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.

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A181971 Triangle read by rows: T(n,0) = 1, T(n,n) = floor((n+3)/2) and T(n,k) = T(n-1,k-1) + T(n-1,k), 0 < k < n.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 4, 5, 3, 1, 5, 9, 8, 3, 1, 6, 14, 17, 11, 4, 1, 7, 20, 31, 28, 15, 4, 1, 8, 27, 51, 59, 43, 19, 5, 1, 9, 35, 78, 110, 102, 62, 24, 5, 1, 10, 44, 113, 188, 212, 164, 86, 29, 6, 1, 11, 54, 157, 301, 400, 376, 250, 115, 35, 6, 1, 12, 65, 211, 458, 701, 776, 626, 365, 150, 41, 7
Offset: 0

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Author

Reinhard Zumkeller, Jul 09 2012

Keywords

Comments

Another variant of Pascal's triangle;
row sums: A081254; central terms: T(2*n,n) = A128082(n+1);
T(n,0) = 1;
T(n,1) = n + 1 for n > 0;
T(n,2) = A000096(n-1) for n > 1;
T(n,3) = A105163(n-2) for n > 2;
T(n,n-2) = A005744(n-1) for n > 1;
T(n,n-1) = A024206(n) for n > 0;
T(n,n) = A008619(n+1).

Examples

			The triangle begins:
.  0:                              1
.  1:                           1     2
.  2:                        1     3     2
.  3:                     1     4     5     3
.  4:                  1     5     9     8     3
.  5:               1     6    14    17    11     4
.  6:            1     7    20    31    28    15     4
.  7:         1     8    27    51    59    43    19     5
.  8:      1     9    35    78   110   102    62    24     5
.  9:   1    10    44   113   188   212   164    86    29     6.
		

Crossrefs

Programs

  • Haskell
    a181971 n k = a181971_tabl !! n !! k
    a181971_row n = a181971_tabl !! n
    a181971_tabl = map snd $ iterate f (1, [1]) where
       f (i, row) = (1 - i, zipWith (+) ([0] ++ row) (row ++ [i]))
    
  • Mathematica
    T[n_ /; n >= 0, k_ /; k >= 0] := T[n, k] = If[n == k, Quotient[n + 3, 2], If[k == 0, 1, If[n > k, T[n - 1, k - 1] + T[n - 1, k]]]];
    Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 12 2021 *)
  • PARI
    {T(n,k)=if(n==k,(n+3)\2,if(k==0,1,if(n>k,T(n-1,k-1)+T(n-1,k))))}
    for(n=0,12,for(k=0,n,print1(T(n,k),","));print("")) \\ Paul D. Hanna, Jul 18 2012

A014430 Subtract 1 from Pascal's triangle, read by rows.

Original entry on oeis.org

1, 2, 2, 3, 5, 3, 4, 9, 9, 4, 5, 14, 19, 14, 5, 6, 20, 34, 34, 20, 6, 7, 27, 55, 69, 55, 27, 7, 8, 35, 83, 125, 125, 83, 35, 8, 9, 44, 119, 209, 251, 209, 119, 44, 9, 10, 54, 164, 329, 461, 461, 329, 164, 54, 10, 11, 65, 219, 494, 791, 923, 791, 494, 219, 65, 11
Offset: 0

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Author

Keywords

Comments

Each value of the sequence (T(x,y)) is equal to the sum of all values in Pascal's Triangle that are in the rectangle defined by the tip (0,0) and the position (x,y). - Florian Kleedorfer (florian.kleedorfer(AT)austria.fm), May 23 2005
To clarify T(n,k) and A129696: We subtract I = Identity matrix from Pascal's triangle to obtain the beheaded variant, A074909. Then take column sums starting from the top of A074909 to get triangle A014430. Row sums of the inverse of triangle T(n,k) gives the Bernoulli numbers, A027641/A026642. Alternatively, triangle T(n,k) as an infinite lower triangular matrix * [the Bernoulli numbers as a vector] = [1, 1, 1, ...]. Given the B_n version starting (1, 1/2, 1/6, ...) triangle T(n,k) * the B_n vector [1, 1/2, 1/6, 0, -1/30, ...] = the triangular numbers. - Gary W. Adamson, Mar 13 2012
From R. J. Mathar, Apr 25 2016: (Start)
If regarded as a symmetric array of the form
1 2 3 4 5 ...
2 5 9 14 20 ...
3 9 19 34 55 ...
4 14 34 69 125 ...
5 20 55 125 251 ...
6 27 83 209 461 ...
7 35 119 329 791 ...
8 44 164 494 1286 ...
9 54 219 714 2001 ...
it contains the rows (and columns) A000096, A062748, A063258, A062988, A124089, ..., A035927 and so on and counts the multisets of digits of numbers in base b>=2 with d>=1 digits (equivalent to the comment in A035927). (End)
Proof of Florian Kleedorfer's formula: Take sums of the columns of the rectangle - these are all binomial coefficients by the Hockey Stick Identity. Note the locations of these coefficients: They form a row going almost all the way to the edge, only missing the 1 - apply the Hockey Stick Identity again. - James East, Jul 03 2020

Examples

			Triangle begins:
  1;
  2,  2;
  3,  5,  3;
  4,  9,  9,   4;
  5, 14, 19,  14,   5;
  6, 20, 34,  34,  20,  6;
  7, 27, 55,  69,  55, 27,  7;
  8, 35, 83, 125, 125, 83, 35, 8;
		

Crossrefs

Triangle with zeros: A014473.
Cf. A000295 (row sums).

