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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A184957 Triangle read by rows: T(n,k) (n >= 1, 1 <= k <= n) is the number of compositions of n into k parts the first of which is >= all the other parts.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 4, 4, 1, 1, 1, 3, 6, 7, 5, 1, 1, 1, 4, 8, 11, 11, 6, 1, 1, 1, 4, 11, 17, 19, 16, 7, 1, 1, 1, 5, 13, 26, 32, 31, 22, 8, 1, 1, 1, 5, 17, 35, 54, 56, 48, 29, 9, 1, 1, 1, 6, 20, 48, 82, 102, 93, 71, 37, 10, 1, 1, 1, 6, 24, 63, 120, 172, 180, 148, 101, 46, 11, 1, 1, 1, 7, 28, 81, 170, 272, 331, 302, 227, 139, 56, 12, 1, 1
Offset: 1

Views

Author

N. J. A. Sloane, Feb 27 2011

Keywords

Comments

If the final diagonal is omitted, this gives the triangular array visible in A156041 and A186807.

Examples

			Triangle begins:
  [1],
  [1, 1],
  [1, 1, 1],
  [1, 2, 1, 1],
  [1, 2, 3, 1, 1],
  [1, 3, 4, 4, 1, 1],
  [1, 3, 6, 7, 5, 1, 1],
  [1, 4, 8, 11, 11, 6, 1, 1],
  [1, 4, 11, 17, 19, 16, 7, 1, 1],
  [1, 5, 13, 26, 32, 31, 22, 8, 1, 1],
  [1, 5, 17, 35, 54, 56, 48, 29, 9, 1, 1],
  ...
		

Crossrefs

Cf. A156040, A156041, A186807, A079500 (row sums).

Programs

  • Maple
    # The following Maple program is a modification of Alois P. Heinz's program for A156041
    b:= proc(n, i, m) option remember;
           if n<0 then 0 elif n=0 then 1 elif i=1 then
          `if`(n<=m, 1, 0) else add(b(n-k, i-1, m), k=0..m) fi
        end:
    A:= (n, k)-> add(b(n-m, k-1, m), m=ceil(n/k)..n):
    [seq([seq(A(d-k, k), k=1..d)], d=1..14)];
  • Mathematica
    Map[Select[#,#>0&]&,Drop[nn=11;CoefficientList[Series[Sum[x^i/(1-y(x-x^(i+1))/(1-x)),{i,1,nn}],{x,0,nn}],{x,y}],1]]//Grid (* Geoffrey Critzer, Jul 15 2013 *)

Formula

T(n,k) = A156041(n-k,k).
O.g.f.: Sum_{i>=1} x^i/(1 - y*(x - x^(i+1))/(1-x)). - Geoffrey Critzer, Jul 15 2013

A186537 G.f.: Sum( x^k/(1-2*x+x^k), k=1..oo).

Original entry on oeis.org

0, 1, 2, 4, 7, 12, 20, 34, 58, 101, 178, 318, 574, 1046, 1920, 3548, 6593, 12312, 23092, 43480, 82154, 155716, 295984, 564050, 1077400, 2062311, 3955186, 7598756, 14622318, 28179338, 54379520, 105071498, 203254164, 393607534, 763001000, 1480458656, 2875091021, 5588152920, 10869906136
Offset: 0

Views

Author

N. J. A. Sloane, Feb 23 2011

Keywords

Comments

This arose while studying the properties of A079500.

Crossrefs

First differences give A079500.

Programs

  • Maple
    add( x^k/(1-2*x+x^k), k=1..61); series(%,x,60); seriestolist(%);
    # second Maple program:
    b:= proc(n, m) option remember; `if`(n=0, 1,
          `if`(m=0, add(b(n-j, j), j=1..n),
          add(b(n-j, min(n-j, m)), j=1..min(n, m))))
        end:
    a:= proc(n) a(n):= `if`(n=0, 0, b(n-1, 0)+a(n-1)) end:
    seq(a(n), n=0..40);  # Alois P. Heinz, May 01 2014
  • Mathematica
    b[n_, m_] := b[n, m] = If[n == 0, 1, If[m == 0, Sum[b[n-j, j], {j, 1, n}], Sum[b[n-j, Min[n-j, m]], {j, 1, Min[n, m]}]]]; a[n_] := If[n == 0, 0, b[n-1, 0] + a[n-1]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, May 05 2014, after Alois P. Heinz *)

