, 21, 8, 1, 1, 7, 21, 38, 56, 51, 42, 28, 9, 1, 1, 8, 27, 56, 85, 102, 84, 64, 36, 10, 1, 1, 9, 34, 80, 130, 172, 175, 134, 93, 45, 11, 1, 1, 10, 42, 108, 200

Offset: 1

Emeric Deutsch, Aug 15 2010

			T(7,4)=8 because we have (1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1), (1,1,1,4), (1,1,4,1), (1,4,1,1), and (4,1,1,1).
Triangle starts:
  1;
  1, 1;
  0, 2, 1;
  1, 1, 3, 1;
  1, 2, 3, 4, 1;
  1, 4, 4, 6, 5, 1;

P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 33-51 (see Table 5).
R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 33-43.

P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.

Cf. A011973, A180177, A180179, A180180, A180181, A180182, A180183.

p := 3: T := proc (n, k) options operator, arrow: sum((-1)^(k-j)*binomial(k, j)*binomial(n-p*k+p*j-1, j-1), j = (p*k-n)/(p-1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form
p := 3: g := z/(1-z)-z^p: G := t*g/(1-t*g): Gser := simplify(series(G, z = 0, 15)): for n to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 13 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form
with(combinat): m := 3: T := proc (n, k) local ct, i: ct := 0: for i to numbcomp(n, k) do if member(m, composition(n, k)[i]) = false then ct := ct+1 else end if end do: ct end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

p = 3; max = 14; g = z/(1-z) - z^p; G = t*g/(1-t*g); Gser = Series[G, {z, 0, max+1}]; t[n_, k_] := SeriesCoefficient[Gser, {z, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2014, after Maple *)

Number of compositions of n without p's and having k parts = Sum_{j=(pk-n)/(p-1)..k} (-1)^(k-j)*binomial(k,j)*binomial(n-pk+pj-1, j-1).

For a given p, the g.f. of the number of compositions without p's is G(t,z) = t*g(z)/(1-t*g(z)), where g(z) = z/(1-z) - z^p; here z marks sum of parts and t marks number of parts.

A180179 Triangle read by rows: T(n,k) is the number of compositions of n without 4's and having k parts; 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 0, 3, 3, 1, 1, 2, 6, 4, 1, 1, 3, 7, 10, 5, 1, 1, 4, 9, 16, 15, 6, 1, 1, 6, 12, 23, 30, 21, 7, 1, 1, 6, 19, 32, 50, 50, 28, 8, 1, 1, 7, 24, 50, 76, 96, 77, 36, 9, 1, 1, 8, 30, 72, 120, 162, 168, 112, 45, 10, 1, 1, 9, 36, 99, 185, 267, 315, 274, 156, 55, 11, 1, 1, 10, 45

Offset: 1

Emeric Deutsch, Aug 15 2010

			T(7,3)=9 because we have (5,1,1), (1,5,1), (1,1,5), (3,3,1), (3,1,3), (1,3,3), (3,2,2), (2,3,2), and (2,2,3).
Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  0,  3,  3,  1;
  1,  2,  6,  4,  1;
  1,  3,  7, 10,  5,  1;

P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 33-51 (see Table 6).
R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 33-43.

P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.

Cf. A011973, A180177, A180178, A180180, A180181, A180182, A180183.

p := 4: T := proc (n, k) options operator, arrow: sum((-1)^(k-j)*binomial(k, j)*binomial(n-p*k+p*j-1, j-1), j = (p*k-n)/(p-1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form
p := 4: g := z/(1-z)-z^p: G := t*g/(1-t*g): Gser := simplify(series(G, z = 0, 15)): for n to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 13 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form
with(combinat): m := 4: T := proc (n, k) local ct, i: ct := 0: for i to numbcomp(n, k) do if member(m, composition(n, k)[i]) = false then ct := ct+1 else end if end do: ct end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

p = 4; max = 14; g = z/(1-z) - z^p; G = t*g/(1-t*g); Gser = Series[G, {z, 0, max+1}]; t[n_, k_] := SeriesCoefficient[Gser, {z, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2014, after Maple *)

Number of compositions of n without p's and having k parts = Sum_{j=(pk-n)/(p-1)..k} (-1)^(k-j)*binomial(k,j)*binomial(n-pk+pj-1, j-1).

