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 1-8 of 8 results.

A107112 Row sums of A107111, viewed as a number triangle.

Original entry on oeis.org

1, 2, 5, 15, 54, 228, 1106, 6069, 37212, 252287, 1873928, 15127191, 131788320, 1231650843, 12283477562, 130137706957, 1458823402224, 17242265101407, 214202486482516, 2789239633559763, 37974772474452740, 539356744121554370
Offset: 0

Views

Author

Paul Barry, May 12 2005

Keywords

Formula

a(n)=sum{k=0..n, sum{j=0..k, binomial((n-k)(k+1), k-j)*binomial(k+j, j)}/(k+1)}

A107113 Diagonal sums of A107111, viewed as a number triangle.

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 24, 52, 125, 307, 809, 2190, 6249, 18348, 56016, 176272, 571113, 1906091, 6517283, 22879944, 82144248, 301617448, 1131401427, 4327790902, 16886037207, 67073179408, 271242546501, 1115489848809, 4662462489619
Offset: 0

Views

Author

Paul Barry, May 12 2005

Keywords

Formula

a(n)=sum{k=0..floor(n/2), sum{j=0..k, binomial((n-2k)(k+1), k-j)*binomial(k+j, j)}/(k+1)}

A007297 Number of connected graphs on n labeled nodes on a circle with straight-line edges that don't cross.

Original entry on oeis.org

1, 1, 4, 23, 156, 1162, 9192, 75819, 644908, 5616182, 49826712, 448771622, 4092553752, 37714212564, 350658882768, 3285490743987, 30989950019532, 294031964658430, 2804331954047160, 26870823304476690, 258548658860327880
Offset: 1

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Author

Keywords

Comments

Apart from the initial 1, reversion of g.f. for A162395 (squares with signs): see A263843.

Examples

			G.f. = x*(1 + x + 4*x^2 + 23*x^3 + 156*x^4 + 1162*x^5 + 9192*x^6 + 75819*x^7 + ...).
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A162395, A000290. 4th row of A107111. Row sums of A089434.
See A263843 for a variant.
Cf. A000108 (non-crossing set partitions), A001006, A001187, A054726 (non-crossing graphs), A054921, A099947, A194560, A293510, A323818, A324167, A324169, A324173.

Programs

  • Maple
    A007297:=proc(n) if n = 1 then 1 else add(binomial(3*n - 3, n + j)*binomial(j - 1, j - n + 1), j = n - 1 .. 2*n - 3)/(n - 1); fi; end;
  • Mathematica
    CoefficientList[ InverseSeries[ Series[(x-x^2)/(1+x)^3, {x, 0, 20}], x], x] // Rest (* From Jean-François Alcover, May 19 2011, after PARI prog. *)
    Table[Binomial[3n, 2n+1] Hypergeometric2F1[1-n, n, 2n+2, -1]/n, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 25 2015 *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse((x-x^2)/(1+x)^3+O(x^(n+2))),n+1)) /* Ralf Stephan */

Formula

Apart from initial term, g.f. is the series reversion of (x-x^2)/(1+x)^3 (A162395). See A263843. - Vladimir Kruchinin, Feb 08 2013
G.f.: (g-z)/z, where g=-1/3+(2/3)*sqrt(1+9z)*sin((1/3)*arcsin((2+27z+54z^2)/2/(1+9*z)^(3/2))). - Emeric Deutsch, Dec 02 2002
a(n) = (1/n)*Sum_{k=0..n} binomial(3n, n-k-1)*binomial(n+k-1, k). - Paul Barry, May 11 2005
a(n) = 4^(n-1)*(Gamma(3*n/2-1)/Gamma(n/2+1)/Gamma(n) -Gamma((3*n-1)/2)/ Gamma( (n+1)/2)/Gamma(n+1)). - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
a(n) = 4^n * binomial(3*n/2, n/2) / (9*n-6) - 4^(n-1) * binomial(3*(n-1)/2, (n-1)/2 ) / n. - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
D-finite with recurrence: n*(n-1)*(3*n-4)*a(n) +36*(n-1)*a(n-1) -12*(3*n-8)*(3*n-1)*(3*n-7)*a(n-2)=0. - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
a(n) = (1/n)*Sum_{k=0..n} C(3n, k)*C(2n-k-2, n-1). - Paul Barry, Sep 27 2005
a(n) ~ (2-sqrt(3)) * 6^n * 3^(n/2) / (sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Mar 17 2014
a(n) = binomial(3*n,2*n+1)*hypergeom([1-n,n], [2*n+2], -1)/n. - Vladimir Reshetnikov, Oct 25 2015
a(n) = 2*A078531(n) - A085614(n+1). - Vladimir Reshetnikov, Apr 24 2016

