A173075 -id:A173075 - cp's OEIS frontend

A132044 Triangle T(n,k) = binomial(n, k) - 1 with T(n,0) = T(n,n) = 1, read by rows.

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 1, 1, 4, 9, 9, 4, 1, 1, 5, 14, 19, 14, 5, 1, 1, 6, 20, 34, 34, 20, 6, 1, 1, 7, 27, 55, 69, 55, 27, 7, 1, 1, 8, 35, 83, 125, 125, 83, 35, 8, 1, 1, 9, 44, 119, 209, 251, 209, 119, 44, 9, 1

Offset: 0

Gary W. Adamson, Aug 08 2007

Row sums = A132045: (1, 2, 3, 6, 13, 28, 59, ...).

The triangle sequences having the form T(n,k,q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 have the row sums Sum_{k=0..n} T(n,k,q) = 2^(n-2)*q^n + 2^n - (n-1) - (5/4)*[n=0] -(q/2)*[n=1]. - G. C. Greubel, Feb 12 2021

			First few rows of the triangle:
  1;
  1, 1;
  1, 1,  1;
  1, 2,  2,  1;
  1, 3,  5,  3,  1;
  1, 4,  9,  9,  4,  1;
  1, 5, 14, 19, 14,  5,  1;
  1, 6, 20, 34, 34, 20,  6, 1;
  1, 7, 27, 55, 69, 55, 27, 7, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A007318, A103451, A132045.

Cf. this sequence (q=0), A173075 (q=1), A173046 (q=2), A173047 (q=3).

T:= func< n,k,q | k eq 0 or k eq n select 1 else Binomial(n,k) + q^n*Binomial(n-2,k-1) -1 >;
[T(n,k,0): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 12 2021

T[n_, k_]:= If[k==0 || k==n, 1, Binomial[n, k] - 1];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* Roger L. Bagula, Feb 08 2010 *)

def T(n,k,q): return 1 if (k==0 or k==n) else binomial(n,k) + q^n*binomial(n-2,k-1) -1
flatten([[T(n,k,0) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 12 2021

T(n, k) = A007318(n,k) + A103451(n,k) - A000012(n,k), an infinite lower triangular matrix.
T(n, k) = binomial(n, k) - 1, with T(n,0) = T(n,n) = 1. - Roger L. Bagula, Feb 08 2010
From G. C. Greubel, Feb 12 2021: (Start)
T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 0.
Sum_{k=0..n} T(n, k, 0) = 2^n - (n-1) - [n=0]. (End)

A167763 Pendular triangle (p=0), read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), otherwise T(n,k) = T(n,n-1-k) + p*T(n-1,k), for n >= k <= 0, with T(n,0) = 1 and T(n,n) = 0^n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 7, 4, 1, 0, 1, 5, 12, 12, 5, 1, 0, 1, 6, 18, 30, 18, 6, 1, 0, 1, 7, 25, 55, 55, 25, 7, 1, 0, 1, 8, 33, 88, 143, 88, 33, 8, 1, 0, 1, 9, 42, 130, 273, 273, 130, 42, 9, 1, 0, 1, 10, 52, 182, 455, 728, 455, 182, 52, 10, 1, 0

Offset: 0

Philippe Deléham, Nov 11 2009

See A118340 for definition of pendular triangles and pendular sums.

The last five rows in the example section appear in the table on p. 8 of Getzler. Cf. also A173075. - Tom Copeland, Jan 22 2020

			Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,  0;
  1,  3,  3,  1,  0;
  1,  4,  7,  4,  1,  0;
  1,  5, 12, 12,  5,  1,  0; ...

G. C. Greubel, Rows n = 0..100 of the triangle, flattened
E. Getzler, The semi-classical approximation for modular operads, arXiv:alg-geom/9612005, 1996.

Cf. this sequence (p=0), A118340 (p=1), A118345 (p=2), A118350 (p=3).