Programs

  • Haskell
    a014430 n k = a014430_tabl !! n !! k
    a014430_row n = a014430_tabl !! n
    a014430_tabl = map (init . tail) $ drop 2 a014473_tabl
    -- Reinhard Zumkeller, Apr 10 2012
    
  • Magma
    [Binomial(n+2,k+1)-1: k in [0..n], n in [0..13]]; // G. C. Greubel, Feb 25 2023
    
  • Mathematica
    Table[Sum[Sum[Binomial[m, j], {m, j, j+(n-k)}], {j,0,k}], {n,0,10}, {k, 0,n}]//Flatten (* Michael De Vlieger, Sep 01 2020 *)
    Table[Binomial[n+2,k+1] -1, {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 25 2023 *)
  • SageMath
    flatten([[binomial(n+2,k+1)-1 for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Feb 25 2023

Formula

T(n, k) = T(n-1, k) + T(n-1, k-1) + 1, T(0, 0)=1. - Ralf Stephan, Jan 23 2005
G.f.: 1 / ((1-x)*(1-x*y)*(1-x*(1+y))). - Ralf Stephan, Jan 24 2005
T(n, k) = Sum_{j=0..k} Sum_{m=j..j+(n-k)} binomial(m, j). - Florian Kleedorfer (florian.kleedorfer(AT)austria.fm), May 23 2005
T(n, k) = binomial(n+2, k+1) - 1. - G. C. Greubel, Feb 25 2023

Extensions

More terms from Erich Friedman
Offset fixed by Reinhard Zumkeller, Apr 10 2012

A124927 Triangle read by rows: T(n,0)=1, T(n,k)=2*binomial(n,k) if k>0 (0<=k<=n).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 6, 2, 1, 8, 12, 8, 2, 1, 10, 20, 20, 10, 2, 1, 12, 30, 40, 30, 12, 2, 1, 14, 42, 70, 70, 42, 14, 2, 1, 16, 56, 112, 140, 112, 56, 16, 2, 1, 18, 72, 168, 252, 252, 168, 72, 18, 2, 1, 20, 90, 240, 420, 504, 420, 240, 90, 20, 2, 1, 22, 110, 330, 660, 924, 924, 660, 330, 110, 22, 2
Offset: 0

Views

Author

Gary W. Adamson, Nov 12 2006

Keywords

Comments

Pascal triangle with all entries doubled except for the first entry in each row. A028326 with first column replaced by 1's. Row sums are 2^(n+1)-1.
From Paul Barry, Sep 19 2008: (Start)
Reversal of A129994. Diagonal sums are A001595. T(2n,n) is A100320.
Binomial transform of matrix with 1,2,2,2,... on main diagonal, zero elsewhere. (End)
This sequence is jointly generated with A210042 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+v(n-1,x) +1 and v(n,x)=x*u(n-1,x)+x*v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle given by (1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 25 2012

Examples

			Triangle starts:
  1;
  1,  2;
  1,  4,  2;
  1,  6,  6,  2;
  1,  8, 12,  8,  2;
  1, 10, 20, 20, 10, 2;
(1, 0, 0, 1, 0, 0, ...) DELTA (0, 2, -1, 0, 0, ...) begins:
  1;
  1,  0;
  1,  2,  0;
  1,  4,  2,  0;
  1,  6,  6,  2,  0;
  1,  8, 12,  8,  2, 0;
  1, 10, 20, 20, 10, 2, 0. - _Philippe Deléham_, Mar 25 2012
		

Crossrefs

Cf. A000225.
Cf. A074909.

Programs

  • Haskell
    a124927 n k = a124927_tabl !! n !! k
    a124927_row n = a124927_tabl !! n
    a124927_tabl = iterate
       (\row -> zipWith (+) ([0] ++ reverse row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Mar 04 2012
    
  • Magma
    [k eq 0 select 1 else 2*Binomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    T:=proc(n,k) if k=0 then 1 else 2*binomial(n,k) fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    (* First program *)
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
    v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A210042 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A124927 *) (* Clark Kimberling, Mar 17 2012 *)
    (* Second program *)
    Table[If[k==0, 1, 2*Binomial[n, k]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 10 2019 *)
  • PARI
    T(n,k) = if(k==0,1, 2*binomial(n,k)); \\ G. C. Greubel, Jul 10 2019
    
