A118340 -id:A118340 - cp's OEIS frontend

A168030 Variant of pendular triangle A118340.

1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0

Offset: 0

Philippe Deléham, Nov 17 2009

Replaced the sums (f(a,b) = a + b) by the operators f(a,b) = a^2 -a*b + b^2 in the construction of triangle in A118340.

			Triangle begins as:
  1;
  1, 0;
  1, 1, 0;
  1, 0, 1, 0;
  1, 1, 0, 1, 0;
  1, 0, 1, 1, 1, 0;
  1, 1, 1, 0, 0, 1, 0;
  1, 0, 0, 0, 0, 1, 1, 0;
  1, 1, 0, 1, 1, 1, 0, 1, 0;
  1, 0, 1, 0, 0, 1, 1, 1, 1, 0;
  1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0;

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

Cf. A118340, A168148 (row sums).

function t(n, k) // t = A118340
  if k lt 0 or k gt n then return 0;
  elif k eq 0 then return 1;
  elif n gt 2*k then return t(n, n-k) + t(n-1, k);
  else return t(n, n-k-1) + t(n-1, k);
  end if; return t;
end function;
T:= func< n,k | t(n,k) mod 2 >; // A168030
[T(n,k): k in [0..n], n in [0..15]];

t[n_, k_, p_]:= t[n, k, p]= If[k<0 || k>n, 0, If[k==0, 1, If[n<=2*k, t[n,n-k-1,p] +p*t[n-1,k,p], t[n,n-k,p] +t[n-1,k, p]]]]; (* A118340 *)
T[n_, k_, p_]:= Mod[t[n,k,p], 2]; (* A168030 *)
Table[T[n,k,1], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 12 2023 *)

@CachedFunction
def t(n, k): # t = A118340
    if (k<0 or k>n): return 0
    elif (k==0): return 1
    elif (n>2*k): return t(n, n-k) + t(n-1, k)
    else: return t(n, n-k-1) + t(n-1, k)
def A168030(n,k): return t(n,k)%2
flatten([[A168030(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Jan 12 2023

From G. C. Greubel, Jan 12 2023: (Start)
T(n, k) = A118340(n, k) mod 2.
Sum_{k=0..n} T(n, k) = A168148(n). (End)

A118345 Pendular triangle, 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), else T(n,k) = T(n,n-1-k) + 2*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, 5, 1, 0, 1, 4, 11, 6, 1, 0, 1, 5, 18, 30, 7, 1, 0, 1, 6, 26, 70, 40, 8, 1, 0, 1, 7, 35, 121, 201, 51, 9, 1, 0, 1, 8, 45, 184, 487, 286, 63, 10, 1, 0, 1, 9, 56, 260, 873, 1445, 386, 76, 11, 1, 0, 1, 10, 68, 350, 1375, 3592, 2147, 502, 90, 12, 1, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

See A118340 for definition of pendular triangles and pendular sums.

			Row 6 equals the pendular sums of row 5:
  [1,  4, 11,  6,  1,  0], where the pendular sums proceed as follows:
  [1, __, __, __, __, __]: T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1]: T(6,5) = T(6,0) + 2*T(5,5) = 1 + 2*0 = 1;
  [1,  5, __, __, __,  1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  7,  1]: T(6,4) = T(6,1) + 2*T(5,4) = 5 + 2*1 = 7;
  [1,  5, 18, __,  7,  1]: T(6,2) = T(6,4) + T(5,2) = 7 + 11 = 18;
  [1,  5, 18, 30,  7,  1]: T(6,3) = T(6,2) + 2*T(5,3) = 18 + 2*6 = 30;
  [1,  5, 18, 30,  7,  1, 0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  5,   1,    0;
  1,  4, 11,   6,    1,    0;
  1,  5, 18,  30,    7,    1,    0;
  1,  6, 26,  70,   40,    8,    1,   0;
  1,  7, 35, 121,  201,   51,    9,   1,  0;
  1,  8, 45, 184,  487,  286,   63,  10,  1,  0;
  1,  9, 56, 260,  873, 1445,  386,  76, 11,  1, 0;
  1, 10, 68, 350, 1375, 3592, 2147, 502, 90, 12, 1, 0; ...
Central terms are T(2*n,n) = A118346(n);
semi-diagonals form successive self-convolutions of the central terms:
T(2*n+1,n) = A118347(n) = [A118346^2](n),
T(2*n+2,n) = A118348(n) = [A118346^3](n).

