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 1-3 of 3 results.

A005157 Number of totally symmetric plane partitions that fit in an n X n X n box.

Original entry on oeis.org

1, 2, 5, 16, 66, 352, 2431, 21760, 252586, 3803648, 74327145, 1885102080, 62062015500, 2652584509440, 147198472495020, 10606175914819584, 992340657705109416, 120567366227960791040, 19023173201224270401428, 3897937005297330777227264
Offset: 0

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Keywords

Comments

Also, number of 2-dimensional shifted complexes on n+1 nodes. [Klivans]
Also the number of totally symmetric partitions which fit in an (n-1)-dimensional box with side length 4 (for n>0). - Graham H. Hawkes, Jan 11 2014
Suppose we index this sequence slightly differently. Let the elements of a partition be represented by points rather than boxes, as in a Ferrers diagram. In this case, a 1 X 1 X 1 (closed) box would fit 8 points -- one at each vertex of the box, and we use the convention that a 0 X 0 X 0 (closed) box contains exactly one point. Using this indexing, the sequence begins (offset is still 0) 2,5,16,... rather than 1,2,5,... If we use the same method of indexing for all other dimensions, then we have the following remarkable result: The number of totally symmetric partitions which fit inside a d-dimensional box with side length n is equal to the number of totally symmetric partitions which fit inside an n-dimensional box of side length d. - Graham H. Hawkes, Jan 11 2014
For two other contexts where this sequence arises, see the Knuth (2019) link (noncrossing paths among the 2(2^n-1) paths defined in that note; independent sets of paths among the first 2^n-1 of those). - N. J. A. Sloane, Feb 09 2019, based on email from Don Knuth.

Examples

			a(2) = 5 because we have: void, 1, 21/1, 22/21, and 22/22.
		

References

  • D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; Eq. (6.8), p. 198 (corrected).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of A214564.

Programs

  • Maple
    A005157 := proc(n) local i,j; mul(mul((i+j+n-1)/(i+2*j-2),j=i..n),i=1..n); end;
  • Mathematica
    Table[Product[(i+j+k-1)/(i+j+k-2),{i,n},{j,i,n},{k,j,n}],{n,0,20}] (* Harvey P. Dale, Jul 17 2011 *)
  • PARI
    A005157(n)=prod(i=1,n,prod(j=i,n,(i+j+n-1)/(i+2*j-2))) \\ M. F. Hasler, Sep 26 2018

Formula

a(n) = Product_{i=1..n} Product_{j=i..n} Product_{k=j..n} (i+j+k-1)/(i+j+k-2). - Paul Barry, May 13 2008
a(n) ~ exp(1/72) * GAMMA(1/3)^(2/3) * n^(7/72) * 3^(3*n*(n+1)/4 + 11/72) / (A^(1/6) * Pi^(1/3) * 2^(n*(2*n+1)/2 + 13/24)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Mar 01 2015
a(n) = sqrt(A323848(n+1,n)) for n >= 1. [proof by Nikolai Beluhov; see Knuth (2019) link] - Alois P. Heinz, Feb 10 2019
Apparently, a(n) = Sum_{k=0..n} A184173(n,k). - Alois P. Heinz, Feb 11 2019
Conjectures: if p == 1 (mod 6) is prime then a(p) == 2^((p+5)/6) (mod p^2); if p == 5 (mod 6) is prime then a(p) == 2^((p+1)/6) (mod p^2) (checked up to p = 1009). - Peter Bala, Feb 17 2023

A323849 Irregular triangle read by rows: T(n,d) (n >= 1, 0 <= d <= 2n-2) = number of n X n integer-valued matrices M such that M_{1,1}=0, M_{n,n}=d, and M_{(i+1),j} = M_{i,j} + (0 or 1), M_{i,(j+1)} = M_{i,j} + (0 or 1).

Original entry on oeis.org

1, 1, 4, 1, 1, 18, 44, 18, 1, 1, 68, 615, 1236, 615, 68, 1, 1, 250, 7313, 46812, 84910, 46812, 7313, 250, 1, 1, 922, 85801, 1592348, 8241540, 14024408, 8241540, 1592348, 85801, 922, 1, 1, 3430, 1030330, 54926890, 759337545, 3397542544, 5530983756, 3397542544, 759337545, 54926890, 1030330, 3430, 1
Offset: 1

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Author

N. J. A. Sloane, Feb 07 2019

Keywords

Examples

			Triangle begins:
  n\d 0   1     2       3       4        5       6       7     8   9 10
  1   1
  2   1   4     1
  3   1  18    44      18       1
  4   1  68   615    1236     615       68       1
  5   1 250  7313   46812   84910    46812    7313     250     1
  6   1 922 85801 1592348 8241540 14024408 8241540 1592348 85801 922  1
  ...
		