  • Sage
    def T(n, k):
        if (k==0): return 1
        else: return 2*binomial(n,k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 10 2019

Formula

T(n,0) = 1; for n>0: T(n,n) = 2, T(n,k) = T(n-1,k) + T(n-1,n-k), 1Reinhard Zumkeller, Mar 04 2012
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,0) = 1, T(1,1) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2012
G.f.: (1-x+x*y)/((-1+x)*(x*y+x-1)). - R. J. Mathar, Aug 11 2015

Extensions

Edited by N. J. A. Sloane, Nov 24 2006

A302971 Triangle read by rows: T(n,k) is the numerator of R(n,k) defined implicitly by the identity Sum_{i=0..l-1} Sum_{j=0..m} R(m,j)*(l-i)^j*i^j = l^(2*m+1) holding for all l,m >= 0.

Original entry on oeis.org

1, 1, 6, 1, 0, 30, 1, -14, 0, 140, 1, -120, 0, 0, 630, 1, -1386, 660, 0, 0, 2772, 1, -21840, 18018, 0, 0, 0, 12012, 1, -450054, 491400, -60060, 0, 0, 0, 51480, 1, -11880960, 15506040, -3712800, 0, 0, 0, 0, 218790, 1, -394788954, 581981400, -196409840, 8817900, 0, 0, 0, 0, 923780, 1, -16172552880, 26003271294, -10863652800, 1031151660, 0, 0, 0, 0, 0, 3879876
Offset: 0

Views

Author

Kolosov Petro, Apr 16 2018

Keywords

Examples

			Triangle begins:
------------------------------------------------------------------------
k=   0          1         2         3    4     5      6      7       8
------------------------------------------------------------------------
n=0: 1;
n=1: 1,         6;
n=2: 1,         0,       30;
n=3: 1,       -14,        0,      140;
n=4: 1,      -120,        0,        0, 630;
n=5: 1,     -1386,      660,        0,   0, 2772;
n=6: 1,    -21840,    18018,        0,   0,    0, 12012;
n=7: 1,   -450054,   491400,   -60060,   0,    0,     0, 51480;
n=8: 1, -11880960, 15506040, -3712800,   0,    0,     0,     0, 218790;
		

Crossrefs

Items of second row are the coefficients in the definition of A287326.
Items of third row are the coefficients in the definition of A300656.
Items of fourth row are the coefficients in the definition of A300785.
T(n,n) gives A002457(n).
Denominators of R(n,k) are shown in A304042.
Row sums return A000079(2n+1) - 1.

Programs

  • Maple
    R := proc(n, k) if k < 0 or k > n then return 0 fi; (2*k+1)*binomial(2*k, k);
    if n = k then % else -%*add((-1)^j*R(n, j)*binomial(j, 2*k+1)*
    bernoulli(2*j-2*k)/(j-k), j=2*k+1..n) fi end: T := (n, k) -> numer(R(n, k)):
    seq(print(seq(T(n, k), k=0..n)), n=0..12);
    # Numerical check that S(m, n) = n^(2*m+1):
    S := (m, n) -> add(add(R(m, j)*(n-k)^j*k^j, j=0..m), k=0..n-1):
    seq(seq(S(m, n) - n^(2*m+1), n=0..12), m=0..12); # Peter Luschny, Apr 30 2018
  • Mathematica
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    T[n_, k_] := Numerator[R[n, k]];
    (* Print Fifteen Initial rows of Triangle A302971 *)
    Column[ Table[ T[n, k], {n, 0, 15}, {k, 0, n}], Center]
  • PARI
    T(n, k) = if ((n>k) || (n<0), 0, if (k==n, (2*n+1)*binomial(2*n, n), if (2*n+1>k, 0, if (n==0, 1, (2*n+1)*binomial(2*n, n)*sum(j=2*n+1, k+1, T(j, k)*binomial(j, 2*n+1)*(-1)^(j-1)/(j-n)*bernfrac(2*j-2*n))))));
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(numerator(T(k,n)), ", ")); print); \\ Michel Marcus, Apr 27 2018

Formula

Recurrence given by Max Alekseyev (see the MathOverflow link):
R(n, k) = 0 if k < 0 or k > n.
R(n, k) = (2k+1)*binomial(2k, k) if k = n.
R(n, k) = (2k+1)*binomial(2k, k)*Sum_{j=2k+1..n} R(n, j)*binomial(j, 2k+1)*(-1)^(j-1)/(j-k)*Bernoulli(2j-2k), otherwise.
T(n, k) = numerator(R(n, k)).