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

Cf. A118346, A118347, A118348, A118349, A118340.

Cf. A167763 (p=0), A118340 (p=1), this sequence (p=2), A118350 (p=3).

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,2): 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(n2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)+2*T(n-1,k)))))

@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,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021

T(2*n+m,n) = [A118346^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A118346.

A118350 Pendular triangle, 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), else T(n,k) = T(n,n-1-k) + 3*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, 6, 1, 0, 1, 4, 13, 7, 1, 0, 1, 5, 21, 42, 8, 1, 0, 1, 6, 30, 96, 54, 9, 1, 0, 1, 7, 40, 163, 325, 67, 10, 1, 0, 1, 8, 51, 244, 770, 445, 81, 11, 1, 0, 1, 9, 63, 340, 1353, 2688, 583, 96, 12, 1, 0, 1, 10, 76, 452, 2093, 6530, 3842, 740, 112, 13, 1, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

See definition of pendular triangle and pendular sums at A118340.

			Row 6 equals the pendular sums of row 5,
  [1,  4, 13,  7,  1,  0], where the sums proceed as follows:
  [1, __, __, __, __, __]: T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1]: T(6,5) = T(6,0) + 3*T(5,5) = 1 + 3*0 = 1;
  [1,  5, __, __, __,  1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  8,  1]: T(6,4) = T(6,1) + 3*T(5,4) = 5 + 3*1 = 8;
  [1,  5, 21, __,  8,  1]: T(6,2) = T(6,4) + T(5,2) = 8 + 13 = 21;
  [1,  5, 21, 42,  8,  1]: T(6,3) = T(6,2) + 3*T(5,3) = 21 + 3*7 = 42;
  [1,  5, 21, 42,  8,  1, 0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  6,   1,    0;
  1,  4, 13,   7,    1,     0;
  1,  5, 21,  42,    8,     1,     0;
  1,  6, 30,  96,   54,     9,     1,    0;
  1,  7, 40, 163,  325,    67,    10,    1,   0;
  1,  8, 51, 244,  770,   445,    81,   11,   1,   0;
  1,  9, 63, 340, 1353,  2688,   583,   96,  12,   1,  0;
  1, 10, 76, 452, 2093,  6530,  3842,  740, 112,  13,  1, 0;
  1, 11, 90, 581, 3010, 11760, 23286, 5230, 917, 129, 14, 1, 0; ...
Central terms are T(2*n,n) = A118351(n);
semi-diagonals form successive self-convolutions of the central terms:
T(2*n+1,n) = A118352(n) = [A118351^2](n),
T(2*n+2,n) = A118353(n) = [A118351^3](n).

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

Cf. A118351, A118352, A118353, A118354.

Cf. A167763 (p=0), A118340 (p=1), A118345 (p=2), this sequence (p=3).

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,3): 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(n2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)+3*T(n-1,k)))))

@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,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021

T(2*n+m,n) = [A118351^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of the central terms A118351.

A119369 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0)=T(n+1,2n)=1 and T(n+1,2n+1)=T(n+1,2n+2)=0.

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 6, 9, 7, 3, 1, 0, 0, 1, 4, 10, 20, 30, 23, 11, 4, 1, 0, 0, 1, 5, 15, 36, 70, 104, 81, 40, 16, 5, 1, 0, 0, 1, 6, 21, 58, 133, 253, 374, 293, 149, 63, 22, 6, 1, 0, 0, 1, 7, 28, 87, 226, 501, 938, 1380, 1087, 564, 248, 93, 29, 7, 1, 0, 0

Offset: 0

Paul D. Hanna, May 16 2006

The diagonals may be generated by iterated convolutions of a base sequence B with the sequence C of central terms. The g.f. B(x) of the base sequence satisfies: B = 1 + x*B^2 + x^2*(B^2 - B); the g.f. C(x) of the central terms satisfies: C(x) = 1/(1+x - x*B(x)).