For a given p, the g.f. of the number of compositions without p's is G(t,z) = t*g(z)/(1-t*g(z)), where g(z) = z/(1-z) - z^p; here z marks sum of parts and t marks number of parts.

A180180 Triangle read by rows: T(n,k) is the number of compositions of n without 5's and having k parts; 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 0, 4, 6, 4, 1, 1, 3, 10, 10, 5, 1, 1, 4, 12, 20, 15, 6, 1, 1, 5, 15, 31, 35, 21, 7, 1, 1, 6, 19, 44, 65, 56, 28, 8, 1, 1, 8, 24, 60, 106, 120, 84, 36, 9, 1, 1, 8, 33, 80, 160, 222, 203, 120, 45, 10, 1, 1, 9, 40, 111, 230, 372, 420, 322, 165, 55, 11, 1

Offset: 1

Emeric Deutsch, Aug 15 2010

			T(8,2)=5 because we have (1,7),(7,1),(2,6),(6,2),and (4,4).
Triangle starts:
 1;
 1,  1;
 1,  2,  1;
 1,  3,  3,  1;
 0,  4,  6,  4,  1;
 1,  3, 10, 10,  5,  1;

P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 33-51 (see Table 7).
R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 33-43.

P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.

Cf. A011973, A180177, A180178, A180179, A180181, A180182, A180183.

p:= 5: T := proc (n, k) options operator, arrow: sum((-1)^(k-j)*binomial(k, j)*binomial(n-p*k+p*j-1, j-1), j = (p*k-n)/(p-1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; yields sequence in triangular form
p:=5: g:=z/(1-z)-z^(p): G:=t*g/(1-t*g): Gser:=simplify(series(G,z=0,15)): for n from 1 to 13 do P[n]:=sort(coeff(Gser,z,n)); # yields sequence in triangular form
with(combinat): m := 5: T := proc (n, k) local ct, i: ct := 0: for i to numbcomp(n, k) do if member(m, composition(n, k)[i]) = false then ct := ct+1 else end if end do: ct end proc: for n to 12 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

p = 5; max = 14; g = z/(1-z) - z^p; G = t*g/(1-t*g); Gser = Series[G, {z, 0, max+1}]; t[n_, k_] := SeriesCoefficient[Gser, {z, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2014, after Maple *)

Number of compositions of n without p's and having k parts = Sum_{j=(pk-n)/(p-1)..k} (-1)^(k-j)*binomial(k,j)*binomial(n-pk+pj-1, j-1).

For a given p, the g.f. of the number of compositions without p's is G(t,z) = t*g(z)/(1-t*g(z)), where g(z) = z/(1-z) - z^p; here z marks sum of parts and t marks number of parts.

A180181 Triangle read by rows: T(n,k) is the number of compositions of n without 6's and having k parts; 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 0, 5, 10, 10, 5, 1, 1, 4, 15, 20, 15, 6, 1, 1, 5, 18, 35, 35, 21, 7, 1, 1, 6, 22, 52, 70, 56, 28, 8, 1, 1, 7, 27, 72, 121, 126, 84, 36, 9, 1, 1, 8, 33, 96, 190, 246, 210, 120, 45, 10, 1, 1, 10, 40, 125, 280, 432, 455, 330, 165, 55, 11, 1

Offset: 1

Emeric Deutsch, Aug 15 2010

			T(8,2)=5 because we have (1,7), (7,1), (3,5), (5,3), and (4,4).
Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  6,  4,  1;
  0,  5, 10, 10,  5,  1;
  1,  4, 15, 20, 15,  6,  1;

P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 33-51 (see Table 8).
R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 33-43.

P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.

Cf. A011973, A180177, A180178, A180179, A180180, A180182, A180183.

p := 6: T := proc (n, k) options operator, arrow: sum((-1)^(k-j)*binomial(k, j)*binomial(n-p*k+p*j-1, j-1), j = (p*k-n)/(p-1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form
p := 6: g := z/(1-z)-z^p: G := t*g/(1-t*g): Gser := simplify(series(G, z = 0, 15)): for n to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 13 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form
with(combinat): m := 6: T := proc (n, k) local ct, i: ct := 0: for i to numbcomp(n, k) do if member(m, composition(n, k)[i]) = false then ct := ct+1 else end if end do: ct end proc: for n to 12 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

p = 6; max = 14; g = z/(1-z) - z^p; G = t*g/(1-t*g); Gser = Series[G, {z, 0, max+1}]; t[n_, k_] := SeriesCoefficient[Gser, {z, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2014, after Maple *)

Number of compositions of n without p's and having k parts = Sum_{j=(pk-n)/(p-1)..k} (-1)^(k-j)*binomial(k,j)*binomial(n-pk+pj-1, j-1).