Extensions

Better description from Philippe Flajolet, Apr 20 2000
More terms from James Sellers, Aug 21 2000
Definition revised and initial a(1)=1 added by N. J. A. Sloane, Nov 05 2015 at the suggestion of Axel Boldt. Some of the formulas may now need to be adjusted slightly.

A143689 a(n) = (3*n^2 - n + 2)/2.

Original entry on oeis.org

1, 2, 6, 13, 23, 36, 52, 71, 93, 118, 146, 177, 211, 248, 288, 331, 377, 426, 478, 533, 591, 652, 716, 783, 853, 926, 1002, 1081, 1163, 1248, 1336, 1427, 1521, 1618, 1718, 1821, 1927, 2036, 2148, 2263, 2381, 2502, 2626, 2753, 2883, 3016, 3152, 3291
Offset: 0

Views

Author

Gary W. Adamson, Aug 29 2008

Keywords

Comments

Equals left border of triangle A033292.
Equals binomial transform of [1, 1, 3, 0, 0, 0, ...].
A242357(a(n)) = 1. - Reinhard Zumkeller, May 11 2014
These might be called "trisected pentagonal numbers": A figurate pentagonal number is composed of three triangles, of which the central one is the largest, and the removal of the triangular frame (3*n) of the central triangle trisects the figure. This is reflected in the formula a(n) = A000326(n+1) - 3*n. See illustration in links. - John Elias, May 27 2022

Crossrefs

a(n) = A000326(n+1) - 3n. Third column of A107111.

Programs

Formula

a(n) = A000326(n+1) - 3*n. (A000326 are the pentagonal numbers.)
a(n) = (3*n^2 - n + 2)/2 = A027599(n+1)/2. - R. J. Mathar, Sep 03 2008
a(n) = a(n-1) + 3*n - 2 (with a(0)=1). - Vincenzo Librandi, Nov 25 2010
a(n) = 2*a(n-1) - a(n-2) + 3.
O.g.f.: (1-x+3*x^2)/((1-x)^3). - Eric Werley, Jun 27 2011
a(n) = A104249(-n). - Bruno Berselli, Jul 08 2015
a(n) = binomial(n,2) + n^2 + 1 = A152947(n+1) + A000290(n). - Franck Maminirina Ramaharo, Mar 01 2018
E.g.f.: exp(x)*(2 + 2*x + 3*x^2)/2. - Stefano Spezia, Apr 19 2025

Extensions

Index of A000326 in definition, formula and example corrected by R. J. Mathar, Sep 03 2008

A365754 Expansion of (1/x) * Series_Reversion( x*(1-x)/(1+x)^4 ).

Original entry on oeis.org

1, 5, 36, 305, 2833, 27916, 286632, 3033513, 32858595, 362515725, 4059475368, 46021411644, 527163783916, 6092053249160, 70939443268112, 831558454663449, 9804617762941095, 116201796106426543, 1383557994261012100, 16541672701743657545, 198510770031798279825
Offset: 0

Views

Author

Seiichi Manyama, Sep 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+k, k)*binomial(4*(n+1), n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(4*(n+1),n-k).
a(n) = (1/(n+1)) * [x^n] ( (1+x)^4 / (1-x) )^(n+1). - Seiichi Manyama, Feb 17 2024

A365755 Expansion of (1/x) * Series_Reversion( x*(1-x)/(1+x)^5 ).

Original entry on oeis.org

1, 6, 52, 530, 5919, 70098, 864784, 10994490, 143042020, 1895316632, 25487708844, 346976558318, 4772478619146, 66222166440780, 925880434336320, 13030945427540170, 184467676431001644, 2624828100099166536, 37521220349342729680, 538573138719587026440
Offset: 0

Views

Author

Seiichi Manyama, Sep 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+k, k)*binomial(5*(n+1), n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(5*(n+1),n-k).