			To obtain row 4, pendular sums of row 3 are carried out as follows.
  [1, 2, 3, 2, 1, 0, 0]: given row 3;
  [1, _, _, _, _, _, _]: start with T(4,0) = T(3,0) = 1;
  [1, _, _, _, _, _, 1]: T(4,6) = T(4,0) + T(3,6) = 1 + 0 = 1;
  [1, 3, _, _, _, _, 1]: T(4,1) = T(4,6) + T(3,1) = 1 + 2 = 3;
  [1, 3, _, _, _, 3, 1]: T(4,5) = T(4,1) + T(3,5) = 3 + 0 = 3;
  [1, 3, 6, _, _, 3, 1]: T(4,2) = T(4,5) + T(3,2) = 3 + 3 = 6;
  [1, 3, 6, _, 7, 3, 1]: T(4,4) = T(4,2) + T(3,4) = 6 + 1 = 7;
  [1, 3, 6, 9, 7, 3, 1]: T(4,3) = T(4,4) + T(3,3) = 7 + 2 = 9;
  [1, 3, 6, 9, 7, 3, 1, 0, 0]: complete row 4 by appending two zeros.
Triangle begins:
  1;
  1, 0,  0;
  1, 1,  1,  0,   0;
  1, 2,  3,  2,   1,   0,   0;
  1, 3,  6,  9,   7,   3,   1,    0,    0;
  1, 4, 10, 20,  30,  23,  11,    4,    1,   0,   0;
  1, 5, 15, 36,  70, 104,  81,   40,   16,   5,   1,  0,  0;
  1, 6, 21, 58, 133, 253, 374,  293,  149,  63,  22,  6,  1, 0, 0;
  1, 7, 28, 87, 226, 501, 938, 1380, 1087, 564, 248, 93, 29, 7, 1, 0, 0;
Central terms are:
  C = A119371 = [1, 0, 1, 2, 7, 23, 81, 293, 1087, 4110, ...].
Lower diagonals start:
  D1 = A119372 = [1, 1, 3, 9, 30, 104, 374, 1380, 5197, ...];
  D2 = A119373 = [1, 2, 6, 20, 70, 253, 938, 3546, 13617, ...].
  Diagonals above central terms (ignoring leading zeros) start:
  U1 = A119375 = [1, 3, 11, 40, 149, 564, 2166, 8420, ...];
  U2 = A119376 = [1, 4, 16, 63, 248, 980, 3894, 15563, ...].
There exists the base sequence:
  B = A119370 = [1, 1, 2, 6, 19, 64, 225, 816, 3031, 11473, ...]
which generates all diagonals by convolutions with central terms:
  D2 = B * D1 = B^2 * C
  U2 = B * U1 = B^2 * C"
where C" = [1, 2, 7, 23, 81, 293, 1087, ...]
are central terms not including the initial [1,0].

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

Cf. A119370, A119371, A119372, A119373, A119374, A119375, A119376.

Variants: A118340, A118345, A118350.

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1
    elif k<0 or k>2*(n-1) then 0
    elif n=2 and k<3 then 1
    else T(n-1, k) + `if`(kG. C. Greubel, Mar 16 2021

T[n_, k_]:= T[n, k]= If[n==0 && k==0, 1, If[k<0 || k>2*(n-1), 0, If[n==2 && k<3, 1, T[n-1, k] +If[kG. C. Greubel, Mar 16 2021 *)

T(n,k)= if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, T(n-1,k) + if(k

@CachedFunction
def T(n, k):
    if (n==0 and k==0): return 1
    elif (k<0 or k>2*(n-1)): return 0
    elif (n==2 and k<3): return 1
    else: return T(n-1, k) + ( T(n, 2*n-k-1) if kG. C. Greubel, Mar 16 2021

Sum_{k=0..2*n} T(n, k) = A119372(n). - G. C. Greubel, Mar 16 2021

A122445 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + 2*T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0) = T(n+1,2n) = 1 and T(n+1,2n+1) = T(n+1,2n+2) = 0.

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 6, 10, 8, 3, 1, 0, 0, 1, 4, 10, 22, 36, 28, 12, 4, 1, 0, 0, 1, 5, 15, 39, 83, 135, 107, 47, 17, 5, 1, 0, 0, 1, 6, 21, 62, 155, 324, 525, 418, 189, 72, 23, 6, 1, 0, 0, 1, 7, 28, 92, 259, 629, 1298, 2094, 1676, 773, 305, 104, 30, 7, 1, 0, 0

Offset: 0

Paul D. Hanna, Sep 07 2006

The diagonals may be generated by iterated convolutions of a base sequence B with the sequence C of central terms. The g.f. B(x) of the base sequence satisfies: B = 1 + x*B^2 + 2x^2*(B^2 - B); the g.f. C(x) of the central terms satisfies: C(x) = 1/(1+x - xB(x)).

