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 1-7 of 7 results.

A183202 Triangle, read by rows, where T(n,k) equals the sum of (n-k) terms in row n of triangle A131338 starting at position nk - k(k-1)/2, with the main diagonal formed from the row sums.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 3, 3, 5, 4, 6, 10, 9, 14, 5, 10, 22, 34, 29, 43, 6, 15, 40, 84, 122, 100, 143, 7, 21, 65, 169, 334, 463, 367, 510, 8, 28, 98, 300, 738, 1390, 1851, 1426, 1936, 9, 36, 140, 489, 1426, 3345, 6043, 7767, 5839, 7775, 10, 45, 192, 749, 2510, 6990, 15735
Offset: 0

Views

Author

Paul D. Hanna, Dec 30 2010

Keywords

Examples

			Triangle begins:
1;
1,1;
2,1,2;
3,3,3,5;
4,6,10,9,14;
5,10,22,34,29,43;
6,15,40,84,122,100,143;
7,21,65,169,334,463,367,510;
8,28,98,300,738,1390,1851,1426,1936;
9,36,140,489,1426,3345,6043,7767,5839,7775;
10,45,192,749,2510,6990,15735,27374,34097,25094,32869; ...
The rows are derived from triangle A131338 by summing terms in the following manner:
(1);
(1),(1);
(1+1),(1),(2);
(1+1+1),(1+2),(3),(5);
(1+1+1+1),(1+2+3),(4+6),(9),(14);
(1+1+1+1+1),(1+2+3+4),(5+7+10),(14+20),(29),(43);
(1+1+1+1+1+1),(1+2+3+4+5),(6+8+11+15),(20+27+37),(51+71),(100),(143); ...
where row n of triangle A131338 consists of n '1's followed by the partial sums of the prior row.
		

Crossrefs

Cf. A131338, A098568, A098569 (row sums), A183203 (antidiagonal sums).

Programs

  • PARI
    {A131338(n, k)=if(k>n*(n+1)/2||k<0,0,if(k<=n,1,sum(i=0, k-n,A131338(n-1,i))))}
    {T(n,k)=if(n==k,A131338(n,n*(n+1)/2),sum(j=n*k-k*(k-1)/2,n*k-k*(k-1)/2+n-k-1,A131338(n,j)))}

A098568 Triangle of triangular binomial coefficients, read by rows, where column k has the g.f.: 1/(1-x)^((k+1)*(k+2)/2) for k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 21, 10, 1, 1, 15, 56, 55, 15, 1, 1, 21, 126, 220, 120, 21, 1, 1, 28, 252, 715, 680, 231, 28, 1, 1, 36, 462, 2002, 3060, 1771, 406, 36, 1, 1, 45, 792, 5005, 11628, 10626, 4060, 666, 45, 1, 1, 55, 1287, 11440, 38760, 53130, 31465, 8436
Offset: 0

Views

Author

Paul D. Hanna, Sep 15 2004

Keywords

Comments

The row sums form A098569: {1,2,5,14,43,143,510,1936,7775,32869,...}. How do the terms of row k tend to be distributed as k grows?
Remarkably, column k of the matrix inverse (A121434) equals signed column k of the triangular matrix power: A107876^(k*(k+1)/2) for k >= 0. - Paul D. Hanna, Aug 25 2006
Surprisingly, the row sums (A098569) equal the row sums of triangle A131338. - Paul D. Hanna, Aug 30 2007
Number of sequences S = s(1)s(2)...s(n) such that S contains m 0's, for 1<=j<=n, s(j)Frank Ruskey, Apr 15 2011
As a rectangular array read by antidiagonals R(n,k) (n>=2, k>=0) is the number of labeled graphs on n nodes that have exactly k arcs where multiple arcs are allowed to connect distinct vertex pairs. R(n,k) = C(C(n,2)+k-1,k). See example below. - Geoffrey Critzer, Nov 12 2011