Formula

G.f.: -(1+x^2+ 1/(x-1) )/(1-x)*( 1 + x*(x-1)^3*(1-x+x^3)/( Q(0)- x*(x-1)^3*(1-x+x^3)) ), where Q(k) = (x+1)*(2*x-1)*(1-x)^2 + x^(k+2)*(x+x^2+x^3-2*x^4-1 - x^(k+3) + x^(k+5)) - x*(-1+2*x-x^(k+3))*(1-2*x+x^2+x^(k+4)-x^(k+5))*(-1+4*x-5*x^2+2*x^3 - x^(k+2)- x^(k+5) + 2*x^(k+3) - x^(2*k+5) + x^(2*k+6))/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Dec 14 2013

A320270 Number of unlabeled balanced semi-binary rooted trees with n nodes.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 6, 7, 10, 13, 19, 25, 35, 46, 65, 88, 124, 171, 242, 334, 470, 653, 921, 1287, 1822, 2565, 3640, 5144, 7311, 10360, 14734, 20918, 29781, 42361, 60389, 86069, 122893, 175479, 250922, 358863
Offset: 1

Views

Author

Gus Wiseman, Oct 08 2018

Keywords

Comments

An unlabeled rooted tree is semi-binary if all out-degrees are <= 2, and balanced if all leaves are the same distance from the root. The number of semi-binary trees with n nodes is equal to the number of binary trees with n+1 leaves; see A001190.

Examples

			The a(1) = 1 through a(7) = 6 balanced semi-binary rooted trees:
  o  (o)  (oo)   ((oo))   (((oo)))   ((o)(oo))    ((oo)(oo))
          ((o))  (((o)))  ((o)(o))   ((((oo))))   (((o)(oo)))
                          ((((o))))  (((o)(o)))   (((((oo)))))
                                     (((((o)))))  ((((o)(o))))
                                                  (((o))((o)))
                                                  ((((((o))))))
		

Crossrefs

Programs

  • Mathematica
    saur[n_]:=If[n==1,{{}},Join@@Table[Select[Union[Sort/@Tuples[saur/@ptn]],SameQ@@Length/@Position[#,{}]&],{ptn,Select[IntegerPartitions[n-1],Length[#]<=2&]}]];
    Table[Length[saur[n]],{n,20}]

A096202 Number of coverings of {1...n} by translation of a single set.

Original entry on oeis.org

1, 2, 3, 6, 11, 22, 45, 92, 188, 382, 791, 1632, 3357, 6922, 14289, 29542, 61013, 126142, 260664, 538850, 1113372, 2300954, 4752279, 9814226, 20257082, 41798206, 86204773, 177729712, 366231907, 754356336, 1553063269, 3196028942, 6573883225, 13515943986, 27775807554
Offset: 1

Views

Author

Jon Wild, Jul 27 2004

Keywords

Comments

The number of sets (up to translation) that with their translations can cover {1...n} in at least one way is given by A079500(n). For example, for n = 5 the 8 sets are {1}, {1,2}, {1,3}, {1,2,3}, {1,2,4}, {1,3,4}, {1,2,3,4}, {1,2,3,4,5}. - Andrew Howroyd, Nov 06 2019

Examples

			a(5)=11 because the following are the 11 coverings of {1...5}, each one of which only uses a single set and its translations:
   {{1},{2},{3},{4},{5}}
   {{1,2},{3,4},{4,5}}
   {{1,2},{2,3},{3,4},{4,5}}
   {{1,2},{2,3},{4,5}}
   {{1,3},{2,4},{3,5}}
   {{1,2,3},{2,3,4},{3,4,5}}
   {{1,2,3},{3,4,5}}
   {{1,2,4},{2,3,5}}
   {{1,3,4},{2,4,5}}
   {{1,2,3,4},{2,3,4,5}}
   {{1,2,3,4,5}}
		

