A118345 -id:A118345 - cp's OEIS frontend

A118346 Central terms of pendular triangle A118345.

1, 1, 5, 30, 201, 1445, 10900, 85128, 682505, 5585115, 46461437, 391743850, 3340361700, 28755475180, 249572076200, 2181469638880, 19186562661273, 169677521094215, 1507881643936015, 13458730170115778, 120599648894147185

Offset: 0

Paul D. Hanna, Apr 26 2006

nonn

Also, g.f. A(x) = (1/x)*series_reversion of x/(1 + x*GF(A005572)), where GF(A005572) is the g.f. of A005572, which is the number of walks on cubic lattice starting and finishing on the xy plane and never going below it.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A118345, A118347, A118348, A118349.

R:=PowerSeriesRing(Rationals(), 30);
[1] cat Coefficients(R!( Reversion( x/((1+x)*(1+4*x+x^2)) ) )); // G. C. Greubel, Mar 17 2021

CoefficientList[1 +InverseSeries[Series[x/((1+x)*(1+4*x+x^2)), {x,0,30}]], x] (* G. C. Greubel, Mar 17 2021 *)

{a(n) = polcoeff(serreverse( x*(1-2*x+sqrt((1-2*x)*(1-6*x)+x*O(x^n)))/(2*(1-2*x)))/x, n)}

def A118346_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( ( x/((1+x)*(1+4*x+x^2)) ).reverse() ).list()
a=A118346_list(31); [1]+a[1:] # G. C. Greubel, Mar 17 2021

G.f.: A=A(x) satisfies A = 1 - 2*x*A + 2*x*A^2 + x*A^3.
G.f.: A(x) = 1 + series_reversion( x/((1+x)*(1+4*x+x^2)) ).
G.f.: A(x) = (1/x)*series_reversion( x*(1-2*x + sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) ).
For n>0: a(n) = (1/n)*Sum_{j=0..n} Sum_{i=0..n-1} ( binomial(n,j) * binomial(j,i) * binomial(n-j,2*j-n-i-1) * 5^(2*n-3*j+2*i+1) ). -Vladimir Kruchinin, Dec 26 2010

A118347 Semi-diagonal (one row below central terms) of pendular triangle A118345 and equal to the self-convolution of the central terms (A118346).

Original entry on oeis.org

1, 2, 11, 70, 487, 3592, 27600, 218566, 1771367, 14621410, 122495659, 1038934480, 8903129300, 76970244560, 670507216168, 5879770542870, 51861650744071, 459804626981158, 4095433894576785, 36628711884398086, 328824295880947471

Offset: 0

Paul D. Hanna, Apr 26 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A118345, A118346, A118348, A118349.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (Reversion( x*(1-2*x +Sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) )/x)^2 )); // G. C. Greubel, Mar 17 2021

CoefficientList[(InverseSeries[Series[x*(1-2*x +Sqrt[(1-2*x)*(1-6*x)])/(2*(1-2*x)), {x, 0, 30}]]/x)^2, x] (* G. C. Greubel, Mar 17 2021 *)

{a(n) = polcoeff( (serreverse(x*(1-2*x+sqrt((1-2*x)*(1-6*x)+x*O(x^n)))/(2*(1-2*x)))/x)^2,n)}

def A118347_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (( x*(1-2*x +sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) ).reverse()/x)^2 ).list()
A118347_list(31) # G. C. Greubel, Mar 17 2021

G.f.: ( series_inverse( x*(1-2*x +sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) )/x )^2.

A118348 Semi-diagonal (two rows below central terms) of pendular triangle A118345 and equal to the self-convolution cube of the central terms (A118346).

Original entry on oeis.org

1, 3, 18, 121, 873, 6606, 51728, 415629, 3407391, 28388847, 239675406, 2045980440, 17629939980, 153142537440, 1339599358944, 11789960853293, 104327344928619, 927627432162129, 8283625668834238, 74259685465582569, 668054892245119353

Offset: 0

Paul D. Hanna, Apr 26 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A118345, A118346, A118347, A118349.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (Reversion( x*(1-2*x +Sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) )/x)^3 )); // G. C. Greubel, Mar 17 2021

CoefficientList[(InverseSeries[Series[x*(1-2*x +Sqrt[(1-2*x)*(1-6*x)])/(2*(1-2*x)), {x, 0, 30}]]/x)^3, x] (* G. C. Greubel, Mar 17 2021 *)

{a(n) = polcoeff( (serreverse(x*(1-2*x+sqrt((1-2*x)*(1-6*x)+x*O(x^n)))/(2*(1-2*x)))/x)^3, n)}

def A118347_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (( x*(1-2*x +sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) ).reverse()/x)^3 ).list()
A118347_list(31) # G. C. Greubel, Mar 17 2021

G.f.: ( series_inverse( x*(1-2*x +sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) )/x )^3.

