A127716 -id:A127716 - cp's OEIS frontend

A127715 Rightmost border of triangle A127714: a(n) = A127714(n, (n+1)*(n+2)/2 - 1).

1, 1, 3, 14, 86, 645, 5662, 56632, 633545, 7820115, 105401961, 1538517351, 24157841308, 405778641302, 7256711524298, 137607733327779, 2757147167433326, 58188668578629283, 1289965088962987961

Offset: 0

Paul D. Hanna, Jan 24 2007

nonn

			a(3) = 14 = the upper left term of M^3.

Paul D. Hanna, Table of n, a(n), n = 0..100.

Cf. A127714, A127716, A033184

From Gary W. Adamson, Aug 02 2011: (Start)
a(n) = upper left term of M^n, M is an infinite square production matrix in which nonzero terms = A033184 with the first "1" deleted, as follows: M =
   1,  1,  0,  0, 0, 0, ...
   2,  2,  1,  0, 0, 0, ...
   5,  5,  3,  1, 0, 0, ...
  14, 14,  9,  4, 1, 0, ...
  42, 42, 28, 14, 5, 1, ...
  ... (End)

A127714 Triangle, read by rows of (n+1)(n+2)/2 terms, generated by the following rule. Start with a single '1' in row n=0; from then on, obtain row n from row n-1 by inserting zeros in row n-1 at positions: {(j+1)n - j*(j-1)/2 | j=0..n} and then take partial sums.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 1, 3, 5, 5, 8, 11, 11, 14, 14, 14, 1, 4, 9, 14, 14, 22, 33, 44, 44, 58, 72, 72, 86, 86, 86, 1, 5, 14, 28, 42, 42, 64, 97, 141, 185, 185, 243, 315, 387, 387, 473, 559, 559, 645, 645, 645, 1, 6, 20, 48, 90, 132, 132, 196, 293, 434, 619, 804, 804

Offset: 0

Paul D. Hanna, Jan 24 2007

			Triangle begins:
  1;
  1, 1, 1;
  1, 2, 2, 3, 3, 3;
  1, 3, 5, 5, 8, 11, 11, 14, 14, 14;
  1, 4, 9, 14, 14, 22, 33, 44, 44, 58, 72, 72, 86, 86, 86;
  1, 5, 14, 28, 42, 42, 64, 97, 141, 185, 185, 243, 315, 387, 387, 473, 559, 559, 645, 645, 645;
  1, 6, 20, 48, 90, 132, 132, 196, 293, 434, 619, 804, 804, 1047, 1362, 1749, 2136, 2136, 2609, 3168, 3727, 3727, 4372, 5017, 5017, 5662, 5662, 5662;
  ...
Obtain row n from row n-1 by inserting zeros in row n-1 at positions:
[n,2*n,3*n-1,4*n-3,5*n-6,6*n-10,...,(j+1)*n - j*(j-1)/2,... | j=0..n],
and then take partial sums; illustrated by the following examples.
Obtain row 3 from row 2 by inserting zeros at positions [3,6,8,9],
and then take partial sums:
[1, 2, 2, 0, 3, 3, 0, 3, 0, 0];
[1, 3, 5, 5, 8,11,11,14,14,14];
Obtain row 4 from row 3 by inserting zeros at positions [4,8,11,13,14],
and then take partial sums:
[1, 3, 5, _5, _0, _8, 11, 11, _0, 14, 14, _0, 14, _0, _0];
[1, 4, 9, 14, 14, 22, 33, 44, 44, 58, 72, 72, 86, 86, 86].

Cf. A127715, A127716; A127632, A000108, A009766; variants: A127452, A127420.

{T(n,k)=local(t);if(n<0 || k<0 || k>(n+1)*(n+2)/2-1,0, t=(sqrtint((2*n+3)^2-8*(k+1))-1)\2; if(k==0,1,if(issquare((2*n+3)^2-8*(k+1)),T(n,k-1),T(n,k-1)+T(n-1,k-n+t))))} {/* for(n=0,8,for(k=0,(n+1)*(n+2)/2-1,print1(T(n,k),","));print("")) */}

T(n,n) = A000108(n);
A009766 (Catalan's triangle) forms lower left sub-triangle;
T(n+1,2*n+1) = A127632(n), where g.f. of A127632 is: 2/(1+sqrt(2*sqrt(1-4*x)-1)).
T(n,n*(n+1)/2) = A127716(n).
T(n,(n+1)*(n+2)/2-1) = A127715(n).

cp's OEIS Frontend

A127715 Rightmost border of triangle A127714: a(n) = A127714(n, (n+1)*(n+2)/2 - 1).

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Formula

A127714 Triangle, read by rows of (n+1)(n+2)/2 terms, generated by the following rule. Start with a single '1' in row n=0; from then on, obtain row n from row n-1 by inserting zeros in row n-1 at positions: {(j+1)n - j*(j-1)/2 | j=0..n} and then take partial sums.

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PARI

Formula

cp's OEIS Frontend

A127715 Rightmost border of triangle A127714: a(n) = A127714(n, (n+1)*(n+2)/2 - 1).

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

A127714 Triangle, read by rows of (n+1)*(n+2)/2 terms, generated by the following rule. Start with a single '1' in row n=0; from then on, obtain row n from row n-1 by inserting zeros in row n-1 at positions: {(j+1)*n - j*(j-1)/2 | j=0..n} and then take partial sums.

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula

A127714 Triangle, read by rows of (n+1)(n+2)/2 terms, generated by the following rule. Start with a single '1' in row n=0; from then on, obtain row n from row n-1 by inserting zeros in row n-1 at positions: {(j+1)n - j*(j-1)/2 | j=0..n} and then take partial sums.