Cf. A001764, A006013, A006629, A093951, A102893, A173075.

function T(n,k,p)
  if k lt 0 or n lt k then return 0;
  elif k eq 0 then return 1;
  elif k eq n then return 0;
  elif n gt 2*k then return T(n,n-k,p) + T(n-1,k,p);
  else return T(n,n-k-1,p) + p*T(n-1,k,p);
  end if;
  return T;
end function;
[T(n,k,0): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2021

T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 17 2021 *)

{T(n,k)=if(k==0,1,if(n==k,0,if(n>2*k,binomial(n+k+1,k)*(n-2*k+1)/(n+k+1),T(n,n-1-k))))} \\ Paul D. Hanna, Nov 12 2009

@CachedFunction
def T(n, k, p):
    if (k<0 or n2*k): return T(n,n-k,p) + T(n-1,k,p)
    else: return T(n, n-k-1, p) + p*T(n-1, k, p)
flatten([[T(n, k, 0) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021

T(2n+m) = [A001764^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A001764.
If n > 2k, T(n,k) = binomial(n+k+1,k)*(n-2k+1)/(n+k+1), else T(n,k) = T(n,n-1-k), with conditions: T(n,0)=1 for n>=0 and T(n,n)=0 for n > 0. - Paul D. Hanna, Nov 12 2009
Sum_{k=0..n} T(n, k, p=0) = A093951(n). - G. C. Greubel, Feb 17 2021

A173046 Triangle T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 10, 10, 1, 1, 19, 37, 19, 1, 1, 36, 105, 105, 36, 1, 1, 69, 270, 403, 270, 69, 1, 1, 134, 660, 1314, 1314, 660, 134, 1, 1, 263, 1563, 3895, 5189, 3895, 1563, 263, 1, 1, 520, 3619, 10835, 18045, 18045, 10835, 3619, 520, 1, 1, 1033, 8236, 28791, 57553, 71931, 57553, 28791, 8236, 1033, 1

Offset: 0

Roger L. Bagula, Feb 08 2010

The triangle sequences having the form T(n,k,q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 have the row sums Sum_{k=0..n} T(n,k,q) = 2^(n-2)*q^n + 2^n - (n-1) - (5/4)*[n=0] -(q/2)*[n=1]. - G. C. Greubel, Feb 16 2021

			Triangle begins as:
  1;
  1,    1;
  1,    5,    1;
  1,   10,   10,     1;
  1,   19,   37,    19,     1;
  1,   36,  105,   105,    36,     1;
  1,   69,  270,   403,   270,    69,     1;
  1,  134,  660,  1314,  1314,   660,   134,     1;
  1,  263, 1563,  3895,  5189,  3895,  1563,   263,    1;
  1,  520, 3619, 10835, 18045, 18045, 10835,  3619,  520,    1;
  1, 1033, 8236, 28791, 57553, 71931, 57553, 28791, 8236, 1033, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A132044 (q=0), A173075 (q=1), this sequence (q=2), A173047 (q=3).

Cf. A000302, A132045.

T:= func< n,k,q | k eq 0 or k eq n select 1 else Binomial(n,k) + q^n*Binomial(n-2,k-1) -1 >;
[T(n,k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 16 2021

T[n_, m_, q_]:= If[k==0 || k==n, 1, Binomial[n, k] +(q^n)*Binomial[n-2, k-1] -1];
Table[T[n,k,2], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Feb 16 2021 *)

def T(n,k,q): return 1 if (k==0 or k==n) else binomial(n,k) + q^n*binomial(n-2,k-1) -1
flatten([[T(n,k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 16 2021

T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 2.