Examples

			G.f.s of columns: 1/(1-x), 1/(1-x)^3, 1/(1-x)^6, 1/(1-x)^10, 1/(1-x)^15, ...
Rows begin:
  1;
  1,  1;
  1,  3,    1;
  1,  6,    6,     1;
  1, 10,   21,    10,      1;
  1, 15,   56,    55,     15,      1;
  1, 21,  126,   220,    120,     21,      1;
  1, 28,  252,   715,    680,    231,     28,     1;
  1, 36,  462,  2002,   3060,   1771,    406,    36,     1;
  1, 45,  792,  5005,  11628,  10626,   4060,   666,    45,    1;
  1, 55, 1287, 11440,  38760,  53130,  31465,  8436,  1035,   55,  1;
  1, 66, 2002, 24310, 116280, 230230, 201376, 82251, 16215, 1540, 66, 1; ...
From _Frank Ruskey_, Apr 15 2011: (Start)
In reference to comment about s(1)s(2)...s(n) above,
   a(4,2) = 6 = |{0012, 0013, 0023, 0101, 0103, 0120}|  and
   a(4,3) = 6 = |{0001, 0002, 0003, 0010, 0020, 0100}|. (End)
From _Geoffrey Critzer_, Nov 12 2011: (Start)
In reference to comment about multigraphs above,
  1,    1,    1,    1,    1,     1,     ...  2 nodes
  1,    3,    6,    10,   15,    21,    ...  3 nodes
  1,    6,    21,   56,   126,   252,   ...  .
  1,    10,   55,   220,  715,   2002,  ...  .
  1,    15,   120,  680,  3060,  11628, ...  .
  1,    21,   231,  1771, 10626, 58130, ...  . (End)
		

Crossrefs

Cf. A098569. A290428 (unlabeled graphs).
Cf. A121434 (inverse); variants: A122175, A122176, A122177; A107876.
Cf. A131338.

Programs

  • Mathematica
    t[n_, k_] = Binomial[(k+1)*(k+2)/2 + n-k-1, n-k]; Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]] (* Jean-François Alcover, Jul 18 2011 *)
  • PARI
    {T(n,k)=binomial((k+1)*(k+2)/2+n-k-1,n-k)}
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

Formula

T(n, k) = binomial((k+1)*(k+2)/2 + n-k-1, n-k).

A121690 G.f.: A(x) = Sum_{k>=0} x^k * (1+x)^(k*(k+1)/2).

Original entry on oeis.org

1, 1, 2, 4, 10, 27, 81, 262, 910, 3363, 13150, 54135, 233671, 1053911, 4951997, 24177536, 122381035, 640937746, 3466900453, 19337255086, 111057640382, 655892813805, 3978591077096, 24760700544301, 157941950878839
Offset: 0

Views

Author

Paul D. Hanna, Aug 15 2006

Keywords

Comments

a(n) is the number of length n permutations that simultaneously avoid the bivincular patterns (123,{2},{}) and (132,{},{2}). - Christian Bean, Jun 03 2015

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[k*(k+1)/2,n-k],{k,0,n}],{n,0,30}] (* Vaclav Kotesovec, Jun 03 2015 *)
  • PARI
    a(n) = sum(k=0,n, binomial(k*(k+1)/2, n-k))
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, x^m*prod(k=1, m, (1 - x*(1+x)^(2*k-2))/(1 - x*(1+x)^(2*k-1) + x*O(x^n)))), n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 21 2018