Crossrefs

Programs

  • PARI
    covers(all, v)={
      my(u=vector(#v+1)); for(i=1, #v, u[i+1]=bitor(u[i], v[i]));
      my(recurse(k,b) = if(bitnegimply(b,u[k+1]), 0, if(k==0, 1, my(t=bitnegimply(b,v[k])); if(t==b, 2*self()(k-1, b), self()(k-1, b) + self()(k-1, t)) )));
      recurse(#v, all)
    }
    a(n)={sum(i=2^(n-1), 2^n-1, covers(2^n-1, vector(valuation(i,2)+1, j, i>>(j-1))))} \\ Andrew Howroyd, Nov 06 2019

Extensions

a(14)-a(32) from Andrew Howroyd, Nov 06 2019
a(33)-a(35) from Jinyuan Wang, Jun 09 2021

A320172 Number of series-reduced balanced rooted identity trees whose leaves are integer partitions whose multiset union is an integer partition of n.

Original entry on oeis.org

1, 2, 5, 9, 19, 38, 79, 163, 352, 750, 1633, 3558, 7783, 17020, 37338, 81920, 180399, 398600, 885101, 1975638, 4435741, 10013855, 22726109, 51807432, 118545425, 272024659, 625488420, 1440067761, 3317675261, 7644488052, 17610215982, 40547552277, 93298838972, 214516498359, 492844378878
Offset: 1

Views

Author

Gus Wiseman, Oct 07 2018

Keywords

Comments

A rooted tree is series-reduced if every non-leaf node has at least two branches, and balanced if all leaves are the same distance from the root. In an identity tree, all branches directly under any given node are different.

Examples

			The a(1) = 1 through a(5) = 19 rooted identity trees:
  (1)  (2)   (3)        (4)         (5)
       (11)  (21)       (22)        (32)
             (111)      (31)        (41)
             ((1)(2))   (211)       (221)
             ((1)(11))  (1111)      (311)
                        ((1)(3))    (2111)
                        ((1)(21))   (11111)
                        ((2)(11))   ((1)(4))
                        ((1)(111))  ((2)(3))
                                    ((1)(31))
                                    ((1)(22))
                                    ((2)(21))
                                    ((3)(11))
                                    ((1)(211))
                                    ((11)(21))
                                    ((2)(111))
                                    ((1)(1111))
                                    ((11)(111))
                                    ((1)(2)(11))
		

Crossrefs

Programs

  • Mathematica
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    mps[set_]:=Union[Sort[Sort/@(#/.x_Integer:>set[[x]])]&/@sps[Range[Length[set]]]];
    gig[m_]:=Prepend[Join@@Table[Union[Sort/@Select[Sort/@Tuples[gig/@mtn],UnsameQ@@#&]],{mtn,Select[mps[m],Length[#]>1&]}],m];
    Table[Sum[Length[Select[gig[y],SameQ@@Length/@Position[#,_Integer]&]],{y,Sort /@IntegerPartitions[n]}],{n,8}]
  • PARI
    WeighT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v,n,(-1)^(n-1)/n))))-1,-#v)}
    seq(n)={my(u=vector(n, n, numbpart(n)), v=vector(n)); while(u, v+=u; u=WeighT(u)-u); v} \\ Andrew Howroyd, Oct 25 2018

Extensions

Terms a(13) and beyond from Andrew Howroyd, Oct 25 2018

A382312 Irregular triangle read by rows: T(n,k) is the number of compositions of n with k records.

Original entry on oeis.org

1, 0, 1, 0, 2, 0, 3, 1, 0, 5, 3, 0, 8, 8, 0, 14, 17, 1, 0, 24, 36, 4, 0, 43, 72, 13, 0, 77, 143, 36, 0, 140, 281, 90, 1, 0, 256, 550, 213, 5, 0, 472, 1073, 484, 19, 0, 874, 2093, 1068, 61, 0, 1628, 4079, 2308, 177, 0, 3045, 7950, 4912, 476, 1, 0, 5719, 15498, 10328, 1217, 6
Offset: 0

Views

Author

John Tyler Rascoe, Mar 21 2025

Keywords

Comments

A record in a composition is a part that is greater than all parts before it, reading left to right. The first part of any nonempty composition is considered a record.