References

  • D. E. Knuth, Email to N. J. A. Sloane, Feb 06 2019.

Crossrefs

Columns k=0-2 give: A000012, A115112, A252869.
T(n,n-1) gives A306372.
Cf. A323848.

Formula

T(n,1) = binomial(2n,n) - 2 = A115112(n).
The triangle is symmetric: T(n,d) = T(n,2n-2-d).

Extensions

Edited by Alois P. Heinz, Feb 11 2019

A306322 Number of n X n integer matrices (m_{i,j}) such that m_{1,1}=0, m_{n,n}=2, and all rows, columns, and falling diagonals are (weakly) monotonic without jumps of 2.

Original entry on oeis.org

1, 0, 0, 25, 386, 4657, 54219, 642815, 7852836, 98755951, 1273299491, 16761968919, 224508932229, 3051075581019, 41979207169125, 583745779595077, 8192478969914858, 115908383594664493, 1651636256584103013, 23685002515500875105, 341589590792856093329
Offset: 0

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Author

Alois P. Heinz, Feb 07 2019

Keywords

Crossrefs

Main diagonal of A323846.
Column d=2 of A323848.

Programs

  • Mathematica
    Nara[i_, j_] := 1/(i+j-1)*Binomial[i+j-1, i]*Binomial[i+j-1, i-1];
    Prepend[Table[2*Sum[Sum[Nara[i, j], {i, n}] + (n-j-1)*Nara[j, n], {j, n}] - 2*Binomial[2*n, n] + Nara[n, n] + 3, {n, 100}], 1] (* Manuel Kauers and Christoph Koutschan, Mar 02 2023 *)

Formula

From Manuel Kauers and Christoph Koutschan, Mar 02 2023: (Start)
a(n) = 2*Sum_{j=1..n} (Sum_{i=1..n} N'(i, j) + (n-j-1)*N'(j, n)) - 2*binomial(2*n, n) + N'(n, n) + 3 for n>0, where N'(n, k) = (binomial(n+k-1, n-1)*binomial(n+k-1, n))/(n+k-1) denotes the Narayana number N(n+k-1, k).
Recurrence: -2*(n+3)*(n+4)^2*(2*n+7)*(118125*n^10 + 1308375*n^9 + 6016950*n^8 + 14827410*n^7 + 20875365*n^6 + 15986367*n^5 + 4449768*n^4 - 2342808*n^3 - 2279152*n^2 - 660240*n - 64000)*a(n+4) + (n+3)*(10040625*n^13 + 210555000*n^12 + 1942194375*n^11 + 10361592450*n^10 + 35325144315*n^9 + 80085358620*n^8 + 121180651809*n^7 + 117919810482*n^6 + 64349576684*n^5 + 7017979960*n^4 - 14571577344*n^3 - 9566235392*n^2 - 2428639744*n - 221347840)*a(n+3) + (-42170625*n^14 - 981642375*n^13 - 10209053025*n^12 - 62559627795*n^11 - 250621464735*n^10 - 687475711989*n^9 - 1311094658043*n^8 - 1718884004625*n^7 - 1471227292164*n^6 - 691541238960*n^5 - 14462120192*n^4 + 188403075920*n^3 + 108128100864*n^2 + 25779317504*n + 2257059840)*a(n+2) + 2*(2*n+3)*(10040625*n^13 + 215870625*n^12 + 2048259375*n^11 + 11279217825*n^10 + 39828085965*n^9 + 93825035775*n^8 + 147951032109*n^7 + 150478534491*n^6 + 86482913102*n^5 + 11547320420*n^4 - 18788310824*n^3 - 12713618176*n^2 - 3178474112*n - 272670720)*a(n+1) - 8*n*(2*n-1)*(2*n+1)*(2*n+3)*(118125*n^10 + 2489625*n^9 + 23107950*n^8 + 124239510*n^7 + 427851585*n^6 + 984186117*n^5 + 1527319428*n^4 + 1572814284*n^3 + 1022652512*n^2 + 375620224*n + 58236160)*a(n) = 0. (End)
a(n) ~ 25 * 2^(4*n - 3) / (9*Pi*n^2) . - Vaclav Kotesovec, Mar 08 2023
Showing 1-3 of 3 results.