cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

Paul D. Hanna, Apr 26 2006

Keywords

Comments

See definition of pendular triangle and pendular sums at A118340.

Examples

			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).

Links

Crossrefs

Cf. A118351, A118352, A118353, A118354.
Cf. A167763 (p=0), A118340 (p=1), A118345 (p=2), this sequence (p=3).

Programs

  • Magma
    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
  • Mathematica
    T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 17 2021 *)
  • PARI
    T(n,k)=if(n2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)+3*T(n-1,k)))))
  • Sage
    @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

Formula

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.

A118354 Convolution triangle, read by rows, where diagonals are successive self-convolutions of A118351.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 6, 0, 1, 3, 13, 42, 0, 1, 4, 21, 96, 325, 0, 1, 5, 30, 163, 770, 2688, 0, 1, 6, 40, 244, 1353, 6530, 23286, 0, 1, 7, 51, 340, 2093, 11760, 57612, 208659, 0, 1, 8, 63, 452, 3010, 18636, 105681, 523446, 1918314, 0, 1, 9, 76, 581, 4125, 27441, 170580, 973953, 4864795, 17994264, 0
Offset: 0

Views

Author

Paul D. Hanna, Apr 26 2006

Keywords

Comments

A118351 equals the central terms of pendular triangle A118350 and the lower diagonals of this triangle form the semi-diagonals of the triangle A118350.

Examples

			Show: T(n,k) = T(n-1,k) - 3*T(n-1,k-1) + 3*T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - 3*T(7,3) + 3*T(8,3) + T(9,3)
or: 2093 = 1353 - 3*244 + 3*340 + 452.
Triangle begins:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  6,   0;
  1, 3, 13,  42,    0;
  1, 4, 21,  96,  325,     0;
  1, 5, 30, 163,  770,  2688,      0;
  1, 6, 40, 244, 1353,  6530,  23286,      0;
  1, 7, 51, 340, 2093, 11760,  57612, 208659,       0;
  1, 8, 63, 452, 3010, 18636, 105681, 523446, 1918314,        0;
  1, 9, 76, 581, 4125, 27441, 170580, 973953, 4864795, 17994264, 0; ...

Links

Crossrefs

Cf. A118350, A118351, A118352, A118353.
Row sums: A151616.

Programs

  • Mathematica
    T[n_, k_]:= T[n, k]= If[k==0, 1, If[k==n, 0, T[n-1, k] -3*T[n-1, k-1] +3*T[n, k-1] +T[n+1, k-1]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2021 *)
  • PARI
    {T(n,k)=polcoeff((serreverse(x*(1-3*x+sqrt((1-3*x)*(1-7*x)+x*O(x^k)))/2/(1-3*x))/x)^(n-k),k)}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
  • Sage
    @CachedFunction
def T(n, k):
    if (k==0): return 1
    elif (k==n): return 0
    else: return T(n-1, k) - 3*T(n-1, k-1) + 3*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, Feb 18 2021

Formula

Since g.f. G=G(x) of A118351 satisfies: G = 1 - 3*x*G + 3*x*G^2 + x*G^3 then
T(n,k) = T(n-1,k) - 3*T(n-1,k-1) + 3*T(n,k-1) + T(n+1,k-1).
Recurrence involving antidiagonals:
T(n,k) = T(n-1,k) + Sum_{j=1..k} [4*T(n-1+j,k-j) - 3*T(n-2+j,k-j)] for n>k>=0.

A118351 Central terms of pendular triangle A118350.

Original entry on oeis.org

1, 1, 6, 42, 325, 2688, 23286, 208659, 1918314, 17994264, 171542460, 1657212768, 16188521454, 159634359415, 1586932321578, 15886925400954, 160026976985205, 1620715748715648, 16493797802077032, 168583560794745684
Offset: 0

Views

Author

Paul D. Hanna, Apr 26 2006

Keywords

Links

Crossrefs

Cf. A118350, A118352, A118353, A118354.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30);
[1] cat Coefficients(R!( Reversion( x/((1+x)*(1+5*x+x^2)) ) )); // G. C. Greubel, Feb 18 2021
  • Mathematica
    T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 18 2021 *)
Join[{1}, Rest@CoefficientList[InverseSeries[Series[ x/((1+x)*(1+5*x+x^2)), {x,0,30}]], x]] (* G. C. Greubel, Feb 18 2021 *)
  • PARI
    {a(n)=polcoeff((serreverse(x*(1-3*x+sqrt((1-3*x)*(1-7*x)+x*O(x^n)))/2/(1-3*x))/x),n)}
for(n=0,30,print1(a(n),", "))
  • PARI
    {a(n)=polcoeff(1 + serreverse( x/((1+x)*(1+5*x+x^2 +x*O(x^n)))),n)}
for(n=0,30,print1(a(n),", "))
  • Sage
    def S_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (x/((1+x)*(1+5*x+x^2))).reverse() ).list()
a=S_list(31); [1]+a[1:] # G. C. Greubel, Feb 18 2021

Formula

G.f. A=A(x) satisfies: A = 1 - 3*x*A + 3*x*A^2 + x*A^3.
G.f.: 1 + Series_Reversion( x/((1+x)*(1+5*x+x^2)) ).
G.f.: (1/x)*Series_Reversion( x*(1-3*x+sqrt((1-3*x)*(1-7*x)))/2/(1-3*x) ).
For n>0: a(n) = 1/n*sum(j=0..n, C(n,j) *sum(i=0..(n-1), C(j,i)*C(n-j,2*j-n-i-1) *6^(2*n-3*j+2*i+1))). - Vladimir Kruchinin, Dec 26 2010
a(n) ~ s^(3/2) / (3*sqrt(2*Pi*(1 + 3*s + 3*s^2)) * n^(3/2) * r^(n+1)), where s = 2*sin(Pi/6 + arctan(sqrt(7)/3)/3) - 1, r = 2*s/(9 - 12*sin(Pi/6 - 2*arctan(sqrt(7)/3)/3)). - Vaclav Kotesovec, Feb 18 2021

A118353 Semi-diagonal (two rows below central terms) of pendular triangle A118350 and equal to the self-convolution cube of the central terms (A118351).

Original entry on oeis.org

1, 3, 21, 163, 1353, 11760, 105681, 973953, 9154821, 87428388, 845894700, 8273978100, 81682757317, 812829371205, 8144563709391, 82104333340467, 832125695906313, 8473862660311392, 86661931504395228, 889705959333345756
Offset: 0

Views

Author

Paul D. Hanna, Apr 26 2006

Keywords

Links

Crossrefs

Cf. A118350, A118351, A118352, A118354.

Programs

  • Mathematica
    T[n_, k_]:= T[n, k]= If[k==0, 1, If[k==n, 0, T[n-1, k] - 3*T[n-1, k-1] + 3*T[n, k-1] + T[n+1, k-1] ]];
Table[T[n, n-3], {n, 3, 30}] (* G. C. Greubel, Feb 18 2021 *)
  • PARI
    my(x='x+O('x^33)); Vec((serreverse(x*(1-3*x+sqrt((1-3*x)*(1-7*x)))/2/(1-3*x))/x)^3)
  • Sage
    @CachedFunction
def T(n, k):
    if (k<0 or nG. C. Greubel, Feb 18 2021
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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