Examples

			Triangle begins:
    k=0    1    2   3  4
 n= 0 1;
 n= 1 0,   1;
 n= 2 0,   2;
 n= 3 0,   3,   1;
 n= 4 0,   5,   3;
 n= 5 0,   8,   8;
 n= 6 0,  14,  17,  1;
 n= 7 0,  24,  36,  4;
 n= 8 0,  43,  72, 13;
 n= 9 0,  77, 143, 36;
 n=10 0, 140, 281, 90, 1;
 ...
The composition (2,1,1,2,4,2,1,5,7) has 4 records.
                 ^       ^     ^ ^
T(4,1) = 5 counts: (4), (3,1), (2,2), (2,1,1), (1,1,1,1).
T(4,2) = 3 counts: (1,1,2), (1,2,1), (1,1,3).
		

Crossrefs

Cf. A002024 (row lengths), A011782 (row sums), A079500 (column k=1), A336482, A352525.

Programs

  • Maple
    b:= proc(n, m) option remember; expand(`if`(n=0, 1, add(
          b(n-j, max(m, j))*`if`(j>m, x, 1), j=1..n)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
    seq(T(n), n=0..16);  # Alois P. Heinz, Mar 28 2025
  • PARI
    T_xy(max_row) = {my(N=max_row+1, x='x+O('x^N), h=prod(i=1,N,1+y*x^i*(1-x)/(1-2*x+x^(i+1)))); vector(N, n, Vecrev(polcoeff(h, n-1)))}
    T_xy(12)

Formula

G.f.: Product_{i>0} (1 + y*x^i * (1 - x)/(1 - 2*x + x^(i+1))).
Sum_{k>0} T(n,k)*k = A336482(n).

A057892 Negabinary numbral addition table read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 2, 6, 2, 3, 3, 3, 3, 4, 0, 12, 0, 4, 5, 5, 13, 13, 5, 5, 6, 26, 6, 2, 6, 26, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 4, 0, 4, 24, 4, 0, 4, 8, 9, 9, 1, 1, 25, 25, 1, 1, 9, 9, 10, 14, 10, 6, 26, 30, 26, 6, 10, 14, 10, 11, 11, 11, 11, 27, 27, 27, 27, 11, 11, 11, 11, 12, 8, 52, 8, 12, 24, 4, 24
Offset: 0

Views

Author

Marc LeBrun, Sep 25 2000

Keywords

Comments

Every negabinary numbral appears infinitely often (since every signed integer can be represented as a sum of two signed integers in infinitely many ways).

Examples

			a(4)=6 since a(4) corresponds to the table entry for [1]+[1]=1+1=2=4-2=[6].
a(24)=2 since a(24) corresponds to the table entry for [3]+[3]=(-1)+(-1)=-2=[2]. - _Sean A. Irvine_, Jul 11 2022
		

Crossrefs

Extensions

a(24) and a(84) corrected and title clarified by Sean A. Irvine, Jul 11 2022

A171682 Number of compositions of n with the smallest part in the first position.

Original entry on oeis.org

1, 2, 3, 6, 10, 20, 37, 72, 140, 275, 540, 1069, 2118, 4206, 8365, 16659, 33204, 66231, 132179, 263913, 527119, 1053113, 2104428, 4205987, 8407382, 16807410, 33603024, 67187111, 134343790, 268638648, 537198557, 1074270342, 2148336463, 4296343787, 8592156886, 17183457812, 34365534564
Offset: 1

Views

Author

Vladeta Jovovic, Dec 15 2009

Keywords

Comments

First differences of A097939.

Examples

			The a(6)=20 such compositions of 6 are
[ 1]  [ 1 1 1 1 1 1 ]
[ 2]  [ 1 1 1 1 2 ]
[ 3]  [ 1 1 1 2 1 ]
[ 4]  [ 1 1 1 3 ]
[ 5]  [ 1 1 2 1 1 ]
[ 6]  [ 1 1 2 2 ]
[ 7]  [ 1 1 3 1 ]
[ 8]  [ 1 1 4 ]
[ 9]  [ 1 2 1 1 1 ]
[10]  [ 1 2 1 2 ]
[11]  [ 1 2 2 1 ]
[12]  [ 1 2 3 ]
[13]  [ 1 3 1 1 ]
[14]  [ 1 3 2 ]
[15]  [ 1 4 1 ]
[16]  [ 1 5 ]
[17]  [ 2 2 2 ]
[18]  [ 2 4 ]
[19]  [ 3 3 ]
[20]  [ 6 ]
- _Joerg Arndt_, Jan 01 2013.
		