A304042 Triangle read by rows: T(n,k) is the denominator of R(n,k) defined implicitly by the identity Sum_{i=0..l-1} Sum_{j=0..m} R(m,j)*(l-i)^j*i^j = l^(2*m+1) holding for all l,m >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Kolosov Petro, May 05 2018

Keywords

Examples

			Triangle begins:
-----------------------------------------------------
k=    0  1  2  3  4  5  6  7  8  9  10 11 12 13 14 15
-----------------------------------------------------
n=0:  1;
n=1:  1, 1;
n=2:  1, 1, 1;
n=3:  1, 1, 1, 1;
n=4:  1, 1, 1, 1, 1;
n=5:  1, 1, 1, 1, 1, 1;
n=6:  1, 1, 1, 1, 1, 1, 1;
n=7:  1, 1, 1, 1, 1, 1, 1, 1;
n=8:  1, 1, 1, 1, 1, 1, 1, 1, 1;
n=9:  1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=10: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=11: 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1;
n=12: 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=13: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=14: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=15: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
		

Crossrefs

Programs

  • Mathematica
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    T[n_, k_] := Denominator[R[n, k]];
    (* Print Fifteen Initial rows of Triangle A304042 *)
    Column[ Table[ T[n, k], {n, 0, 15}, {k, 0, n}], Center]
  • PARI
    up_to = 1274; \\ = binomial(50+1,2)-1
    A304042aux(n, k) = if((k<0)||(k>n),0,(k+k+1)*binomial(2*k, k)*if(k==n,1,sum(j=k+k+1,n, A304042aux(n, j)*binomial(j, k+k+1)*((-1)^(j-1))/(j-k)*bernfrac(2*(j-k)))));
    A304042tr(n, k) = denominator(A304042aux(n, k));
    A304042list(up_to) = { my(v = vector(up_to), i=0); for(n=0,oo, for(k=0,n, if(i++ > up_to, return(v)); v[i] = A304042tr(n,k))); (v); };
    v304042 = A304042list(1+up_to);
    A304042(n) = v304042[1+n]; \\ Antti Karttunen, Nov 07 2018

Formula

Recurrence given by Max Alekseyev (see the MathOverflow link):
R(n, k) = 0 if k < 0 or k > n.
R(n, k) = (2k+1)*binomial(2k, k) if k = n.
R(n, k) = (2k+1)*binomial(2k, k)*Sum_{j=2k+1..n} R(n, j)*binomial(j, 2k+1)*(-1)^(j-1)/(j-k)*Bernoulli(2j-2k), otherwise.
T(n, k) = denominator(R(n, k)).

A014631 Numbers in order in which they appear in Pascal's triangle.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 10, 15, 20, 7, 21, 35, 8, 28, 56, 70, 9, 36, 84, 126, 45, 120, 210, 252, 11, 55, 165, 330, 462, 12, 66, 220, 495, 792, 924, 13, 78, 286, 715, 1287, 1716, 14, 91, 364, 1001, 2002, 3003, 3432, 105, 455, 1365, 5005, 6435, 16, 560, 1820, 4368, 8008, 11440
Offset: 1

Views

Author

Keywords

Comments

A permutation of the natural numbers. - Robert G. Wilson v, Jun 12 2014
In Pascal's triangle a(n) occurs the first time in row A265912(n). - Reinhard Zumkeller, Dec 18 2015

Crossrefs

Cf. A034868, A119629 (inverse), A265912.

Programs

  • Haskell
    import Data.List (nub)
    a014631 n = a014631_list !! (n-1)
    a014631_list = 1 : (nub $ concatMap tail a034868_tabf)
    -- Reinhard Zumkeller, Dec 19 2015
    
  • Mathematica
    lst = {1}; t = Flatten[Table[Binomial[n, m], {n, 16}, {m, Floor[n/2]}]]; Do[ If[ !MemberQ[lst, t[[n]]], AppendTo[lst, t[[n]] ]], {n, Length@t}]; lst (* Robert G. Wilson v *)
    DeleteDuplicates[Flatten[Table[Binomial[n,m],{n,20},{m,0,Floor[n/2]}]]] (* Harvey P. Dale, Apr 08 2013 *)
  • Python
    from itertools import count, islice
    def A014631_gen(): # generator of terms
        s, c =(1,), set()
        for i in count(0):
            for d in s:
                if d not in c:
                    yield d
                    c.add(d)
            s=(1,)+tuple(s[j]+s[j+1] for j in range(len(s)-1)) + ((s[-1]<<1,) if i&1 else ())
    A014631_list = list(islice(A014631_gen(),30)) # Chai Wah Wu, Oct 17 2023

Extensions

More terms from Erich Friedman
Offset changed by Reinhard Zumkeller, Dec 18 2015

A325191 Number of integer partitions of n such that the difference between the length of the minimal triangular partition containing and the maximal triangular partition contained in the Young diagram is 1.