For a given p, the g.f. of the number of compositions without p's is G(t,z) = t*g(z)/(1-t*g(z)), where g(z) = z/(1-z) - z^p; here z marks sum of parts and t marks number of parts.

A180183 Triangle read by rows: T(n,k) is the number of compositions of n without 8's and having k parts; 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 0, 7, 21, 35, 35, 21, 7, 1, 1, 6, 28, 56, 70, 56, 28, 8, 1, 1, 7, 33, 84, 126, 126, 84, 36, 9, 1, 1, 8, 39, 116, 210, 252, 210, 120, 45, 10, 1, 1, 9, 46, 153, 325, 462, 462, 330, 165, 55, 11, 1

Offset: 1

Emeric Deutsch, Aug 15 2010

			T(10,2)=7 because we have (1,9),(9,1),(3,7),(7,3),(4,6),(6,4), and (5,5).
Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  6,  4,  1;
  1,  5, 10, 10,  5,  1;
  1,  6, 15, 20, 15,  6,  1;
  0,  7, 21, 35, 35, 21,  7,  1;
  1,  6, 28, 56, 70, 56, 28,  8,  1;

P. Chinn and S. Heubach, Compositions of n with no occurrence of k, Congressus Numerantium, 164 (2003), pp. 33-51.
R.P. Grimaldi, Compositions without the summand 1, Congressus Numerantium, 152, 2001, 33-43.

P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.

Cf. A011973, A180177, A180178, A180179, A180180, A180181, A180182.

p:= 8: T := proc (n, k) options operator, arrow: sum((-1)^(k-j)*binomial(k, j)*binomial(n-p*k+p*j-1, j-1), j = (p*k-n)/(p-1) .. k) end proc: for n to 13 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form
p := 8: g := z/(1-z)-z^p: G := t*g/(1-t*g): Gser := simplify(series(G, z = 0, 15)): for n to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 13 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form
with(combinat): m := 8: T := proc (n, k) local ct, i: ct := 0: for i to numbcomp(n, k) do if member(m, composition(n, k)[i]) = false then ct := ct+1 else end if end do: ct end proc: for n to 12 do seq(T(n, k), k = 1 .. n) end do; # yields sequence in triangular form

p = 8; max = 14; g = z/(1-z) - z^p; G = t*g/(1-t*g); Gser = Series[G, {z, 0, max+1}]; t[n_, k_] := SeriesCoefficient[Gser, {z, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2014, after Maple *)

Number of compositions of n without p's and having k parts = Sum_{j=(pk-n)/(p-1)..k} (-1)^(k-j)*binomial(k,j)*binomial(n-pk+pj-1, j-1).

For a given p, the g.f. of the number of compositions without p's is G(t,z) = t*g(z)/(1-t*g(z)), where g(z) = z/(1-z) - z^p; here z marks sum of parts and t marks number of parts.

A371967 Irregular triangle T(r,w) read by rows: number of ways of placing w non-attacking wazirs on a 3 X r board.

Original entry on oeis.org

1, 1, 3, 1, 1, 6, 8, 2, 1, 9, 24, 22, 6, 1, 1, 12, 49, 84, 61, 18, 2, 1, 15, 83, 215, 276, 174, 53, 9, 1, 1, 18, 126, 442, 840, 880, 504, 158, 28, 2, 1, 21, 178, 792, 2023, 3063, 2763, 1478, 472, 93, 12, 1, 1, 24, 239, 1292, 4176, 8406, 10692, 8604, 4374, 1416, 297, 38, 2, 1, 27, 309

Offset: 0

R. J. Mathar, Apr 14 2024

			The triangle starts with r>=0 rows and w>=0 wazirs as
  1 ;
  1 3 1 ;
  1 6 8 2  ;
  1 9 24 22 6 1 ;
  1 12 49 84 61 18 2  ;
  1 15 83 215 276 174 53 9 1 ;
  1 18 126 442 840 880 504 158 28 2  ;
  1 21 178 792 2023 3063 2763 1478 472 93 12 1 ;
  1 24 239 1292 4176 8406 10692 8604 4374 1416 297 38 2  ;
  1 27 309 1969 7731 19591 32716 36257 26674 13035 4264 945 142 15 1 ;
  ...