Crossrefs

Cf. A079500.

Programs

  • Mathematica
    nn=37;Drop[CoefficientList[Series[Sum[x^i/(1-x^i/(1-x)),{i,1,nn}],{x,0,nn}],x],1]  (* Geoffrey Critzer, Mar 12 2013 *)
  • PARI
    N=66; x='x+O('x^N);
    gf= (1-x) * sum(k=1,N, x^k/(1-x-x^k) );
    Vec(gf)
    /* Joerg Arndt, Jan 01 2013 */

Formula

G.f.: (1-x) * Sum_{k>=1} x^k/(1-x-x^k). [Joerg Arndt, Jan 01 2013]
a(n) ~ 2^(n-2). - Vaclav Kotesovec, Sep 10 2014
G.f.: Sum_{n>=1} q^n/(1-Sum_{k>=n} q^k). - Joerg Arndt, Jan 03 2024

Extensions

Added more terms, Joerg Arndt, Jan 01 2013

A306269 Regular triangle read by rows where T(n,k) is the number of unlabeled balanced rooted semi-identity trees with n >= 1 nodes and depth 0 <= k < n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 0, 1, 2, 2, 1, 1, 1, 0, 1, 3, 3, 2, 1, 1, 1, 0, 1, 3, 4, 3, 2, 1, 1, 1, 0, 1, 5, 6, 5, 3, 2, 1, 1, 1, 0, 1, 5, 9, 7, 5, 3, 2, 1, 1, 1, 0, 1, 7, 12, 12, 8, 5, 3, 2, 1, 1, 1, 0, 1, 8, 17, 17, 13, 8, 5, 3, 2, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Feb 01 2019

Keywords

Comments

A rooted tree is a semi-identity tree if the non-leaf branches of the root are all distinct and are themselves semi-identity trees. It is balanced if all leaves are the same distance from the root.

Examples

			Triangle begins:
  1
  0  1
  0  1  1
  0  1  1  1
  0  1  1  1  1
  0  1  2  1  1  1
  0  1  2  2  1  1  1
  0  1  3  3  2  1  1  1
  0  1  3  4  3  2  1  1  1
  0  1  5  6  5  3  2  1  1  1
  0  1  5  9  7  5  3  2  1  1  1
  0  1  7 12 12  8  5  3  2  1  1  1
  0  1  8 17 17 13  8  5  3  2  1  1  1
  0  1 10 25 26 20 14  8  5  3  2  1  1  1
  0  1 12 34 39 31 21 14  8  5  3  2  1  1  1
The postpositive terms of row 9 {3, 4, 3, 2} count the following trees:
  ((ooooooo))   (((oooooo)))    ((((ooooo))))    (((((oooo)))))
  ((o)(ooooo))  (((o)(oooo)))   ((((o)(ooo))))   (((((o)(oo)))))
  ((oo)(oooo))  (((oo)(ooo)))   ((((o))((oo))))
                (((o))((ooo)))
		

Crossrefs

Programs

  • Mathematica
    ubk[n_,k_]:=Select[Join@@Table[Select[Union[Sort/@Tuples[ubk[#,k-1]&/@ptn]],UnsameQ@@DeleteCases[#,{}]&],{ptn,IntegerPartitions[n-1]}],SameQ[k,##]&@@Length/@Position[#,{}]&];
    Table[Length[ubk[n,k]],{n,1,10},{k,0,n-1}]