			To obtain row 4, pendular sums of row 3 are carried out as follows.
  [1, 2, 3,  2, 1, 0, 0]: given row 3;
  [1, _, _, __, _, _, _]: start with T(4,0) = T(3,0) = 1;
  [1, _, _, __, _, _, 1]: T(4,6) = T(4,0) + 2*T(3,6) = 1 + 2*0 = 1;
  [1, 3, _, __, _, _, 1]: T(4,1) = T(4,6) + 1*T(3,1) = 1 + 1*2 = 3;
  [1, 3, _, __, _, 3, 1]: T(4,5) = T(4,1) + 2*T(3,5) = 3 + 2*0 = 3;
  [1, 3, 6, __, _, 3, 1]: T(4,2) = T(4,5) + 1*T(3,2) = 3 + 1*3 = 6;
  [1, 3, 6, __, 8, 3, 1]: T(4,4) = T(4,2) + 2*T(3,4) = 6 + 2*1 = 8;
  [1, 3, 6, 10, 8, 3, 1]: T(4,3) = T(4,4) + 1*T(3,3) = 8 + 1*2 = 10;
  [1, 3, 6, 10, 8, 3, 1,0,0]: complete row 4 by appending two zeros.
Triangle begins:
  1;
  1, 0,  0;
  1, 1,  1,  0,   0;
  1, 2,  3,  2,   1,   0,   0;
  1, 3,  6, 10,   8,   3,   1,   0,   0;
  1, 4, 10, 22,  36,  28,  12,   4,   1,  0,  0;
  1, 5, 15, 39,  83, 135, 107,  47,  17,  5,  1, 0, 0;
  1, 6, 21, 62, 155, 324, 525, 418, 189, 72, 23, 6, 1, 0, 0;
Central terms are:
  C = A122447 = [1, 0, 1, 2, 8, 28, 107, 418, 1676, 6848, ...].
Lower diagonals start:
  D1 = A122448 = [1, 1, 3, 10, 36, 135, 525, 2094, 8524, ...];
  D2 = A122449 = [1, 2, 6, 22, 83, 324, 1298, 5302, 22002, ...].
Diagonals above central terms (ignoring leading zeros) start:
  U1 = A122450 = [1, 3, 12, 47, 189, 773, 3208, 13478, 57222, ...];
  U2 = A122451 = [1, 4, 17, 72, 305, 1300, 5576, 24068, 104510, ...].
There exists the base sequence:
  B = A122446 = [1, 1, 2, 7, 24, 88, 336, 1321, 5316, 21788, ...]
which generates all diagonals by convolutions with central terms:
  D2 = B * D1 = B^2 * C
  U2 = B * U1 = B^2 * C"
where C" = [1, 2, 8, 28, 107, 418, 1676, 6848, 28418, ...]
are central terms not including the initial [1,0].

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

Cf. A122446, A122447 (central terms), A122452 (row sums).
Diagonals: A122448, A122449, A122450, A122451.
Variants: A118340, A118345, A118350, A119369.

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1
    elif k<0 or k>2*(n-1) then 0
    elif n=2 and k<3 then 1
    else T(n-1, k) + `if`(kG. C. Greubel, Mar 16 2021

T[n_, k_]:= T[n, k]= If[n==0 && k==0, 1, If[k<0 || k>2*(n-1), 0, If[n==2 && k<3, 1, T[n-1, k] + If[kG. C. Greubel, Mar 16 2021 *)

{T(n,k)= if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, if(k

@CachedFunction
def T(n, k):
    if (n==0 and k==0): return 1
    elif (k<0 or k>2*(n-1)): return 0
    elif (n==2 and k<3): return 1
    else: return T(n-1, k) + ( T(n, 2*n-k-1) if kG. C. Greubel, Mar 16 2021

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

A118343 Triangle, read by rows, where diagonals are successive self-convolutions of A108447.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 4, 0, 1, 3, 9, 20, 0, 1, 4, 15, 48, 113, 0, 1, 5, 22, 85, 282, 688, 0, 1, 6, 30, 132, 519, 1762, 4404, 0, 1, 7, 39, 190, 837, 3330, 11488, 29219, 0, 1, 8, 49, 260, 1250, 5516, 22135, 77270, 199140, 0, 1, 9, 60, 343, 1773, 8461, 37404, 151089, 532239, 1385904, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

A108447 equals the central terms of pendular triangle A118340 and the diagonals of this triangle form the semi-diagonals of the triangle A118340. Row sums equal A054727, the number of forests of rooted trees with n nodes on a circle without crossing edges.