Sum_{k=0..n} T(n, k, 2) = 4^(n-1) + 2^n - (n-1) - (5/4)*[n=0] = A000302(n-1) + A132045(n) - (5/4)*[n=0]. - [n=1]. - G. C. Greubel, Feb 16 2021

Edited by G. C. Greubel, Feb 16 2021

A173047 Triangle T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 29, 29, 1, 1, 84, 167, 84, 1, 1, 247, 738, 738, 247, 1, 1, 734, 2930, 4393, 2930, 734, 1, 1, 2193, 10955, 21904, 21904, 10955, 2193, 1, 1, 6568, 39393, 98470, 131289, 98470, 39393, 6568, 1, 1, 19691, 137816, 413426, 689030, 689030, 413426, 137816, 19691, 1

Offset: 0

Roger L. Bagula, Feb 08 2010

The triangle sequences having the form T(n,k,q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 have the row sums Sum_{k=0..n} T(n,k,q) = 2^(n-2)*q^n + 2^n - (n-1) - (5/4)*[n=0] -(q/2)*[n=1]. - G. C. Greubel, Feb 16 2021

			Ttiangle begins as:
  1;
  1,     1;
  1,    10,      1;
  1,    29,     29,      1;
  1,    84,    167,     84,      1;
  1,   247,    738,    738,    247,      1;
  1,   734,   2930,   4393,   2930,    734,      1;
  1,  2193,  10955,  21904,  21904,  10955,   2193,      1;
  1,  6568,  39393,  98470, 131289,  98470,  39393,   6568,     1;
  1, 19691, 137816, 413426, 689030, 689030, 413426, 137816, 19691, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A132044 (q=0), A173075 (q=1), A173046 (q=2), this sequence (q=3).

T:= func< n,k,q | k eq 0 or k eq n select 1 else Binomial(n,k) + q^n*Binomial(n-2,k-1) -1 >;
[T(n,k,3): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 16 2021

T[n_, k_, q_]:= If[k==0 || k==n, 1, Binomial[n, k] +(q^n)*Binomial[n-2, k-1] -1];
Table[T[n,k,3], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Feb 16 2021 *)

def T(n,k,q): return 1 if (k==0 or k==n) else binomial(n,k) + q^n*binomial(n-2,k-1) -1
flatten([[T(n,k,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 16 2021

T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 3.

Sum_{k=0..n} T(n, k, 3) = (1/4)*(6^n + 2^(n+2) - 4*(n-1) - 5*[n=0] - 6*[n=1]). - G. C. Greubel, Feb 16 2021

Edited by G. C. Greubel, Feb 16 2021

A173076 Triangle T(n, k, q) = binomial(n, k) - 1 + q^(floor(n/2))*binomial(n-2, k-1) with T(n, 0, q) = T(n, n, q) = 1 and q = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 7, 13, 7, 1, 1, 8, 21, 21, 8, 1, 1, 13, 46, 67, 46, 13, 1, 1, 14, 60, 114, 114, 60, 14, 1, 1, 23, 123, 295, 389, 295, 123, 23, 1, 1, 24, 147, 419, 685, 685, 419, 147, 24, 1, 1, 41, 300, 1015, 2001, 2491, 2001, 1015, 300, 41, 1

Offset: 0

Roger L. Bagula, Feb 09 2010

			Triangle begins as:
  1;
  1,  1;
  1,  3,   1;
  1,  4,   4,    1;
  1,  7,  13,    7,    1;
  1,  8,  21,   21,    8,    1;
  1, 13,  46,   67,   46,   13,    1;
  1, 14,  60,  114,  114,   60,   14,    1;
  1, 23, 123,  295,  389,  295,  123,   23,   1;
  1, 24, 147,  419,  685,  685,  419,  147,  24,  1;
  1, 41, 300, 1015, 2001, 2491, 2001, 1015, 300, 41, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A132044 (q=0), A173075 (q=1), this sequence (q=2), A173077 (q=3).