A144406 Rectangular array A read by upward antidiagonals: entry A(n,k) in row n and column k gives the number of compositions of k in which no part exceeds n, n>=1, k>=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 4, 5, 1, 1, 1, 2, 4, 7, 8, 1, 1, 1, 2, 4, 8, 13, 13, 1, 1, 1, 2, 4, 8, 15, 24, 21, 1, 1, 1, 2, 4, 8, 16, 29, 44, 34, 1, 1, 1, 2, 4, 8, 16, 31, 56, 81, 55, 1, 1, 1, 2, 4, 8, 16, 32, 61, 108, 149, 89, 1, 1, 1, 2, 4, 8, 16, 32, 63, 120, 208, 274, 144, 1
Offset: 1

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 29 2008

Keywords

Comments

Polynomial expansion as antidiagonal of p(x,n) = (x-1)/(x^n*(-x+(2*x-1)/x^n)). Based on the Pisot general polynomial type q(x,n) = x^n - (x^n-1)/(x-1) (the original name of the sequence).
Row sums are 1, 2, 3, 5, 8, 14, ... (A079500).
Conjecture: Since the array row sequences successively tend to A000079, the absolute values of nonzero differences between two successive row sequences tend to A045623 = {1,2,5,12,28,64,144,320,704,1536,...}, as k -> infinity. - L. Edson Jeffery, Dec 26 2013

Examples

			Array A begins:
  {1, 1, 1, 1, 1,  1,  1,  1,   1,   1,   1, ...}
  {1, 1, 2, 3, 5,  8, 13, 21,  34,  55,  89, ...}
  {1, 1, 2, 4, 7, 13, 24, 44,  81, 149, 274, ...}
  {1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, ...}
  {1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, ...}
  {1, 1, 2, 4, 8, 16, 32, 63, 125, 248, 492, ...}
  {1, 1, 2, 4, 8, 16, 32, 64, 127, 253, 504, ...}
  {1, 1, 2, 4, 8, 16, 32, 64, 128, 255, 509, ...}
  {1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, ...}
  ... - _L. Edson Jeffery_, Dec 26 2013
As a triangle:
  {1},
  {1, 1},
  {1, 1, 1},
  {1, 1, 2, 1},
  {1, 1, 2, 3, 1},
  {1, 1, 2, 4, 5, 1},
  {1, 1, 2, 4, 7, 8, 1},
  {1, 1, 2, 4, 8, 13, 13, 1},
  {1, 1, 2, 4, 8, 15, 24, 21, 1},
  {1, 1, 2, 4, 8, 16, 29, 44, 34, 1},
  {1, 1, 2, 4, 8, 16, 31, 56, 81, 55, 1},
  {1, 1, 2, 4, 8, 16, 32, 61, 108, 149, 89, 1},
  {1, 1, 2, 4, 8, 16, 32, 63, 120, 208, 274, 144, 1},
  {1, 1, 2, 4, 8, 16, 32, 64, 125, 236, 401, 504, 233, 1},
  {1, 1, 2, 4, 8, 16, 32, 64, 127, 248, 464, 773, 927, 377, 1}
		

Crossrefs

Same as A048887 but with a column of 1's added on the left (the number of compositions of 0 is defined to be equal to 1).
Array rows (with appropriate offsets) are A000012, A000045, A000073, A000078, A001591, A001592, etc.

Programs

  • Mathematica
    g[x_, n_] = x^(n) - (x^n - 1)/(x - 1);
    h[x_, n_] = FullSimplify[ExpandAll[x^(n)*g[1/x, n]]];
    f[t_, n_] := 1/h[t, n];
    a = Table[CoefficientList[Series[f[t, m], {t, 0, 30}], t], {m, 1, 31}];
    b = Table[Table[a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}];
    Flatten[b] (* Triangle version *)
    Grid[Table[CoefficientList[Series[(1 - x)/(1 - 2 x + x^(n + 1)), {x, 0, 10}], x], {n, 1, 10}]] (* Array version - L. Edson Jeffery, Jul 18 2014 *)

Formula

t(n,m) = antidiagonal_expansion of p(x,n) where p(x,n) = (x-1)/(x^n*(-x+(2*x-1)/x^n)).
G.f. for array A: (1-x)/(1 - 2*x + x^(n+1)), n>=1. - L. Edson Jeffery, Dec 26 2013

Extensions

Definition changed by L. Edson Jeffery, Jul 18 2014
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