			Show: T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - T(7,3) + T(8,3) + T(9,3)
or 837 = 519 - 132 + 190 + 260.
Triangle begins:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  4,   0;
  1, 3,  9,  20,    0;
  1, 4, 15,  48,  113,    0;
  1, 5, 22,  85,  282,  688,     0;
  1, 6, 30, 132,  519, 1762,  4404,      0;
  1, 7, 39, 190,  837, 3330, 11488,  29219,      0;
  1, 8, 49, 260, 1250, 5516, 22135,  77270, 199140,       0;
  1, 9, 60, 343, 1773, 8461, 37404, 151089, 532239, 1385904, 0;

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

Cf. A054727 (row sums), A108447, A118340.

T:= proc(n, k) option remember;
      if k<0 or  k>n then 0;
    elif k=0 then 1;
    elif k=n then 0;
    else T(n-1, k) -T(n-1, k-1) +T(n, k-1) +T(n+1, k-1);
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Mar 17 2021

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, 1, If[k==n, 0, T[n-1, k] -T[n-1, k-1] +T[n, k-1] +T[n+1, k-1] ]]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 17 2021 *)

{T(n,k)=polcoeff((serreverse(x*(1-x+sqrt((1-x)*(1-5*x)+x*O(x^k)))/2/(1-x))/x)^(n-k),k)}

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

Since g.f. G=G(x) of A108447 satisfies: G = 1 - x*G + x*G^2 + x*G^3 then T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n,k-1) + T(n+1,k-1). Also, a recurrence involving antidiagonals is: T(n,k) = T(n-1,k) + Sum_{j=1..k} [2*T(n-1+j,k-j) - T(n-2+j,k-j)] for n>k>=0.

Sum_{k=0..n} T(n,k) = [n=0] + A054727(n) = [n=0] + Sum_{j=1..n} binomial(n, j-1)*binomial(3*n-2*j-1, n-j)/(2*n-j). - G. C. Greubel, Mar 17 2021

A118344 Pendular Catalan triangle, 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), else T(n,k) = T(n,n-1-k) - T(n-1,k) - T(n-1,k+1), 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, 2, 1, 0, 1, 4, 5, 3, 1, 0, 1, 5, 9, 5, 4, 1, 0, 1, 6, 14, 14, 9, 5, 1, 0, 1, 7, 20, 28, 14, 14, 6, 1, 0, 1, 8, 27, 48, 42, 28, 20, 7, 1, 0, 1, 9, 35, 75, 90, 42, 48, 27, 8, 1, 0, 1, 10, 44, 110, 165, 132, 90, 75, 35, 9, 1, 0, 1, 11, 54, 154, 275, 297, 132, 165, 110, 44, 10, 1, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

See A118340 for definition of pendular triangles and pendular sums.

			Row 6 equals the pendular sums of row 5:
  [1,  4,  5,  3,  1,  0], where the sums proceed as follows:
  [1, __, __, __, __, __]: T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1]: T(6,5) = T(6,0) - T(5,5) = 1 - 0 = 1;
  [1,  5, __, __, __,  1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  4,  1]: T(6,4) = T(6,1) - T(5,4) - T(5,5) = 5-1-0 = 4;
  [1,  5,  9, __,  4,  1]: T(6,2) = T(6,4) + T(5,2) = 4 + 5 = 9;
  [1,  5,  9,  5,  4,  1]: T(6,3) = T(6,2) - T(5,3) - T(5,4) = 9-3-1 = 5;
  [1,  5,  9,  5,  4,  1,  0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  2,   1,   0;
  1,  4,  5,   3,   1,   0;
  1,  5,  9,   5,   4,   1,   0;
  1,  6, 14,  14,   9,   5,   1,   0;
  1,  7, 20,  28,  14,  14,   6,   1,   0;
  1,  8, 27,  48,  42,  28,  20,   7,   1,  0;
  1,  9, 35,  75,  90,  42,  48,  27,   8,  1,  0;
  1, 10, 44, 110, 165, 132,  90,  75,  35,  9,  1,  0;
  1, 11, 54, 154, 275, 297, 132, 165, 110, 44, 10,  1,  0;
Central terms are Catalan numbers T(2*n,n) = A000108(n);
semi-diagonals form successive self-convolutions of the central terms:
  T(2*n+1,n) = [A000108^2](n),
  T(2*n+2,n) = [A000108^3](n).

