A173120 -id:A173120 - cp's OEIS frontend

A072405 Triangle T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise, read by rows.

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 1, 1, 4, 7, 7, 4, 1, 1, 5, 11, 14, 11, 5, 1, 1, 6, 16, 25, 25, 16, 6, 1, 1, 7, 22, 41, 50, 41, 22, 7, 1, 1, 8, 29, 63, 91, 91, 63, 29, 8, 1, 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1, 1, 10, 46, 129, 246, 336, 336, 246, 129, 46, 10, 1, 1, 11, 56, 175, 375, 582, 672, 582, 375, 175, 56, 11, 1

Offset: 0

Henry Bottomley, Jun 16 2002

Starting 1,0,1,1,1,... this is the Riordan array ((1-x+x^2)/(1-x), x/(1-x)). Its diagonal sums are A006355. Its inverse is A106509. - Paul Barry, May 04 2005

			Rows start as:
  1;
  1, 1;
  1, 1,  1; (key row for starting the recurrence)
  1, 2,  2,  1;
  1, 3,  4,  3,  1;
  1, 4,  7,  7,  4, 1;
  1, 5, 11, 14, 11, 5, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.

Row sums give essentially A003945, A007283, or A042950.
Cf. A072406 for number of odd terms in each row.
Cf. A051597, A096646, A122218 (identical for n > 1).
Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), A173118 (q=2), A173119 (q=3), A173120 (q=-4).

T:= func< n,k | n lt 3 select 1 else Binomial(n,k) - Binomial(n-2,k-1) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 28 2021

t[2, 1] = 1; t[n_, n_] = t[, 0] = 1; t[n, k_] := t[n, k] = t[n-1, k-1] + t[n-1, k]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2013, after Ralf Stephan *)

A072405(n, k) = if(n>2, binomial(n, k)-binomial(n-2, k-1), 1) \\ M. F. Hasler, Jan 06 2024

def T(n,k): return 1 if n<3 else binomial(n,k) - binomial(n-2,k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2021

T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise.
T(n, k) = T(n-1, k-1) + T(n-1, k) starting with T(2, 0) = T(2, 1) = T(2, 2) = 1 and T(n, 0) = T(n, n) = 1.
G.f.: (1-x^2*y) / (1 - x*(1+y)). - Ralf Stephan, Jan 31 2005
From G. C. Greubel, Apr 28 2021: (Start)
Sum_{k=0..n} T(n, k) = (n+1)*[n<3] + 3*2^(n-2)*[n>=3].
T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = -1. (End)

A173117 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 5, 8, 5, 1, 1, 6, 14, 14, 6, 1, 1, 7, 20, 28, 20, 7, 1, 1, 8, 27, 49, 49, 27, 8, 1, 1, 9, 35, 76, 98, 76, 35, 9, 1, 1, 10, 44, 111, 175, 175, 111, 44, 10, 1, 1, 11, 54, 155, 286, 350, 286, 155, 54, 11, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1,  3,  1;
  1,  4,  4,   1;
  1,  5,  8,   5,   1;
  1,  6, 14,  14,   6,   1;
  1,  7, 20,  28,  20,   7,   1;
  1,  8, 27,  49,  49,  27,   8,   1;
  1,  9, 35,  76,  98,  76,  35,   9,  1;
  1, 10, 44, 111, 175, 175, 111,  44, 10,  1;
  1, 11, 54, 155, 286, 350, 286, 155, 54, 11, 1;

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

Cf. A007318 (q=0), A072405 (q= -1), this sequence (q=1), A173118 (q=2), A173119 (q=3), A173120 (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,1], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = 1. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A173118 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 6, 10, 6, 1, 1, 7, 20, 20, 7, 1, 1, 8, 27, 40, 27, 8, 1, 1, 9, 35, 75, 75, 35, 9, 1, 1, 10, 44, 110, 150, 110, 44, 10, 1, 1, 11, 54, 154, 276, 276, 154, 54, 11, 1, 1, 12, 65, 208, 430, 552, 430, 208, 65, 12, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1,  4,  1;
  1,  5,  5,   1;
  1,  6, 10,   6,   1;
  1,  7, 20,  20,   7,   1;
  1,  8, 27,  40,  27,   8,   1;
  1,  9, 35,  75,  75,  35,   9,   1;
  1, 10, 44, 110, 150, 110,  44,  10,  1;
  1, 11, 54, 154, 276, 276, 154,  54, 11,  1;
  1, 12, 65, 208, 430, 552, 430, 208, 65, 12, 1;