Alois P. Heinz, Rows r = 0..165 (first 17 rows from R. J. Mathar)
R. J. Mathar, Bivariate Generating Functions for Non-attacking Wazirs on Rectangular Boards
Jacob A. Siehler, Selections without adjacency on a rectangular grid, arXiv:1409.3869 [math.CO] (2014) Table 2.
Wikipedia, Wazir (chess)

Cf. A051736 (row sums), A035607 (on 2Xr board), A011973 (on 1Xr board), A232833 (on rXr board).
T(n,n) gives A371978.
Row maxima give A371979.
Cf. A007494.

b:= proc(n, l) option remember; `if`(n=0, 1,
      add(`if`(Bits[And](j, l)>0, 0, expand(b(n-1, j)*
      x^add(i, i=Bits[Split](j)))), j=[0, 1, 2, 4, 5]))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..10);  # Alois P. Heinz, Apr 14 2024

b[n_, l_] := b[n, l] = If[n == 0, 1, Sum[If[BitAnd[j, l] > 0, 0, Expand[b[n - 1, j]*x^DigitCount[j, 2, 1]]], {j, {0, 1, 2, 4, 5}}]];
T[n_] := CoefficientList[b[n, 0], x];
Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jun 05 2024, after Alois P. Heinz *)

T(r,0) = 1.
T(r,1) = 3*r.
T(r,2) = A064225(r-1).
T(r,3) = A172229(r).
T(r,4) = 27*r^4/8 -117*r^3/4 +829*r^2/8 -715*r/4 +126. [Siehler Table 3]
T(3,w) = A232833(3,w).
G.f.: (1+x*y) *(1 +x*y +x*y^2 -x^2*y^3)/(1 -x -x*y -x^2*y^3 -2*x^2*y -3*x^2*y^2 -x^3*y^2 +x^3*y^4 +x^4*y^4). - R. J. Mathar, Apr 21 2024

A006921 Diagonals of Pascal's triangle mod 2 interpreted as binary numbers.

Original entry on oeis.org

1, 1, 3, 2, 7, 5, 13, 8, 29, 21, 55, 34, 115, 81, 209, 128, 465, 337, 883, 546, 1847, 1301, 3357, 2056, 7437, 5381, 14087, 8706, 29443, 20737, 53505, 32768, 119041, 86273, 226051, 139778, 472839, 333061, 859405, 526344, 1903901, 1377557, 3606327

Offset: 0

N. J. A. Sloane

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..1000
B. R. Hodgson, Letter to N. J. A. Sloane, Oct. 1991
B. R. Hodgson, On some number sequences related to the parity of binomial coefficients, Fib. Quart., 30 (1992), 35-47.

Cf. A011973, A000079, A047999 (Sierpiński), A007318, A101624.
Cf. A168081, A260022.
Cf. A257971 (first differences).

a006921 = sum . zipWith (*)
                a000079_list . map (flip mod 2) . reverse . a011973_row
-- Reinhard Zumkeller, Jul 14 2015

b2:=(n,k)->binomial(n,k) mod 2;
H:=n->add(b2(n-r,r)*2^( floor(n/2)-r ), r=0..floor(n/2));
[seq(H(n),n=0..30)]; # N. J. A. Sloane, Jul 14 2015

def A006921(n): return sum(int(not r & ~(n-r))*2**(n//2-r) for r in range(n//2+1)) # Chai Wah Wu, Jun 20 2022

a(2*n) = A260022(n); a(2*n+1) = A168081(n+1). - Reinhard Zumkeller, Jul 14 2015

a(n) = Sum_{r=0..n/2} binomial(n-r,r){mod 2} * 2^(floor(n/2)-r). - _N. J. A. Sloane, Jul 14 2015

A065942 Central column of triangle A065941.

