[go: up one dir, main page]

login
Search: a136621 -id:a136621
     Sort: relevance | references | number | modified | created      Format: long | short | data
Parker's partition triangle T(n,k) read by rows (n >= 1 and 0 <= k <= n-1).
+10
15
1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 9, 20, 11, 1, 1, 13, 48, 51, 18, 1, 1, 20, 100, 169, 112, 26, 1, 1, 28, 194, 461, 486, 221, 38, 1, 1, 40, 352, 1128, 1667, 1210, 411, 52, 1, 1, 54, 615, 2517, 4959, 5095, 2761, 720, 73, 1, 1, 75, 1034, 5288, 13241, 18084, 13894, 5850, 1221, 97, 1
OFFSET
1,5
COMMENTS
The entries in row n are the coefficients of q^(k*(n+1)) in the q-binomial coefficient [2n, n], where k runs from 0 to n-1. - James A. Sellers
T(n,k) is the number of partitions of k*(n+1) into at most n parts each no bigger than n (see the links). - Petros Hadjicostas, May 30 2020
Named after the American mathematician Ernest Tilden Parker (1926-1991). - Amiram Eldar, Jun 20 2021
LINKS
Richard K. Guy, Parker's permutation problem involves the Catalan numbers, Preprint, 1992. (Annotated scanned copy)
Richard K. Guy, Parker's permutation problem involves the Catalan numbers, Amer. Math. Monthly, Vol. 100, No. 3 (1993), pp. 287-289.
Wikipedia, E. T. Parker.
EXAMPLE
Triangle T(n,k) (with rows n >= 1 and columns k = 0..n-1) starts:
1;
1, 1;
1, 3, 1;
1, 5, 7 1;
1, 9, 20, 11, 1;
1, 13, 48, 51, 18, 1;
...
MAPLE
b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(t*i
<n, 0, b(n, i-1, t)+b(n-i, min(i, n-i), t-1)))
end:
T:= (n, k)-> b(k*(n+1), n$2):
seq(seq(T(n, k), k=0..n-1), n=1..12); # Alois P. Heinz, May 30 2020
MATHEMATICA
s[n_] := s[n] = Series[Product[(1-q^(2n-k)) / (1-q^(k+1)), {k, 0, n-1}], {q, 0, n^2}];
t[n_, k_] := SeriesCoefficient[s[n], k(n+1)];
Flatten[Table[t[n, k], {n, 1, 12}, {k, 0, n-1}]] (* Jean-François Alcover, Jan 27 2012 *)
b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[n < 0 || t i < n, 0, b[n, i - 1, t] + b[n - i, Min[i, n - i], t - 1]]];
T[n_, k_] := b[k(n+1), n, n];
Table[T[n, k], {n, 1, 12}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Nov 27 2020, after Alois P. Heinz *)
PROG
(PARI) T(n, k) = #partitions(k*(n+1), n, n);
for (n=1, 10, for (k=0, n-1, print1(T(n, k), ", "); ); print(); ); \\ Petros Hadjicostas, May 30 2020
/* Second program, courtesy of G. C. Greubel */
T(n, k) = polcoeff(prod(j=0, n-1, (1-q^(2*n-j))/(1-q^(j+1)) ), k*(n+1) );
vector(12, n, vector(n, k, T(n, k-1))); \\ Petros Hadjicostas, May 31 2020
CROSSREFS
Cf. A000108 (row sums), A136621 (mirror image).
KEYWORD
nonn,tabl,easy,nice
EXTENSIONS
More terms from James A. Sellers
Offset corrected by Alois P. Heinz, May 30 2020
STATUS
approved
Central elements in Parker's partition triangle.
+10
5
1, 3, 20, 169, 1667, 18084, 208960, 2527074, 31630390, 406680465, 5342750699, 71442850111, 969548468960, 13323571588607, 185072895183632, 2594890728951909, 36681505784903758, 522291180086851188, 7484621370716999785, 107876522368295972285, 1562916545414144667559
OFFSET
0,2
LINKS
R. K. Guy, Parker's permutation problem involves the Catalan numbers, Amer. Math. Monthly 100 (1993), 287-289.
FORMULA
a(n) = coefficient of q^((m^2-1)/2) = q(2*n*(n+1)) in the q-binomial coefficient [2*m, m] = [2*(2*n+1), 2*n+1], where m = 2*n+1. [Corrected by Petros Hadjicostas, May 30 2020]
a(n) is the number of partitions of 2*n*(n+1) into at most 2*n+1 parts each no bigger than 2*n+1. - Petros Hadjicostas, May 30 2020
MAPLE
b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(t*i
<n, 0, b(n, i-1, t)+b(n-i, min(i, n-i), t-1)))
end:
a:= n-> b(2*n*(n+1), 2*n+1$2):
seq(a(n), n=0..20); # Alois P. Heinz, May 30 2020
MATHEMATICA
a[n_] := SeriesCoefficient[QBinomial[2(2n+1), 2n+1, q], {q, 0, 2n(n+1)}];
Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Aug 19 2019 *)
CROSSREFS
KEYWORD
easy,nonn,nice
EXTENSIONS
a(18)-a(20) from Alois P. Heinz, May 30 2020
STATUS
approved
Partial sums of the irregular table A136624.
+10
1
1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 3, 5, 2, 1, 1, 3, 8, 4, 2, 1, 1, 3, 9, 10, 4, 2, 1, 1, 3, 9, 17, 8, 4, 2, 1, 1, 3, 9, 23, 16, 8, 4, 2, 1, 1, 3, 9, 27, 28, 14, 8, 4, 2, 1, 1, 3, 9, 28, 43, 26, 14, 8, 4, 2, 1, 1, 3, 9, 28, 60, 41, 24, 14, 8, 4, 2
OFFSET
1,6
COMMENTS
A129176 can also be viewed as partial sums, but are perpendicular to the sequences of A136622.
EXAMPLE
A136624 begins
1
...1
......2...1
..........2...3...3...1
..............2...2...6...7
..................2...2...4
......................2...2
..........................2
therefore this sequence begins
1...1...1...1...1...1...1...1
....1...1...1...1...1...1...1
........2...3...3...3...3...3
............2...5...8...9...9
................2...4..10..17
....................2...4...8
........................2...4
............................2
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Alford Arnold, Jan 29 2008
STATUS
approved

Search completed in 0.006 seconds