A345462 Triangle T(n,k) (n >= 1, 0 <= k <= n-1) read by rows: number of distinct permutations after k steps of the "first transposition" algorithm.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 24, 13, 4, 1, 120, 67, 23, 5, 1, 720, 411, 146, 36, 6, 1, 5040, 2921, 1067, 272, 52, 7, 1, 40320, 23633, 8800, 2311, 456, 71, 8, 1, 362880, 214551, 81055, 21723, 4419, 709, 93, 9, 1, 3628800, 2160343, 825382, 224650, 46654, 7720, 1042, 118, 10, 1
Offset: 1

Views

Author

Olivier Gérard, Jun 20 2021

Keywords

Comments

The first transposition algorithm is: if the permutation is sorted, then exit; otherwise, exchange the first unsorted letter with the letter currently at its index. Repeat.
At each step at least 1 letter (possibly 2) is sorted.
If one counts the steps necessary to reach the identity, this gives the Stirling numbers of the first kind (reversed).

Examples

			Triangle begins:
      1;
      2,     1;
      6,     3,    1;
     24,    13,    4,    1;
    120,    67,   23,    5,   1;
    720,   411,  146,   36,   6,  1;
   5040,  2921, 1067,  272,  52,  7, 1;
  40320, 23633, 8800, 2311, 456, 71, 8, 1;
  ...
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 3 / Sorting and Searching, Addison-Wesley, 1973.

Crossrefs

Cf. A321352, A345461 (same idea for other sorting algorithms).
Cf. A180191 (second column, k=1).
Cf. A107111 a triangle with some common parts.
Cf. A143689 (diagonal T(n,n-3)).

Programs

  • Maple
    b:= proc(n, k) option remember; (k+1)!*
          binomial(n, k)*add((-1)^i/i!, i=0..k+1)/n
        end:
    T:= proc(n, k) option remember;
         `if`(k=0, n!, T(n, k-1)-b(n, n-k+1))
        end:
    seq(seq(T(n, k), k=0..n-1), n=1..10);  # Alois P. Heinz, Aug 11 2021
  • Mathematica
    b[n_, k_] := b[n, k] = (k+1)!*Binomial[n, k]*Sum[(-1)^i/i!, {i, 0, k+1}]/n;
    T[n_, k_] := T[n, k] = If[k == 0, n!, T[n, k-1] - b[n, n-k+1]];
    Table[Table[T[n, k], {k, 0, n - 1}], {n, 1, 10}] // Flatten (* Jean-François Alcover, Mar 06 2022, after Alois P. Heinz *)

Formula

T(n,0) = n!; T(n,n-3) = (3*(n-1)^2 - n + 3)/2.
From Alois P. Heinz, Aug 11 2021: (Start)
T(n,k) = T(n,k-1) - A010027(n,n-k) for k >= 1.
T(n,k) - T(n,k+1) = A123513(n,k).
T(n,0) - T(n,1) = A000255(n-1) for n >= 2.
T(n,1) - T(n,2) = A000166(n) for n >= 3.
T(n,2) - T(n,3) = A000274(n) for n >= 4.
T(n,3) - T(n,4) = A000313(n) for n >= 5. (End)

A366012 a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(n*(n+1),n-k).

Original entry on oeis.org

1, 2, 13, 156, 2833, 70098, 2214280, 85464984, 3906724321, 206648387550, 12425282899588, 837384222603448, 62539219710804627, 5127758187193514824, 457986530357734020432, 44263628968974498793648, 4602969726808566383149761, 512486177498084438210961270, 60827938291895363867587959628
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[1/(n + 1) Sum[Binomial[n + k, n] Binomial[n (n + 1), n - k], {k, 0, n}], {n, 0, 18}]
    Table[SeriesCoefficient[(1/x) InverseSeries[Series[x (1 - x)/(1 + x)^n, {x, 0, n + 1}], x], {x, 0, n}], {n, 0, 18}]

Formula

a(n) = [x^n] (1/x) * Series_Reversion( x * (1 - x) / (1 + x)^n ).
a(n) ~ exp(n + 3/2) * n^(n - 3/2) / sqrt(2*Pi). - Vaclav Kotesovec, Sep 26 2023
Showing 1-8 of 8 results.