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 1-4 of 4 results.

A097939 Sum of the smallest parts of all compositions of n.

Original entry on oeis.org

1, 3, 6, 12, 22, 42, 79, 151, 291, 566, 1106, 2175, 4293, 8499, 16864, 33523, 66727, 132958, 265137, 529050, 1056169, 2109282, 4213710, 8419697, 16827079, 33634489, 67237513, 134424624, 268768414, 537407062, 1074605619, 2148875961, 4297212424, 8593556211, 17185713097, 34369170909
Offset: 1

Views

Author

Vladeta Jovovic, Sep 05 2004

Keywords

Comments

Sums of the antidiagonals of A099238. - Paul Barry, Oct 08 2004

Crossrefs

Programs

  • Maple
    A097939:=n->add(add(binomial(n-r*(k+1)-1,k), k=0..floor((n-r-1)/(r+1))), r=0..n-1): seq(A097939(n), n=1..50); # Wesley Ivan Hurt, Dec 03 2016
    # second Maple Program:
    b:= proc(n, m) option remember; `if`(n=0, m,
          add(b(n-j, min(j, m)), j=1..n))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=1..40);  # Alois P. Heinz, Jul 26 2020
  • Mathematica
    Drop[ CoefficientList[ Series[ Sum[x^k/(1 - x - x^k), {k, 50}], {x, 0, 35}], x], 1] (* Robert G. Wilson v, Sep 08 2004 *)
  • PARI
    N=66; x='x+O('x^N);
    gf= sum(k=1,N, x^k/(1-x-x^k) );
    Vec(gf)
    /* Joerg Arndt, Jan 01 2013 */
    
  • PARI
    {a(n)=polcoeff(sum(m=1,n,x^m*sumdiv(m,d,1/(1-x +x*O(x^n))^d) ),n)}

Formula

G.f.: Sum_{k>=1} x^k/(1-x-x^k).
a(n) = Sum_{r=0..n-1} Sum_{k=0..floor((n-r-1)/(r+1))} binomial(n-r(k+1)-1, k). - Paul Barry, Oct 08 2004
G.f.: (1-x)^2 * Sum_{k>=1} k*x^k/((x^k+x-1)*(x^(k+1)+x-1)). - Vladeta Jovovic, Apr 23 2006
G.f.: Sum_{k>=1} x^k/((1-x)^k*(1-x^k)). - Vladeta Jovovic, Mar 02 2008
G.f.: Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series) where a=1/(1-x). - Joerg Arndt, Jan 30 2011
G.f.: Sum_{n>=1} (a*x)^n/(1-x^n) where a=1/(1-x). - Joerg Arndt, Jan 01 2013
G.f.: Sum_{n>=1} x^n * Sum_{d|n} 1/(1-x)^d. - Paul D. Hanna, Jul 18 2013
a(n) ~ 2^(n-1). - Vaclav Kotesovec, Oct 28 2014

Extensions

More terms from Robert G. Wilson v, Sep 08 2004

A099233 Square array read by antidiagonals associated to sections of 1/(1-x-x^k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 6, 5, 1, 1, 1, 5, 10, 13, 8, 1, 1, 1, 6, 15, 26, 28, 13, 1, 1, 1, 7, 21, 45, 69, 60, 21, 1, 1, 1, 8, 28, 71, 140, 181, 129, 34, 1, 1, 1, 9, 36, 105, 251, 431, 476, 277, 55, 1, 1, 1, 10, 45, 148, 413, 882, 1326, 1252, 595, 89, 1
Offset: 0

Views

Author

Paul Barry, Oct 08 2004

Keywords

Examples

			Rows begin
  1, 1, 1,  1,  1,   1, ...
  1, 1, 2,  3,  5,   8, ...
  1, 1, 3,  6, 13,  28, ...
  1, 1, 4, 10, 26,  69, ...
  1, 1, 5, 15, 45, 140, ...
Row 1 is the 0-section of 1/(1-x-x)   (A000079);
Row 2 is the 1-section of 1/(1-x-x^2) (A000045);
Row 3 is the 2-section of 1/(1-x-x^3) (A000930);
Row 4 is the 3-section of 1/(1-x-x^4) (A003269);
etc.
		

Crossrefs

Sums of antidiagonals are A099236.
Columns include A000217, A008778.
Rows include A000045, A002478, A099234, A099235.
Main diagonal gives A099237.
Cf. A099238.

Formula

Square array T(n, k) = Sum_{j=0..n} binomial(k(n-j), j).
Rows are generated by 1/(1-x(1+x)^k) and satisfy a(n) = Sum_{k=0..n} binomial(n, k)a(n-k-1).

A125104 Triangle read by rows counting compositions (ordered partitions) by minimal part size.