Formula

a(n) = Sum_{k=0..n} C(k*(k+1)/2,n-k).
a(n) = A131338(n+1, n*(n+1)/2 + 1) for n>=0, where triangle A131338 starts with a '1' in row 0 and then for n>0 row n consists of n '1's followed by the partial sums of the prior row. - Paul D. Hanna, Aug 30 2007
From Paul D. Hanna, Apr 24 2010: (Start)
Let q = (1+x), then g.f. A(x) equals the continued fraction:
A(x) = 1/(1 - q*x/(1 - (q^2-q)*x/(1 - q^3*x/(1 - (q^4-q^2)*x/(1 - q^5*x/(1- (q^6-q^3)*x/(1 - q^7*x/(1 - (q^8-q^4)*x/(1 - ...)))))))))
due to an identity of a partial elliptic theta function.
(End)
G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (1 - x*(1+x)^(2*k-2))/(1 - x*(1+x)^(2*k-1)). - Paul D. Hanna, Mar 21 2018
log(a(n)) ~ n*log(n) - 2*n*log(log(n)) - n*(1 - log(2)) + 4*n*log(log(n))/log(n) - 2*n*log(2)/log(n). - Vaclav Kotesovec, Jul 01 2025

A098569 Row sums of the triangle of triangular binomial coefficients given by A098568.

Original entry on oeis.org

1, 2, 5, 14, 43, 143, 510, 1936, 7775, 32869, 145665, 674338, 3251208, 16282580, 84512702, 453697993, 2514668492, 14367066833, 84489482201, 510760424832, 3170267071640, 20182121448815, 131642848217536, 878999194493046, 6003048930287115, 41899203336942661
Offset: 0

Views

Author

Paul D. Hanna, Sep 15 2004, Jun 29 2007

Keywords

Comments

From Lara Pudwell, Oct 23 2008: (Start)
A permutation p avoids a pattern q if it has no subsequence that is order-isomorphic to q. For example, p avoids the pattern 132 if it has no subsequence abc with a < c < b.
Barred pattern avoidance considers permutations that avoid a pattern except in a special case. Given a barred pattern q, we may form two patterns, q1 = the sequence of unbarred letters of q and q2 = the sequence of all letters of q.
A permutation p avoids barred pattern q if every instance of q1 in p is embedded in a copy of q2 in p. In other words, p avoids q1, except in the special case that a copy of q1 is a subsequence of a copy of q2.
For example, if q=5{bar 1}32{bar 4}, then q1=532 and q2 = 51324. p avoids q if every for decreasing subsequence acd of length 3 in p, one can find letters b and e so that the subsequence abcde of p has b < d < c < e < a.
(End)
Also equals the row sums of triangle A131338, which starts with a '1' in row 0 and then for n > 0 row n consists of n '1's followed by the partial sums of the prior row.
Also the number of permutations in S_n avoiding {bar 4}25{bar 1}3 (i.e., every occurrence of 253 is contained in an occurrence of a 42513). - Lara Pudwell, Apr 25 2008 (see the Claesson-Dukes-Kitaev article)
From Frank Ruskey, Apr 17 2011: (Start)
Number of sequences S = s(1)s(2)...s(n) such that
S contains m 0's,
for 1 <= j <= n, s(j) < j and s(j-s(j)) = 0,
for 1 < j <= n, if s(j) positive, then s(j-1) < s(j).
(End)
a(n) is also the number of length n permutations that simultaneously avoid the bivincular patterns (132,{2},{}) and (132,{},{2}). - Christian Bean, Mar 25 2015
a(n) is also the number of length n permutations that simultaneously avoid the bivincular patterns (123,{2},{}) and (123,{},{2}). These are the same as the permutations avoiding {bar 4}23{bar 1}5. - Christian Bean, Jun 03 2015
From Peter R. W. McNamara, Jun 22 2019: (Start)
a(n) is the number of upper-triangular matrices with nonnegative integer entries whose entries sum to n, and whose diagonal entries are all positive.
a(n) is the number of ascent sequences [d(1), d(2), ..., d(n)] A022493 for which d(k) comes from the interval [0, d(k-1)] or equals 1 + max([d(1), d(2), ..., d(k-1)]) = 1 + asc([d(1), d(2), ..., d(k-1)]) where asc(.) counts the ascents of its argument. Such sequences are called "self modified ascent sequences" in Bousquet-Mélou et al.
The elements of a (2+2)-free poset can be partitioned into levels, where all elements at the same level have the same strict down-set. Then a(n) is the number of unlabeled (2+2)-free posets with n elements that contain a chain with exactly one element at each level.
(End)