Original entry on oeis.org

0, 0, 2, 0, 3, 3, 0, 4, 6, 4, 0, 5, 10, 10, 5, 0, 6, 15, 20, 15, 6, 0, 7, 21, 35, 35, 21, 7, 0, 8, 28, 56, 70, 56, 28, 8, 0, 9, 36, 84, 126, 126, 84, 36, 9, 0, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0, 11, 55, 165, 330, 462
Offset: 0

Views

Author

Gus Wiseman, Apr 11 2019

Keywords

Comments

The Heinz numbers of these partitions are given by A325196.
Under the Bulgarian solitaire step, these partitions form cycles of length >= 2. Length >= 2 means not the length=1 self-loop which occurs from the triangular partition when n is a triangular number. See A074909 for self-loops included. - Kevin Ryde, Sep 27 2019

Examples

			The a(2) = 2 through a(12) = 10 partitions (empty columns not shown):
  (2)   (22)   (32)   (322)   (332)   (432)   (4322)   (4332)
  (11)  (31)   (221)  (331)   (422)   (3321)  (4331)   (4422)
        (211)  (311)  (421)   (431)   (4221)  (4421)   (4431)
                      (3211)  (3221)  (4311)  (5321)   (5322)
                              (3311)          (43211)  (5331)
                              (4211)                   (5421)
                                                       (43221)
                                                       (43311)
                                                       (44211)
                                                       (53211)
		

Crossrefs

Programs

  • Mathematica
    otb[ptn_]:=Min@@MapIndexed[#1+#2[[1]]-1&,Append[ptn,0]];
    otbmax[ptn_]:=Max@@MapIndexed[#1+#2[[1]]-1&,Append[ptn,0]];
    Table[Length[Select[IntegerPartitions[n],otb[#]+1==otbmax[#]&]],{n,0,30}]
  • PARI
    a(n) = my(t=ceil(sqrtint(8*n+1)/2), r=n-t*(t-1)/2); if(r==0,0, binomial(t,r)); \\ Kevin Ryde, Sep 27 2019

Formula

Positions of zeros are A000217 = n * (n + 1) / 2.
a(n) = A074909(n) - A010054(n). - Kevin Ryde, Sep 27 2019

A344678 Coefficients for normal ordering of (x + D)^n and for the unsigned, probabilist's (or Chebyshev) Hermite polynomials H_n(x+y).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 3, 1, 1, 4, 6, 6, 12, 4, 3, 6, 1, 1, 5, 10, 10, 30, 10, 15, 30, 5, 15, 10, 1, 1, 6, 15, 15, 60, 20, 45, 90, 15, 90, 60, 6, 15, 45, 15, 1, 1, 7, 21, 21, 105, 35, 105, 210, 35, 315, 210, 21, 105, 315, 105, 7, 105, 105, 21, 1
Offset: 0