Original entry on oeis.org

1, 1, 1, 1, 0, 3, 1, 0, 1, 6, 1, 0, 0, 2, 13, 1, 0, 0, 1, 3, 27, 1, 0, 0, 0, 2, 5, 56, 1, 0, 0, 0, 1, 2, 9, 115, 1, 0, 0, 0, 0, 2, 3, 15, 235, 1, 0, 0, 0, 0, 1, 2, 5, 25, 478, 1, 0, 0, 0, 0, 0, 2, 2, 8, 42, 969, 1, 0, 0, 0, 0, 0, 1, 2, 3, 12, 70, 1959, 1, 0, 0, 0, 0, 0, 0, 2, 2, 5, 18, 116, 3952, 1, 0, 0, 0, 0, 0, 0, 1, 2, 2, 8, 27, 192, 7959, 1, 0, 0, 0, 0, 0, 0, 0, 2, 2, 3, 11, 41, 317, 16007
Offset: 0

Views

Author

Alford Arnold, Nov 28 2006, corrected Nov 28 2006

Keywords

Comments

The diagonals of this array can be generated from Table A099238 as follows: A000079 - A000045 = [1, 2, 4, 8, 16, 32, ...] - [0, 1, 1, 2, 3, 5, ...] = [1, 1, 3, 6, 13, 27, ...] = A099036, A000045 - A000930, A000930 - A003269, A003269 - A003520, etc.

Examples

			Row 4 of the array is (1, 0, 1, 6) because there are six compositions with minimum part of size one: 1111, 31, 13, 211, 121, 112; one of size two: 22; none of size three; and 1 of size four: 4.
Triangle (after 45-degree counterclockwise rotation) begins:
1 1 3 6 13 27 56 115 235 478 969 1959 3952 7959
.1 0 1 2 3 5 9 15 25 42 70 116 192
..1 0 0 1 2 2 3 5 8 12 18 27
...1 0 0 0 1 2 2 2 3 5 8
....1 0 0 0 0 1 2 2 2 2
.....1 0 0 0 0 0 1 2 2
......1 0 0 0 0 0 0 1
.......1 0 0 0 0 0 0
........1 0 0 0 0 0
		

Crossrefs

Extensions

Edited by N. J. A. Sloane, Dec 21 2006
More terms from Vladeta Jovovic, Jul 10 2007

A144409 Antidiagonal expansion of: f(t,n) = If[n == 1, 1/(1 - t), 1/(1 - t^floor(n/2) - t^n)].

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 0, 1, 3, 1, 1, 0, 1, 2, 5, 1, 1, 0, 1, 0, 3, 8, 1, 1, 0, 0, 0, 2, 4, 13, 1, 1, 0, 0, 1, 1, 0, 6, 21, 1, 1, 0, 0, 1, 0, 1, 3, 9, 34, 1, 1, 0, 0, 0, 0, 0, 1, 0, 13, 55, 1, 1, 0, 0, 0, 1, 0, 2, 2, 5, 19, 89, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 28, 144, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 3
Offset: 1

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 30 2008

Keywords

Comments

Row sums are {1, 2, 3, 5, 6, 10, 14, 21, 31, 50, 71, 120, 177, 288, 445}.

Examples

			{1},
{1, 1},
{1, 1, 1},
{1, 1, 2, 1},
{1, 0, 1, 3, 1},
{1, 0, 1, 2, 5, 1},
{1, 0, 1, 0, 3, 8, 1},
{1, 0, 0, 0, 2, 4, 13, 1},
{1, 0, 0, 1, 1, 0, 6, 21, 1},
{1, 0, 0, 1, 0, 1, 3, 9, 34, 1},
{1, 0, 0, 0, 0, 0, 1, 0, 13, 55, 1},
{1, 0, 0, 0, 1, 0, 2, 2, 5, 19, 89, 1},
{1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 28, 144, 1},
{1, 0, 0, 0, 0, 0, 0, 1, 0, 3, 8, 41, 233, 1},
{1, 0, 0, 0, 0, 1, 0, 0, 0, 3, 2, 0, 60, 377, 1}
		

Crossrefs

Cf. A099238.

Programs

  • Mathematica
    f[t_, n_] = If[n == 1, 1/(1 - t), 1/(1 - t^Floor[n/2] - t^n)]; a = Table[Table[SeriesCoefficient[Series[f[t, m], {t, 0, 30}], n], {n, 0, 30}], {m, 1, 31}]; b = Table[Table[a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}] ; Flatten[b]

Formula

f(t,n) = If[n == 1, 1/(1 - t), 1/(1 - t^floor(n/2) - t^n)); t(n,m) = antidiagonal_expansion(f(t,n)).
Showing 1-4 of 4 results.