Examples

			In reference to comment about s(1)s(2)...s(n) above, a(3) = 14 = |{0000, 0001, 0002, 0003, 0010, 0020, 0100, 0012, 0013, 0023, 0101, 0103, 0120, 0123}|. - _Frank Ruskey_, Apr 17 2011
From _Paul D. Hanna_, Aug 24 2025: (Start)
The following array (A131338) illustrates a process that generates these numbers. Start with [1] in row n = 0. For n > 0, form row n by concatenating n 1's with the partial sums of the prior row. The row sums of row n equals a(n) for n >= 0; equivalently, the final term of row n+1 equals a(n). Continuing in this way generates all the terms of this sequence.
  n = 0; [1];
  n = 1: [1, 1];
  n = 2; [1, 1, 1, 2];
  n = 3: [1, 1, 1, 1, 2, 3, 5];
  n = 4: [1, 1, 1, 1, 1, 2, 3, 4, 6, 9, 14];
  n = 5: [1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 20, 29, 43];
  n = 6: [1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 8, 11, 15, 20, 27, 37, 51, 71, 100, 143];
  n = 7: [1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 35, 46, 61, 81, 108, 145, 196, 267, 367, 510];
  ... (End)
		

Crossrefs

Programs

  • Maple
    A098569 := proc(n)
        add( binomial((k+1)*(k+2)/2+n-k-1,n-k),k=0..n) ;
    end proc:
    seq(A098569(n),n=0..40) ; # Georg Fischer, Oct 29 2023
  • Mathematica
    Table[Sum[Binomial[(k+1)*(k+2)/2+n-k-1, n-k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 05 2015 *)
  • PARI
    a(n)=sum(k=0,n,binomial((k+1)*(k+2)/2+n-k-1,n-k))

Formula

a(n) = Sum_{k=0..n} C( (k+1)*(k+2)/2 + n-k-1, n-k).
G.f: Sum_{k>=0} x^k*y^C(k+1,2) where y = 1/(1-x). - Christian Bean, Mar 25 2015
log(a(n)) ~ n*(log(n) - 2*log(log(n)) + log(2) - 1 + 4*log(log(n))/log(n) - 2*log(2)/log(n) - 2/log(n)^2). - Vaclav Kotesovec, Oct 30 2023

Extensions

Offset changed to 0 by Georg Fischer, Oct 29 2023

A182961 Triangle, read by rows, where terms in row n equal the partial sums of row n-1 with 1's inserted at positions [0,n,2n-1,3n-3,4n-6,5n-10,...,n(n+1)/2-1] for n>0, with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 5, 1, 1, 2, 4, 1, 5, 8, 1, 9, 1, 14, 1, 1, 2, 4, 8, 1, 9, 14, 22, 1, 23, 32, 1, 33, 1, 47, 1, 1, 2, 4, 8, 16, 1, 17, 26, 40, 62, 1, 63, 86, 118, 1, 119, 152, 1, 153, 1, 200, 1
Offset: 0

Views

Author

Paul D. Hanna, Dec 31 2010

Keywords

Examples

			This triangle T(n,k), where k=0..n(n+1)/2 in row n>=0, begins:
1;
(1),1;
(1),1,(1),2;
(1),1,2,(1),3,(1),5;
(1),1,2,4,(1),5,8,(1),9,(1),14;
(1),1,2,4,8,(1),9,14,22,(1),23,32,(1),33,(1),47;
(1),1,2,4,8,16,(1),17,26,40,62,(1),63,86,118,(1),119,152,(1),153,(1),200;
(1),1,2,4,8,16,32,(1),33,50,76,116,178,(1),179,242,328,446,(1),447,566,718,(1),719,872,(1),873,(1),1073;
...
where row n is equal to the partial sums of terms in row n-1, with 1's inserted at positions [0,n,2n-1,3n-3,4n-6,5n-10,...,n(n+1)/2-1].
The row sums and rightmost border form sequence A129867, which equals the row sums of triangle A130469.
Triangle A130469 begins:
1;
1, 1;
2, 2, 1;
6, 4, 3, 1;
24, 12, 6, 4, 1;
120, 48, 18, 8, 5, 1;
720, 240, 72, 24, 10, 6, 1; ...
which has the same row sums as this triangle.
		