Views

Author

Tom Copeland, May 26 2021

Keywords

Comments

This irregular triangular array contains the integer coefficients of the normal ordering of the expansions of the differential operator R = (x + D)^n with D = d/dx. This operator is the raising/creation operator for the unsigned, modified Hermite polynomials H_n(x) of A099174; i.e., R H_n(x) = H_{n+1}(x). The lowering/annihilation operator L is D; i.e, L H_n(x) = D H_n(x) = n H_{n-1}(x).
Generalizing to the ladder operators for S_n(x) a general Sheffer polynomial sequence, (L + R)^n 1 = H_n(S.(x)), where the umbral composition is defined by H_n(S.(x)) = (h. + S.(x))^n = Sum_{k = 0..n} binomial(n,k) h_{n-k} S_k(x), where h_n are the Taylor series coefficients of e^{t^2/2}; i.e., e^{h.t} = e^{t^2/2}.
Another interpretation is that the n-th row contains the coefficients of the terms in the normal ordering of the 2^n permutations of two binary symbols given by (L + R)^n under the Leibniz condition LR = RL + 1 equivalent to the commutator relation [L,R] = LR - RL = 1. For example, (x + D)^2 = xx + xD + Dx + DD = x^2 + 2 xD + 1 + D^2.
As in the examples, the coefficients are ordered as those for the terms x^k D^m, ordered first from higher k to lower k and subordinately from lower m to higher m.
The number sequences associated to these polynomials are intimately related to the complete graphs K_n, which have n vertices and (n-1) edges. H_n(x) are the independence polynomials for K_n. The moments, h_n, of the H_n(x) Appell polynomials are the aerated double factorials A001147, the number of perfect matchings in the complete graph K_{n}, zero for odd n. The row lengths, A002620, give the number of maximal strokes on the complete graph K_n. The triangular numbers--the sum of two consecutive row lengths--give the number of edges of K_n. The row sums, A005425, are the number of matchings of the corona K'_n of the complete graph K_n and the complete graph K_1. K_n can be viewed as the projection onto a plane of the edges of the regular (n-1)-dimensional simplex, whose face polynomials are (x+1)^n - 1 (cf. A135278 and A074909).
The coefficients represent the combinatorics of a switchboard problem in which among n subscribers, a subscriber may talk to one of the others, someone on an outside line, or not at all--no conference calls allowed. E.g., the coefficient 12 for H_4(x+y) is the number of ways among 4 subscribers that a pair of subscribers can talk to each other while another is on an outside line and the remaining subscriber is disconnected. The coefficient 3 for the polynomial corresponds to the number of ways two pairs of subscribers can talk among themselves. This can be regarded as a dinner table problem also where a diner may exchange positions with another diner; remain seated; or get up, change plans, and sit back down in the same chair--no more than one exchange per diner.
From Peter Luschny, May 28 2021: (Start)
If we write the monomials of the row polynomials in degree-lexicographic order, we observe: The coefficient triangle appears as a series of concatenated subtriangles. The first one is Pascal's triangle A007318. Appending the rows of triangle A094305 begins in row 2. In row 4, the next triangle starts, which is A344565. This scheme seems to go on indefinitely. [This is now set out in A344911.] (End)
From Tom Copeland, May 29 2021: (Start)
Simplex model: H_n(x+y) = (h. + x + y)^n with h_n the aerated odd double factorials (h_0 = 1, h_1 = 0, h_2 = 1, h_3 = 0, h_4 = 3, h_5 = 0, h_6 = 15, ...), the number of perfect matchings for the n vertices of the (n-1)-Dimensional simplices, i.e., the hypertriangles, or hypertetrahedrons. This multinomial enumerates permutations of three families of objects--vertices labeled with either x or y, or perfect (pair) matchings of the unlabeled vertices for the n vertices of the (n-1)-D simplex. For example, (x+D)^2 = xx + xD + Dx + D^2 = x^2 + 2xD + D^2 + 1 corresponds to H_2(x+y) = (h.+ x + y)^2 = h_0 (x+y)^2 + 2 h_1 (x+y) + h_2 = x^2 + 2xy + y^2 + 1, which, in turn, corresponds to a line segment, the 1-D simplex, with both vertices labeled with x's; or one with an x, the other a y; or both vertices labeled with y's; or one unlabeled matched pair. The exponent k of x^k y^m represents the number of these vertices to which an x is assigned, and m, those with a y assigned. The remaining unlabeled vertices (a sub-simplex) form an independent edge set of (single color) colored edges, i.e., no colored edge is touching another colored edge (a perfect matching for the sub-simplex) nor the vertices assigned an x or y. Accordingly, the coefficients for k=m=0 represent the number of ways to assign a perfect matching to the simplex--zero for simplices with an odd number of vertices and the odd double factorials (1, 3, 15, 105, ...) for the simplices with an even number. For the simplices with an odd number of vertices, the coefficients for k+m=1 are the odd double factorials. (Note the polynomials are invariant upon interchange of the variables x and y.) (End)

Examples

			(x + D)^0 = 1,
(x + D)^1 = x + D,
(x + D)^2 = x^2 + 2 x D + 1 + D^2,
(x + D)^3 = x^3 + 3 x^2 D + 3 x + 3 x D^2 + 3 D + D^3,
(x + D)^4 = x^4 + 4 x^3 D + 6 x^2 + 6 x^2 D^2 + 12 x D + 4 x D^3 + 3 + 6 D^2 + D^4.
(x + D)^5 = x^5 + 5 x^4 D + 10 x^3 + 10 x^3 D^2 + 30 x^2 D + 10 x^2 D^3 + 15 x + 30 x D^2 + 5 x D^4 + 15 D + 10 D^3 + D^5
H_6(x + y) = x^6 + 6 x^5 y + 15 x^4 + 15 x^4 y^2 + 60 x^3 y + 20 x^3 y^3 + 45 x^2 + 90 x^2 y^2 + 15 x^2 y^4 + 90 x y + 60 x y^3 + 6 x y^5 + 15 + 45 y^2 + 15 y^4 + y^6
H_7(x + y) = x^7 + 7 x^6 y + 21 x^5 + 21 x^5 y^2 + 105 x^4 y + 35 x^4 y^3 + 105 x^3 + 210 x^3 y^2 + 35 x^3 y^4 + 315 x^2 y + 210 x^2 y^3 + 21 x^2 y^5 + 105 x + 315 x y^2 + 105 x y^4 + 7 x y^6 + 105 y + 105 y^3 + 21 y^5 + y^7
		