T:= func< n,k,q | k eq 0 or k eq n select 1 else Binomial(n,k) + q^(Floor(n/2))*Binomial(n-2,k-1) -1 >;
[T(n,k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 09 2021

T[n_, k_, q_]:= If[k==0 || k==n, 1, Binomial[n, k] - 1 + q^(Floor[n/2])*Binomial[n-2, k-1]];
Table[T[n, k, 2], {n,0,10}, {k,0,n}]//Flatten

def T(n,k,q): return 1 if (k==0 or k==n) else binomial(n,k) + q^(n//2)*binomial(n-2,k-1) -1
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)])

T(n, k, q) = binomial(n, k) - 1 + q^floor(n/2)*binomial(n-2, k-1) with T(n, 0, q) = T(n, n, q) = 1 and q = 2.

Edited by G. C. Greubel, Jul 09 2021

A173077 Triangle T(n, k, q) = binomial(n, k) - 1 + q^floor(n/2)*binomial(n-2, k-1) with T(n, 0, q) = T(n, n, q) = 1 and q = 3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 12, 23, 12, 1, 1, 13, 36, 36, 13, 1, 1, 32, 122, 181, 122, 32, 1, 1, 33, 155, 304, 304, 155, 33, 1, 1, 88, 513, 1270, 1689, 1270, 513, 88, 1, 1, 89, 602, 1784, 2960, 2960, 1784, 602, 89, 1, 1, 252, 1988, 6923, 13817, 17261, 13817, 6923, 1988, 252, 1

Offset: 0

Roger L. Bagula, Feb 09 2010

			Triangle starts:
  1;
  1,   1;
  1,   4,    1;
  1,   5,    5,    1;
  1,  12,   23,   12,     1;
  1,  13,   36,   36,    13,     1;
  1,  32,  122,  181,   122,    32,     1;
  1,  33,  155,  304,   304,   155,    33,    1;
  1,  88,  513, 1270,  1689,  1270,   513,   88,    1;
  1,  89,  602, 1784,  2960,  2960,  1784,  602,   89,   1;
  1, 252, 1988, 6923, 13817, 17261, 13817, 6923, 1988, 252, 1;
  ...
Row sums: 1, 2, 6, 12, 49, 100, 491, 986, 5433, 10872, 63223, ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A132044 (q=0), A173075 (q=1), A173076 (q=2), this sequence (q=3).

T:= func< n,k,q | k eq 0 or k eq n select 1 else Binomial(n,k) + q^(Floor(n/2))*Binomial(n-2,k-1) -1 >;
[T(n,k,3): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 09 2021

T[n_, k_]:= If[k==0 || k==n, 1, Binomial[n, k] - 1 + 3^Floor[n/2] Binomial[n-2, k- 1]];
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten

def T(n,k,q): return 1 if (k==0 or k==n) else binomial(n,k) + q^(n//2)*binomial(n-2,k-1) -1
flatten([[T(n,k,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 09 2021

T(n, k, q) = binomial(n, k) - 1 + q^floor(n/2)*binomial(n-2, k-1) with T(n, 0, q) = T(n, n, q) = 1 and q = 3.

cp's OEIS Frontend

A132044 Triangle T(n,k) = binomial(n, k) - 1 with T(n,0) = T(n,n) = 1, read by rows.

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A167763 Pendular triangle (p=0), read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), otherwise T(n,k) = T(n,n-1-k) + p*T(n-1,k), for n >= k <= 0, with T(n,0) = 1 and T(n,n) = 0^n.

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A173046 Triangle T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 2, read by rows.

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A173047 Triangle T(n, k, q) = binomial(n, k) + q^n*binomial(n-2, k-1) - 1 with T(n, 0) = T(n, n) = 1 and q = 3, read by rows.

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A173076 Triangle T(n, k, q) = binomial(n, k) - 1 + q^(floor(n/2))*binomial(n-2, k-1) with T(n, 0, q) = T(n, n, q) = 1 and q = 2, read by rows.

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A173077 Triangle T(n, k, q) = binomial(n, k) - 1 + q^floor(n/2)*binomial(n-2, k-1) with T(n, 0, q) = T(n, n, q) = 1 and q = 3, read by rows.

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