Crossrefs

Cf. A129867, A130469; variant: A131338.

Programs

  • PARI
    {T(n,k)=local(A=[1],B); for(m=0,n, t=0;B=[];
    for(j=0,#A-1, if(j==t*m-t*(t+1)/2, t+=1;B=concat(B,1)); B=concat(B,A[j+1]));
    A=Vec( Ser(B)/(1-x+O(x^#B)) ) ); if(k+1>#A, 0, B[k+1])}
    for(n=0,12,for(k=0,n*(n+1)/2,print1(T(n, k), ", ")); print(""))

Formula

Row sums equal A129867;
n-th row sum = 1 + Sum_{k=1..n} k*(n-k+1)!.
T(n,n(n+1)/2) = A129867(n) for n>0, with T(0,0) = 1.

A214403 Triangle, read by rows of terms T(n,k) for k=0..n^2, that starts with a '1' in row 0 with row n>0 consisting of 2*n-1 '1's followed by the partial sums of the prior row.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 3, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 8, 11, 15, 21, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 17, 22, 28, 36, 47, 62, 83, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 19, 24, 30, 37, 45, 55, 68, 85, 107, 135, 171, 218, 280, 363
Offset: 0

Views

Author

Paul D. Hanna, Jul 15 2012

Keywords

Comments

Right border and row sums form A178325.

Examples

			Triangle begins:
  [1];
  [1, 1];
  [1,1,1, 1, 2];
  [1,1,1,1,1, 1,2,3, 4, 6];
  [1,1,1,1,1,1,1, 1,2,3,4,5, 6,8,11, 15, 21];
  [1,1,1,1,1,1,1,1,1, 1,2,3,4,5,6,7, 8,10,13,17,22, 28,36,47, 62, 83];
  ...
Row sums equal the row sums (A178325) of triangle A214398,
where A214398(n, k) = binomial(k^2+n-k-1, n-k):
  1;
  1, 1;
  1, 4, 1;
  1, 10, 9, 1;
  1, 20, 45, 16, 1;
  1, 35, 165, 136, 25, 1;
  1, 56, 495, 816, 325, 36, 1;
  1, 84, 1287, 3876, 2925, 666, 49, 1;
  ...
		

Crossrefs

Programs

  • PARI
    {T(n, k)=if(k>n^2||n<0||k<0, 0, if(n==0,1,if(k<=2*n-1, 1, sum(i=0, k-2*n+1, T(n-1, i)))))}
    for(n=0,10,for(k=0,n^2,print1(T(n,k),", "));print(""))

A183203 Antidiagonal sums of triangle A183202.

Original entry on oeis.org

1, 1, 3, 4, 9, 14, 31, 53, 117, 215, 478, 930, 2092, 4266, 9732, 20645, 47820, 104941, 246931, 558190, 1334462, 3097018, 7521731, 17874931, 44092014
Offset: 0

Views

Author

Paul D. Hanna, Dec 30 2010

Keywords

Crossrefs

Cf. A183202.

Programs

  • PARI
    {A131338(n, k)=if(k>n*(n+1)/2||k<0,0,if(k<=n,1,sum(i=0, k-n,A131338(n-1,i))))}
    {A183202(n,k)=if(n==k,A131338(n,n*(n+1)/2),sum(j=n*k-k*(k-1)/2,n*k-k*(k-1)/2+n-k-1,A131338(n,j)))}
    {a(n)=sum(k=0,n,A183202(n-k,k))}
Showing 1-7 of 7 results.