Original entry on oeis.org

1, 1, 3, 4, 15, 21, 84, 120, 495, 715, 3003, 4368, 18564, 27132, 116280, 170544, 735471, 1081575, 4686825, 6906900, 30045015, 44352165, 193536720, 286097760, 1251677700, 1852482996, 8122425444, 12033222880, 52860229080, 78378960360

Offset: 0

Len Smiley, Nov 29 2001

When viewed as (1,1), (3,4), (15,21), ... this represents a shallow staircase on Pascal's triangle, arranged as a square array. - Paul Barry, Mar 11 2003
Also central column of triangle A011973 (taking rows with odd number of terms only). - John Molokach, Jul 08 2013
Interleaving of A005809 and A045721. - Bruce J. Nicholson, Apr 24 2018

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 15*x^4 + 21*x^5 + 84*x^6 + 120*x^7 + ... - _Michael Somos_, Jun 23 2018

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001 (Chapter 14)

Michael De Vlieger, Table of n, a(n) for n = 0..2416
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271; Corrections.

Cf. A065941 (complete triangle), A047749.

Cf. A005809, A045721.

List([0..40],n->Binomial(n+Int(n/2),n)); # Muniru A Asiru, Apr 28 2018

Array[Binomial[# + Floor[#/2], #] &, 30, 0] (* Michael De Vlieger, Apr 27 2018 *)

a(n) = binomial(n+n\2, n); \\ Altug Alkan, Apr 24 2018

a(n) = binomial(2n-floor((n+1)/2), floor(n/2)).
a(n+1) = Sum_{k=0..ceiling(n/2)} binomial(n+k, k). - Benoit Cloitre, Mar 06 2004
a(n) = binomial(n+floor(n/2), n). - Paul Barry, May 18 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n-1+k, k). - Paul Barry, Jul 06 2004
a(2n-1) = binomial(3n-3,n-1); a(2n) = binomial(3n-2,n-1). - John Molokach, Jul 08 2013
G.f.: A(x) = x*(d/dx)[log(S(x)-1)] = x*[(d/dx) S(x)]/[S(x)-1], where S(x) is the g.f. of A047749. - Vladimir Kruchinin, Jun 12 2014.
Conjecture: 8*n*(n-1)*a(n) -36*(n-1)*(n-3)*a(n-1) +6*(-9*n^2+18*n-14)*a(n-2) +27*(3*n-7)*(3*n-8)*a(n-3)=0. - R. J. Mathar, Jun 13 2014
0 = a(n)*(+281138850*a(n+2) +729089100*a(n+3) -77071527*a(n+4) -134472793*a(n+5)) +a(n+1)*(+15618825*a(n+2) -1650969*a(n+3) -9342280*a(n+4) -1729448*a(n+5)) +a(n+2)*(-19089675*a(n+2) -61394833*a(n+3) +6470716*a(n+4) +14929796*a(n+5)) +a(n+3)*(-1291668*a(n+3) +553572*a(n+4) +246032*a(n+5)) for all n in Z. - Michael Somos, Jun 23 2018

A145574 Array a(n,m) for number of partitions of n>=2 with m parts having no part 1. Hence m=1..floor(n/2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 2, 1, 1, 3, 3, 1, 1, 4, 4, 2, 1, 1, 4, 5, 3, 1, 1, 5, 7, 5, 2, 1, 1, 5, 8, 6, 3, 1, 1, 6, 10, 9, 5, 2, 1, 1, 6, 12, 11, 7, 3, 1, 1, 7, 14, 15, 10, 5, 2, 1, 1, 7, 16, 18, 13, 7, 3, 1, 1, 8, 19, 23, 18, 11, 5, 2, 1, 1, 8, 21, 27, 23, 14, 7, 3, 1, 1, 9, 24, 34, 30

Offset: 2

Wolfdieter Lang and Malin Sjodahl, Mar 06 2009

The row lengths sequence is floor(n/2) = [1,1,2,2,3,3,4,4,...], see A008619(n-1), n>=2.
Obtained from the characteristic partition array A145573 by summing in row n>=2 over entries belonging to like parts number m.
The column sequences give A000012, A004526, A001399, A001400, A001401, A001402, A026813 for m=1..7.

			1;
1;
1, 1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 3, 2, 1;
1, 3, 3, 1;
1, 4, 4, 2, 1;

Alois P. Heinz, Rows n = 2..200, flattened
W. Lang and M. Sjodahl, First 20 rows of the array and row sums.