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

Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), this sequence (q=2), A173119 (q=3), A173120 (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,2], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = 2. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A173119 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 6, 6, 1, 1, 7, 12, 7, 1, 1, 8, 28, 28, 8, 1, 1, 9, 36, 56, 36, 9, 1, 1, 10, 45, 119, 119, 45, 10, 1, 1, 11, 55, 164, 238, 164, 55, 11, 1, 1, 12, 66, 219, 483, 483, 219, 66, 12, 1, 1, 13, 78, 285, 702, 966, 702, 285, 78, 13, 1

Offset: 0

Roger L. Bagula, Feb 10 2010

			Triangle begins as:
  1;
  1,  1;
  1,  5,  1;
  1,  6,  6,   1;
  1,  7, 12,   7,   1;
  1,  8, 28,  28,   8,   1;
  1,  9, 36,  56,  36,   9,   1;
  1, 10, 45, 119, 119,  45,  10,   1;
  1, 11, 55, 164, 238, 164,  55,  11,  1;
  1, 12, 66, 219, 483, 483, 219,  66, 12,  1;
  1, 13, 78, 285, 702, 966, 702, 285, 78, 13, 1;

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

Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), A173118 (q=2), this sequence (q=3), A173120 (q= -4), A173122.

T[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j]*Boole[n>2*j], {j,0,5}]];
Table[T[n,k,3], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 27 2021 *)

@CachedFunction
def T(n,k,q): return 1 if (k==0 or k==n) else q*bool(n==2) + sum( q^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
flatten([[T(n,k,3) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Apr 27 2021

T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3.

Sum_{k=0..n} T(n, k, q) = [n=0] + q*[n=2] + Sum_{j=0..5} q^j*2^(n-2*j)*[n > 2*j] for q = 3. - G. C. Greubel, Apr 27 2021

Edited by G. C. Greubel, Apr 27 2021

A173122 Irregular triangle T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with t(n,0,q) = t(n,n,q) = 1, read by rows.

Original entry on oeis.org

1, 2, 4, 1, 8, 2, 16, 4, 32, 8, 2, 64, 16, 4, 128, 32, 8, 2, 256, 64, 16, 4, 512, 128, 32, 8, 2, 1024, 256, 64, 16, 4, 2048, 512, 128, 32, 8, 2, 4096, 1024, 256, 64, 16, 4, 8192, 2048, 512, 128, 32, 8, 16384, 4096, 1024, 256, 64, 16, 32768, 8192, 2048, 512, 128, 32, 65536, 16384, 4096, 1024, 256, 64

Offset: 0

Roger L. Bagula, Feb 10 2010

			Irregular triangle begins as:
     1;
     2;
     4,   1;
     8,   2;
    16,   4;
    32,   8,  2;
    64,  16,  4;
   128,  32,  8,  2;
   256,  64, 16,  4;
   512, 128, 32,  8, 2;
  1024, 256, 64, 16, 4;

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

Cf. A007318, A072405, A173117, A173118, A173119, A173120.

t[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j] *Boole[n>2*j], {j, 0, 5}]];
T[n_]:= CoefficientList[Series[Sum[t[n,k,q], {k,0,n}], {q,0,n}], q];
Table[T[n], {n, 0, 12}]//Flatten (* modified by G. C. Greubel, Apr 29 2021 *)

@CachedFunction
def t(n, k, x): return 1 if (k==0 or k==n) else x*bool(n==2) + sum( x^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
def s(n,x): return sum( t(n,k,x) for k in (0..n) )
flatten([taylor(s(n,x), x, 0, n).list() for n in (0..12)]) # G. C. Greubel, Apr 29 2021

T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with t(n,0,q) = t(n,n,q) = 1.

More terms and edited by G. C. Greubel, Apr 29 2021

cp's OEIS Frontend

A072405 Triangle T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise, read by rows.

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A173117 Triangle T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1, read by rows.

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A173118 Triangle T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2, read by rows.

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A173119 Triangle T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3, read by rows.

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A173122 Irregular triangle T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with t(n,0,q) = t(n,n,q) = 1, read by rows.

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A173117 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 1, read by rows.

A173118 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 2, read by rows.

A173119 Triangle T(n, k, q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with T(n,0) = T(n,n) = 1 for q = 3, read by rows.

A173122 Irregular triangle T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q[n=2] + Sum_{j=0..5} q^jbinomial(n-2j, k-j)[n>2*j] with t(n,0,q) = t(n,n,q) = 1, read by rows.