Crossrefs

Programs

  • Mathematica
    Last /@ CoefficientRules[#, {x, y}] & /@ Table[Simplify[(-y)^n (-2)^(-n/2) HermiteH[n, (x + 1/y)/Sqrt[-2]]], {n, 0, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bivariate e.g.f. is e^{t^2/2} e^{t(x + y)} = Sum_{n >= 0} H_n(x+y) t^n/n! = e^{t H.(x+y)} = e^{t (x + H.(y))}, as described below.
The coefficient of x^k D^m is n! h_{n-k-m} / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n with h_n, as defined in the comments, aerated A001147.
Row lengths, r(n): 1, 2, 4, 6, 9, 12, 16, 20, 25, ... A002620(n).
Row sums: A005425 = 1, 2, 5, 14, 43, 142, ... .
The recursion H_{n+1}(x+y) = (x+y) H_n(x+y) + n H_{n-1}(x+y) follows from the differential raising and lowering operations of the Hermite polynomials.
The Baker-Campbell-Hausdorff-Dynkin expansion leads to the disentangling relation e^{t (x + D)} = e^{t^2/2} e^{tx} e^{tD} from which the formula above for the coefficients may be derived via differentiation with respect to t.
The row bivariate polynomials P_n(x,y) with y a commutative analog of D, or L, have the e.g.f. e^{-xy} e^{t(x + D)} e^{xy} = e^{-xy} e^{t^2/2} e^{tx} e{tD} e^{xy} = e^{t^2/2} e^{t(x + y)} = e^{t(h. + x + y)} = e^{t (x + H.(y))} = e^{t H.(x +y)}, so P_n(x,y) = H_n(x + y) = (x + H.(y))^n, the Hermite polynomials mentioned in the comments along with the umbral composition. The row sums are H_n(2), listed in A005425. For example, P_3(x,y) = (x + H.(y))^3 = x^3 H_0(y) + 3 x^2 H_1(y) + 3 x H_2(y) + H_3(y) = H_3(x+y) = (x+y)^3 + 3(x+y) = x^3 + 3 x^2 y + 3 x + 3 x y^2 + 3 y + y^2.
Alternatively, P_n(x,y) = H_n(x+y) = (z + d/dz)^n 1 with z replaced by (x+y) after the repeated differentiations since (x + D)^n 1 = H_n(x).
With initial index 1, the lengths r(n) of the rows of nonzero coefficients are the same as those for the polynomials given by 1 + (x+y)^2 + (x+y)^4 + ... + (x+y)^n for n even and for those for (x+y)^1 + (x+y)^3 + ... + (x+y)^n for n odd since the Hermite polynomials are even or odd polynomials. Consequently, r(n)= O(n) = 1 + 2 + 4 + ... n for n odd and r(n) = E(n) = 2 + 4 + ... + n for n even, so O(n) = ((n+1)/2)^2 and E(n) = (n/2)((n/2)+1) = n(n+2)/4 = 2 T(n/2) where T(k) are the triangular numbers defined by T(k) = 0 + 1 + 2 + 3 + ... + k = A000217(k). E(n) corresponds to A002378. Additionally, r(n) + r(n+1) = 1 + 2 + 3 + ... + n+1 = T(n+1).
From Tom Copeland, May 31 2021: (Start)
e^{D^2/2} = e^{h.D}, so e^{D^2/2} x^n = e^{h. D} x^n = (h. + x)^n = H_n(x) and e^{D^2/2} (x+y)^n = e^{h. D} (x+y)^n = (h. + x + y)^n = H_n(x+y).
From the Appell Sheffer polynomial calculus, the umbral compositional inverse of the sequence H_n(x+y), i.e., the sequence HI_n(x+y) such that H_n(HI.(x+y)) = (h. + HI.(x+y))^n = (h. + hi. + x + y)^n = (x+y)^n, is determined by e^{-t^2/2} = e^{hi. t}, so hi_n = -h_n and HI_n(x+y) = (-h. + x + y)^n = (-1)^n (h. - x - y)^n = (-1)^n H_n(-(x+y)). Then H_n(-H.(-(x+y))) = (x+y)^n.
In addition, HI_n(x) = (x - D) HI_{n-1}(x) = (x - D)^n 1 = e^{-D^2/2} x^n = e^{hi. D} x^n = e^{-h. D} x^n with the e.g.f. e^{-t^2/2} e^{xt}.
The umbral compositional inversion property follows from x^n = e^{-D^2/2} e^{D^2/2} x^n = e^{-D^2/2} H_n(x) = H_n(HI.(x)) = e^{D^2/2} e^{-D^2/2} x^n = HI_n(H.(x)). (End)
The umbral relations above reveal that H_n(x+y) = (h. + x + y)^n = Sum_{k = 0..n} binomial(n,k) h_k (x+y)^{n-k}, which gives, e.g., for n = 3, H_3(x+y) = h_0 * (x+y)^3 + 3 h_1 * (x+y)^2 + 3 h_2 * (x+y) + h_3 = (x+y)^3 + 3 (x+y), the n-th through 0th rows of the Pascal matrix embedded within the n-th row of the Pascal matrix modulated by h_k. - Tom Copeland, Jun 01 2021
Varvak gives the coefficients of x^(n-m-k) D^{m-k} as n! / ( 2^k k! (n-k-m)! (m-k)! ), referring to them as the Weyl binomial coefficients, and derives them from rook numbers on Ferrers boards. (No mention of Hermite polynomials nor matchings on simplices are made.) Another combinatorial model and equivalent formula are presented in Blasiak and Flajolet (p. 16). References to much earlier work are given in both papers. - Tom Copeland, Jun 03 2021

A053221 Row sums of triangle A053218.