Cf. A145573, A002865 (row sums).

b:= proc(n, i, t) option remember; `if`(2*t>n or t*i b(n, n, m):
seq(seq(a(n, m), m=1..iquo(n, 2)), n=2..30); # Alois P. Heinz, Oct 18 2012

nn=15; f[list_]:=Select[list,#>0&]; p=Product[1/(1-y x^i), {i, 2, nn}]; Drop[Map[f, CoefficientList[Series[p, {x, 0, nn}], {x, y}]], 1]//Grid  (* Geoffrey Critzer, Sep 23 2012 *)

# Prints the table; cf. A011973.
for n in (2..20): [Partitions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

a(n,m) = sum over entries of A145573(n,k) array which belong to partitions with part number m, for m=1..floor(n/2)). Note that partitions with parts number m>floor(n/2) have always at least one part 1.

G.f.: Product_{i>=2} 1/(1- y*x^i). - Geoffrey Critzer, Sep 23 2012

A173284 Triangle by columns, Fibonacci numbers in every column shifted down twice, for k > 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 5, 2, 1, 8, 3, 1, 13, 5, 2, 21, 8, 3, 1, 34, 13, 5, 2, 1, 55, 21, 8, 3, 1, 89, 34, 13, 5, 2, 1, 144, 55, 21, 8, 3, 1, 233, 89, 34, 13, 5, 2, 1, 377, 144, 55, 21, 8, 3, 1, 610, 233, 89, 34, 13, 5, 2, 1

Offset: 0

Gary W. Adamson, Feb 14 2010

The row sums equal A052952.
Let the triangle = M. Then lim_{n->infinity} M^n = A173285 as a left-shifted vector.
A173284 * [1, 2, 3, ...] = A054451: (1, 1, 4, 5, 12, 17, 33, ...). - Gary W. Adamson, Mar 03 2010
From Johannes W. Meijer, Sep 05 2013: (Start)
Triangle read by rows formed from antidiagonals of triangle A104762.
The diagonal sums lead to A004695. (End)

			First few rows of the triangle:
    1;
    1;
    2,   1;
    3,   1;
    5,   2,  1;
    8,   3,  1;
   13,   5,  2,  1;
   21,   8,  3,  1;
   34,  13,  5,  2,  1;
   55,  21,  8,  3,  1;
   89,  34, 13,  5,  2, 1;
  144,  55, 21,  8,  3, 1;
  233,  89, 34, 13,  5, 2, 1;
  377, 144, 55, 21,  8, 3, 1;
  610, 233, 89, 34, 13, 5, 2, 1;
  ...

Cf. A000045, A004695, A052952, A054451, A173285.

Cf. (Similar triangles) A008315 (Catalan), A011973 (Pascal), A102541 (Losanitsch), A122196 (Fractal), A122197 (Fractal), A128099 (Pell-Jacobsthal), A152198, A152204, A207538, A209634.

T := proc(n, k): if n<0 then return(0) elif k < 0 or k > floor(n/2) then return(0) else combinat[fibonacci](n-2*k+1) fi: end: seq(seq(T(n, k), k=0..floor(n/2)), n=0..14); # Johannes W. Meijer, Sep 05 2013

Triangle by columns, Fibonacci numbers in every column shifted down twice, for k > 0.
From Johannes W. Meijer, Sep 05 2013: (Start)
T(n,k) = A000045(n-2*k+1), n >= 0 and 0 <= k <= floor(n/2).
T(n,k) = A104762(n-k, k). (End)

Term a(15) corrected by Johannes W. Meijer, Sep 05 2013

cp's OEIS Frontend

A180178 Triangle read by rows: T(n,k) is the number of compositions of n without 3's and having k parts; 1 <= k <= n.

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A180180 Triangle read by rows: T(n,k) is the number of compositions of n without 5's and having k parts; 1 <= k <= n.

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A180181 Triangle read by rows: T(n,k) is the number of compositions of n without 6's and having k parts; 1 <= k <= n.

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A180183 Triangle read by rows: T(n,k) is the number of compositions of n without 8's and having k parts; 1 <= k <= n.

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A371967 Irregular triangle T(r,w) read by rows: number of ways of placing w non-attacking wazirs on a 3 X r board.

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A006921 Diagonals of Pascal's triangle mod 2 interpreted as binary numbers.

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