A127420 -id:A127420 - cp's OEIS frontend

A091352 Row sums of triangle A091351, in which the k-th column lists the row sums of the k-th power of A091351 (when considered as a lower triangular matrix).

1, 2, 4, 9, 24, 77, 295, 1329, 6934, 41351, 278680, 2101434, 17574552, 161740316, 1626733108, 17771416521, 209739328924, 2661301094008, 36148700652163, 523597247829867, 8059284921781892, 131408547139817541

Offset: 0

Paul D. Hanna, Jan 02 2004

nonn

Equals column 1 of table A125781. Equals row sums and column 0 (shifted) of triangle A127420. - Paul D. Hanna, Feb 11 2007

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

Cf. A091351, A091353, A091354.

Cf. A125781, A127420.

A127452 Triangle, read by rows of n(n+1)/2 + 1 terms, generated by the recurrence: start with a single '1' in row 0; row n+1 is generated from row n by first inserting zeros at positions {(m+1)(m+2)/2 - 1, m>=0} in row n and then taking the partial sums in reverse order.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 6, 6, 4, 4, 2, 1, 1, 24, 24, 18, 18, 12, 8, 8, 4, 2, 1, 1, 120, 120, 96, 96, 72, 54, 54, 36, 24, 16, 16, 8, 4, 2, 1, 1, 720, 720, 600, 600, 480, 384, 384, 288, 216, 162, 162, 108, 72, 48, 32, 32, 16, 8, 4, 2, 1, 1

Offset: 0

Paul D. Hanna, Jan 15 2007

The first column equals the factorials. Triangle A127420 is generated by a similar recurrence.

			The triangle begins:
1;
1, 1;
2, 2, 1, 1;
6, 6, 4, 4, 2, 1, 1;
24, 24, 18, 18, 12, 8, 8, 4, 2, 1, 1;
120, 120, 96, 96, 72, 54, 54, 36, 24, 16, 16, 8, 4, 2, 1, 1;
720, 720, 600, 600, 480, 384, 384, 288, 216, 162, 162, 108, 72, 48, 32, 32, 16, 8, 4, 2, 1, 1;
...
The recurrence is illustrated by the following examples.
Start with a single '1' in row 0.
To get row 1, insert 0 in row 0 at position 0,
and take partial sums in reverse order:
0,_1;
1,_1;
To get row 2, insert 0 in row 1 at positions [0,2],
and take partial sums in reverse order:
0,_1,_0,_1;
2,_2,_1,_1;
To get row 3, insert 0 in row 2 at positions [0,2,5],
and take partial sums in reverse order:
0,_2,_0,_2,_1,_0,_1;
6,_6,_4,_4,_2,_1,_1;
To get row 4, insert 0 in row 3 at positions [0,2,5,9],
and take partial sums in reverse order:
_0,__6,__0,__6,__4,_0,_4,_2,_1,_0,_1;
24,_24,_18,_18,_12,_8,_8,_4,_2,_1,_1;
etc.
Continuing in this way generates the factorials in the first column.

Paul D. Hanna, Rows n = 0..30, flattened.

Cf. A018927, A127420, A047969, A182961 (variant).

T(n,k)=if(n<0 || k<0,0,if(n==0 && k==0,1, if(k==0, n!, if(issquare(8*k+1),T(n,k-1),T(n,k-1)-T(n-1,k-(sqrtint(8*k+1)+1)\2)))))

T(n,k)=local(t=(sqrtint(8*k+1)-1)\2);(n-t)!*(n-t)^(k-t*(t+1)/2)*(n-t+1)^(t-k+t*(t+1)/2)

Sum_{k=0..n*(n+1)/2} k*T(n,k) = A018927(n+1) = Sum_{k=0..n} k*k!*{(k+1)^(n-k+1)-k^(n-k+1)}.

T(n,k) = (n-t)! * (n-t)^(k - t*(t+1)/2) * (n-t+1)^(t-k + t*(t+1)/2) where t=floor((sqrt(8*k+1)-1)/2). Also, Sum_{j=k*(k+1)/2..(k+1)*(k+2)/2-1} T(n,j) = A047969(n-k,k) = (n-k)!*((n-k+1)^(k+1)-(n-k)^(k+1)).

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

A091352 Row sums of triangle A091351, in which the k-th column lists the row sums of the k-th power of A091351 (when considered as a lower triangular matrix).

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A127452 Triangle, read by rows of n(n+1)/2 + 1 terms, generated by the recurrence: start with a single '1' in row 0; row n+1 is generated from row n by first inserting zeros at positions {(m+1)(m+2)/2 - 1, m>=0} in row n and then taking the partial sums in reverse order.

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PARI

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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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Author

Keywords

Examples

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PARI

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cp's OEIS Frontend

A091352 Row sums of triangle A091351, in which the k-th column lists the row sums of the k-th power of A091351 (when considered as a lower triangular matrix).

Views

Author

Keywords

Comments

Links

Crossrefs

A127452 Triangle, read by rows of n*(n+1)/2 + 1 terms, generated by the recurrence: start with a single '1' in row 0; row n+1 is generated from row n by first inserting zeros at positions {(m+1)*(m+2)/2 - 1, m>=0} in row n and then taking the partial sums in reverse order.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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.

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula

A127452 Triangle, read by rows of n(n+1)/2 + 1 terms, generated by the recurrence: start with a single '1' in row 0; row n+1 is generated from row n by first inserting zeros at positions {(m+1)(m+2)/2 - 1, m>=0} in row n and then taking the partial sums in reverse order.

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.