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

Cf. A000108, A033184, A118340, A026010 (row sums shift left).

T:= proc(n, k) option remember;
      if k<0 or k>n then 0;
    elif k=0 then 1;
    elif k=n then 0;
    elif n>2*k then T(n, n-k) +T(n-1, k);
    else T(n, n-k-1) -T(n-1, k) -T(n-1, k+1);
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Mar 17 2021

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, 1, If[k==n, 0, If[n>2*k, T[n, n-k] +T[n-1, k], T[n, n-k-1] -T[n-1, k] -T[n-1, k+1] ]]]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 17 2021 *)

T(n,k)=if(n2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)-T(n-1,k)-if(n-1>k,T(n-1,k+1)) ))))

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return 1
    elif (k==n): return 0
    elif (n>2*k): return T(n, n-k) +T(n-1, k)
    else: return T(n, n-k-1) -T(n-1, k) -T(n-1, k+1)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 17 2021

T(2*n+m, n) = [A000108^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A000108.

Sum_{k=0..n} T(n,k) = (1/2)*[n=0] + A026010(n-1) = (1/2)*[n=0] + (1/2)^((5 + (-1)^n)/2)*(6*n + 1 + 3*(-1)^n)*Catalan((2*n - 1 + (-1)^n)/4). - G. C. Greubel, Mar 17 2021

A168148 Row sums of triangle in A168030.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 3, 6, 6, 6, 4, 6, 7, 10, 6, 11, 12, 10, 6, 8, 8, 12, 8, 12, 11, 18, 12, 13, 16, 20, 11, 22, 22, 18, 12, 14, 14, 16, 10, 14, 12, 24, 16, 16, 18, 22, 12, 22, 23, 34, 20, 25, 28, 26, 17, 30, 26, 38, 24, 26, 31, 42, 22, 43, 44, 34, 22, 28, 26, 30, 20, 26

Offset: 0

Philippe Deléham, Nov 19 2009

nonn

G. C. Greubel, Table of n, a(n) for n = 0..5000

Cf. A118340, A168030.

t[n_, k_]:= t[n, k]= If[k<0 || k>n, 0, If[k==0, 1, If[n<=2*k, t[n, n-k -1] + t[n-1,k], t[n,n-k] + t[n-1,k]]]]; (* A118340 *)
Table[Sum[Mod[t[n, k], 2], {k,0,n}], {n,0,80}] (* G. C. Greubel, Jan 12 2023 *)

@CachedFunction
def t(n, k): # t = A118340
    if (k<0 or k>n): return 0
    elif (k==0): return 1
    elif (n>2*k): return t(n, n-k) + t(n-1, k)
    else: return t(n, n-k-1) + t(n-1, k)
def A168148(n): return sum( t(n,k)%2 for k in range(n+1))
[A168148(n) for n in range(81)] # G. C. Greubel, Jan 12 2023

From G. C. Greubel, Jan 12 2023: (Start)
a(n) = Sum_{k=0..n} A168030(n, k).
a(n) = Sum_{k=0..n} (A118340(n, k) mod 2). (End)

Terms a(16) onward added by G. C. Greubel, Jan 12 2023

cp's OEIS Frontend

A168030 Variant of pendular triangle A118340.

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A118345 Pendular triangle, 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), else T(n,k) = T(n,n-1-k) + 2*T(n-1,k), for n>=k>=0, with T(n,0) = 1 and T(n,n) = 0^n.

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A118350 Pendular triangle, 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), else T(n,k) = T(n,n-1-k) + 3*T(n-1,k), for n>=k>=0, with T(n,0)=1 and T(n,n)=0^n.

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A119369 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0)=T(n+1,2n)=1 and T(n+1,2n+1)=T(n+1,2n+2)=0.

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A122445 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + 2*T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0) = T(n+1,2n) = 1 and T(n+1,2n+1) = T(n+1,2n+2) = 0.

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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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