A118340 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) + 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, 4, 1, 0, 1, 4, 9, 5, 1, 0, 1, 5, 15, 20, 6, 1, 0, 1, 6, 22, 48, 28, 7, 1, 0, 1, 7, 30, 85, 113, 37, 8, 1, 0, 1, 8, 39, 132, 282, 169, 47, 9, 1, 0, 1, 9, 49, 190, 519, 688, 237, 58, 10, 1, 0, 1, 10, 60, 260, 837, 1762, 1074, 318, 70, 11, 1, 0

Offset: 0

Paul D. Hanna, Apr 25 2006

Definitions. A pendular triangle is a triangle in which row n is generated from the pendular sums of row n-1. Pendular sums of a row are partial sums taken in back-and-forth order, starting with the leftmost term, jumping to the rightmost term, back to the leftmost unused term, then forward to the rightmost unused term, etc.
In each pass, the partial sum is placed in the new row directly under the term most recently used in the sum. Continue in this way until all the terms of the prior row have been used and then complete the new row by appending a zero at the end. Pendular sums are so named because the process resembles a swinging pendulum that slows down on each pass until it eventually comes to rest in the center.
In the simplest case, pendular triangles obey 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) + p*T(n-1,k), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n, for some fixed number p.
In which case the g.f. G=G(x) of the central terms satisfies: G = 1 - p*x*G + p*x*G^2 + x*G^3.
More generally, a pendular triangle is defined 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) + Sum_{j>=1} p(j)*T(n-1,k-1+j), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n.
Remarkably, the g.f. G=G(x) of the central terms satisfies: G = 1 + x*G^3 + Sum_{j>=1} p(j)*x^j*[G^(2*j) - G^(2*j-1)].
Further, the g.f. of the m-th lower semi-diagonal equals G(x)^(m+1) for m>=0, where the m-th semi-diagonal consists of those terms located at m rows below the central terms.
For variants of pendular triangles, the main diagonal may be nonzero, but then the g.f.s of the semi-diagonals are more complex.

			Row 6 equals the pendular sums of row 5:
  [1,  4,  9,  5,  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, __, __,  6,  1], T(6,4) = T(6,1) + T(5,4) = 5 + 1 = 6;
  [1,  5, 15, __,  6,  1], T(6,2) = T(6,4) + T(5,2) = 6 + 9 = 15;
  [1,  5, 15, 20,  6,  1], T(6,3) = T(6,2) + T(5,3) = 15 + 5 = 20;
  [1,  5, 15, 20,  6,  1, 0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  4,   1,    0;
  1,  4,  9,   5,    1,    0;
  1,  5, 15,  20,    6,    1,    0;
  1,  6, 22,  48,   28,    7,    1,    0;
  1,  7, 30,  85,  113,   37,    8,    1,   0;
  1,  8, 39, 132,  282,  169,   47,    9,   1,  0;
  1,  9, 49, 190,  519,  688,  237,   58,  10,  1,  0;
  1, 10, 60, 260,  837, 1762, 1074,  318,  70, 11,  1, 0;
  1, 11, 72, 343, 1250, 3330, 4404, 1568, 413, 83, 12, 1, 0; ...
Central terms are T(2*n,n) = A108447(n);
semi-diagonals form successive self-convolutions of the central terms:
T(2*n+1,n) = A118341(n) = [A108447^2](n),
T(2*n+2,n) = A118342(n) = [A108447^3](n).

G. C. Greubel, Rows n = 0..100 of the triangle, flattened(Rows n = 0..20 from Paul D. Hanna)

Cf. A108447 (central terms), A118341, A118343.
variants: A118344 (Catalan), A118362 (ternary), A118350, A118355.
Cf. A167763 (p=0), this sequence (p=1), A118345 (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,1): 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-1,k) + T(n,n-k), T(n-1,k) + T(n,n-1-k)))))}
for(n=0,12, for(k=0,n, print1(T(n,k),", "));print(""))

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

T(2*n+m,n) = [A108447^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A108447; compare semi-diagonals to the diagonals of convolution triangle A118343.

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

A118349 Convolution triangle, read by rows, where diagonals are successive self-convolutions of A118346.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 5, 0, 1, 3, 11, 30, 0, 1, 4, 18, 70, 201, 0, 1, 5, 26, 121, 487, 1445, 0, 1, 6, 35, 184, 873, 3592, 10900, 0, 1, 7, 45, 260, 1375, 6606, 27600, 85128, 0, 1, 8, 56, 350, 2010, 10672, 51728, 218566, 682505, 0, 1, 9, 68, 455, 2796, 15996, 85182, 415629, 1771367, 5585115, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

A118346 equals the central terms of pendular triangle A118345 and the diagonals of this triangle form the semi-diagonals of the triangle A118345.

			Show: T(n,k) = T(n-1,k) - 2*T(n-1,k-1) + 2*T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - 2*T(7,3) + 2*T(8,3) + T(9,3)
or: 1375 = 873 - 2*184 + 2*260 + 350.
Triangle begins:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  5,   0;
  1, 3, 11,  30,    0;
  1, 4, 18,  70,  201,     0;
  1, 5, 26, 121,  487,  1445,     0;
  1, 6, 35, 184,  873,  3592, 10900,      0;
  1, 7, 45, 260, 1375,  6606, 27600,  85128,       0;
  1, 8, 56, 350, 2010, 10672, 51728, 218566,  682505,       0;
  1, 9, 68, 455, 2796, 15996, 85182, 415629, 1771367, 5585115, 0;

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

Cf. A118346, A118345.

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) -2*T(n-1, k-1) +2*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] -2*T[n-1, k-1] +2*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*(1-2*x+sqrt((1-2*x)*(1-6*x)+x*O(x^k)))/2/(1-2*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) -2*T(n-1, k-1) +2*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 A118346 satisfies: G = 1 - 2*x*G + 2*x*G^2 + x*G^3 then T(n,k) = T(n-1,k) - 2*T(n-1,k-1) + 2*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} [3*T(n-1+j,k-j) - 2*T(n-2+j,k-j)] for n>k>=0.

cp's OEIS Frontend

A118346 Central terms of pendular triangle A118345.

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A118347 Semi-diagonal (one row below central terms) of pendular triangle A118345 and equal to the self-convolution of the central terms (A118346).

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A118348 Semi-diagonal (two rows below central terms) of pendular triangle A118345 and equal to the self-convolution cube of the central terms (A118346).

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A118340 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) + 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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