Original entry on oeis.org

1, 5, 16, 43, 106, 249, 568, 1271, 2806, 6133, 13300, 28659, 61426, 131057, 278512, 589807, 1245166, 2621421, 5505004, 11534315, 24117226, 50331625, 104857576, 218103783, 452984806, 939524069, 1946157028, 4026531811, 8321499106
Offset: 1

Views

Author

Asher Auel, Jan 01 2000

Keywords

Comments

Considered as a vector, the sequence = A074909 * [1, 2, 3, ...], where A074909 is the beheaded Pascal's triangle as a matrix. - Gary W. Adamson, Mar 06 2012
a(n) is the sum of the upper left n X n subarray of A052509 (viewed as an infinite square array). For example (1+1+1) + (1+2+2) + (1+3+4) = 16. - J. M. Bergot, Nov 06 2012
Number of ternary strings of length n that contain at least one 2 and at most one 0. For example, a(3) = 16 since the strings are the 6 permutations of 201, the 3 permutations of 211, the 3 permutations of 220, the 3 permutations of 221, and 222. - Enrique Navarrete, Jul 25 2021

Examples

			a(4) = 4 + 7 + 12 + 20 = 43.
		

Crossrefs

Programs

  • Magma
    [(n+2)*2^(n-1)-n-1: n in [1..50]]; // G. C. Greubel, Sep 03 2018
  • Maple
    A053221 := proc(n) (n+2)*2^(n-1)-n-1 ; end proc: # R. J. Mathar, Sep 02 2011
  • Mathematica
    Table[(n + 2)*2^(n - 1) - n - 1, {n, 29}] (* or *)
    Rest@ CoefficientList[Series[-x (-1 + x + x^2)/((2 x - 1)^2*(x - 1)^2), {x, 0, 29}], x] (* Michael De Vlieger, Sep 22 2017 *)
    LinearRecurrence[{6,-13,12,-4},{1,5,16,43},30] (* Harvey P. Dale, Jun 28 2021 *)
  • PARI
    vector(50,n, (n+2)*2^(n-1)-n-1) \\ G. C. Greubel, Sep 03 2018
    

Formula

a(n) = (n+2)*2^(n-1)-n-1. - Vladeta Jovovic, Feb 28 2003
G.f.: -x*(-1+x+x^2) / ( (2*x-1)^2*(x-1)^2 ). - R. J. Mathar, Sep 02 2011
a(n) = (1/2) * Sum_{k=1..n} Sum_{i=1..n} C(k,i) + C(n,k). - Wesley Ivan Hurt, Sep 22 2017
E.g.f.: exp(x)*(exp(x)-1)*(1+x). - Enrique Navarrete, Jul 25 2021
a(n+1) = 2*a(n) + A006127(n). - Ya-Ping Lu, Jan 01 2024

A208744 Triangle relating to ordered Bell numbers, A000670.

Original entry on oeis.org

1, 1, 2, 1, 3, 9, 1, 4, 18, 52, 1, 5, 30, 130, 375, 1, 6, 45, 260, 1125, 3246, 1, 7, 63, 455, 2625, 11361, 32781, 1, 8, 84, 728, 5250, 30296, 131124, 378344, 1, 9, 108, 1092, 9450, 68166, 393372, 1702548, 4912515, 1, 10, 135, 1560, 15750, 136332, 983430, 5675160, 24562575, 70872610
Offset: 1

Views

Author

Gary W. Adamson, Mar 05 2012

Keywords

Comments

Row sums = A000670 starting (1, 3, 13, 75,...).
Right border = A052882 starting (1, 2, 9, 52, 375,...).
A000670 is the eigensequence of triangle A074909, deleting the first "1".
Triangle A074909 is the "beheaded" Pascal's triangle: (1; 1,2; 1,3,3;...).

Examples

			Row 4 (nonzero terms) = (1, 4, 18, 52) = termwise product of (1, 4, 6, 4) and (1, 1, 3, 13).
First few rows of the triangle:
1;
1, 2;
1, 3, 9;
1, 4, 18, 52;
1, 5, 30, 130, 375;
1, 6, 45, 260, 1125, 3246;
1, 7, 63, 455, 2625, 11361, 32781;
1, 8, 84, 728, 5250, 30296, 131124, 378344;
...
		

Crossrefs

Programs

  • Mathematica
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i + k + r)*(i + r)^(n - r)/(i!*(k - i - r)!), {i, 0, k - r}], {k, r, n}]; Fubini[0, 1] = 1;
    a[n_, k_] := Binomial[n, k] Fubini[k, 1];
    Table[a[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Mar 30 2016 *)

Formula

As infinite lower triangular matrices, A074909 * A000670 as the main diagonal and the rest zeros.
E.g.f. (exp(x) - 1)/(2 - exp(x*t)) = x + (1 + 2*t)*x^2/2! + (1 + 3*t + 9*t^2)*x^3/3! + .... Cf. A154921. - Peter Bala, Aug 31 2016

Extensions

a(36) corrected by Jean-François Alcover, Mar 30 2016
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