A111373 A generalized Pascal triangle.

1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 3, 0, 0, 1, 3, 0, 0, 4, 0, 0, 1, 0, 7, 0, 0, 5, 0, 0, 1, 0, 0, 12, 0, 0, 6, 0, 0, 1, 12, 0, 0, 18, 0, 0, 7, 0, 0, 1, 0, 30, 0, 0, 25, 0, 0, 8, 0, 0, 1, 0, 0, 55, 0, 0, 33, 0, 0, 9, 0, 0, 1, 55, 0, 0, 88, 0, 0, 42, 0, 0, 10, 0, 0, 1, 0, 143, 0, 0, 130, 0, 0, 52, 0, 0, 11, 0, 0, 1

Offset

0,12

Comments

First diagonal is A000012, the all 1's sequence. Second nonzero diagonal is A000027 = n. Third nonzero diagonal is A027379 = n*(n+5)/2 for n>=1, or essentially A000217(n) - 3. Fourth nonzero diagonal is A111396. - Jonathan Vos Post, Nov 10 2005

Row sums are A126042. - Paul Barry, Dec 16 2006

Links

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

I. Bajunaid et al., Function series, Catalan numbers and random walks on trees, Amer. Math. Monthly 112 (2005), 765-785.

Emeric Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.

Formula

Each term is the sum of the two terms above it to the left and two steps to the right.

From Paul Barry, Dec 16 2006: (Start)

Riordan array (g(x^3),x*g(x^3)) where g(x)=(2/sqrt(3x))*sin(asin(sqrt(27x/4))/3), the g.f. of A001764;

Number triangle T(n,k) = C(3*floor((n+2k)/3)-2k,floor((n+2k)/3)-k)*(k+1)/(2*floor((n+2k)/3)-k+ 1)(2*cos(2*pi*(n-k)/3)+1)/3. (End)

Inverse of Riordan array (1/(1+x^3), x/(1+x^3)), A126030. - Paul Barry, Dec 16 2006

G.f. (x*A(x))^k=sum{n>=k, T(n,k)*x^n}, where A(x)=1+x^3*A(x)^3. - Vladimir Kruchinin, Feb 18 2011

From G. C. Greubel, Jul 30 2022: (Start)

T(n, k) = (3/(n-k))*binomial(n, (n-k)/3 -1)*(k+1) for ( (n-k) mod 3 ) = 0, otherwise 0, with T(n, n) = 1.

T(n, n-3) = A000027(n-2), n >= 3.

T(n, n-6) = A027379(n-5), n >= 6.

T(n, n-9) = A111396(n-8), n >= 9.

T(n, n-12) = A167543(n+5), n >= 12.

Sum_{k=0..n} T(n, k) = A126042(n). (End)

Example

Triangle begins:
   1;
   0,  1;
   0,  0,  1;
   1,  0,  0,  1;
   0,  2,  0,  0,  1;
   0,  0,  3,  0,  0,  1;
   3,  0,  0,  4,  0,  0,  1;
   0,  7,  0,  0,  5,  0,  0,  1;
   0,  0, 12,  0,  0,  6,  0,  0,  1;
  12,  0,  0, 18,  0,  0,  7,  0,  0,  1;
   0, 30,  0,  0, 25,  0,  0,  8,  0,  0,  1;
   0,  0, 55,  0,  0, 33,  0,  0,  9,  0,  0,  1;
  55,  0,  0, 88,  0,  0, 42,  0,  0, 10,  0,  0,  1;
Production matrix is
  0, 1;
  0, 0, 1;
  1, 0, 0, 1;
  0, 1, 0, 0, 1;
  0, 0, 1, 0, 0, 1;
  0, 0, 0, 1, 0, 0, 1;
  0, 0, 0, 0, 1, 0, 0, 1;
  0, 0, 0, 0, 0, 1, 0, 0, 1;
  0, 0, 0, 0, 0, 0, 1, 0, 0, 1;

Mathematica

T[n_, k_]= If[k==n, 1, If[Mod[n-k, 3]==0, (3/(n-k))*Binomial[n, (n-k)/3 -1]*(k + 1), 0]];
Table[T[n, k], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Jul 30 2022 *)

Prog

(Magma)
function A111373(n, k)
  if k eq n then return 1;
  elif ((n-k) mod 3) eq 0 then return (3/(n-k))*Binomial(n, Floor((n-k-3)/3))*(k+1);
  else return 0;
  end if; return A111373;
end function;
[A111373(n, k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 30 2022
(SageMath)
def A111373(n, k):
    if(k==n): return 1
    elif ((n-k)%3==0): return (3/(n-k))*binomial(n, (n-k)/3 -1)*(k+1)
    else: return 0
flatten([[A111373(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 30 2022

Crossrefs

First column is A001764. Bears same relation to A001764 as A053121 does to A000108.

Cf. A000012, A000027, A027379, A111396, A126042 (row sums), A167543.

Sequence in context: A212193 A253189 A126030 * A336757 A116376 A165766

Adjacent sequences: A111370 A111371 A111372 * A111374 A111375 A111376

Keyword

nonn,easy,tabl

Author

N. J. A. Sloane, Nov 09 2005

Extensions

More terms from Kerri Sullivan (ksulliva(AT)ashland.edu), Jan 23 2006

Status

approved