Peter Doyle has authored 2 sequences.
A216040
Number of permutations sortable using two parallel stacks.
Original entry on oeis.org
1, 1, 2, 6, 23, 103, 513, 2760, 15741, 93944, 581303, 3704045, 24180340, 161082639, 1091681427, 7508269793, 52302594344, 368422746908, 2620789110712, 18806093326963, 136000505625886, 990406677136685, 7258100272108212
Offset: 0
Up to n = 4, the only permutation that can't be sorted is 2341. This fails because after moving 2 to one stack, you must move 3 to the other stack, and now the 4 will block either the 2 or the 3. (If you use a double-ended queue instead of two stacks, then this permutation becomes sortable; cf. A182216.)
- Daniel Denton and Peter Doyle, Table of n, a(n) for n = 0..100
- Daniel Denton, Methods of computing deque sortable permutations given complete and incomplete information, arXiv:1208.1532 [math.CO], 2012.
- Andrew Elvey-Price and Anthony J. Guttmann, Permutations sortable by deques and by two stacks in parallel, arxiv:1508.02273 [math.CO], 2015-2016.
- Andrew Elvey-Price and Anthony J. Guttmann, Permutations sortable by deques and by two stacks in parallel, European Journal of Combinatorics, 59 (2017), 71-95.
A101842
Triangle read by rows: T(n,k), for k=-n..n-1, is the scaled (by 2^n n!) probability that the sum of n uniform [-1, 1] variables is between k and k+1.
Original entry on oeis.org
1, 1, 1, 3, 3, 1, 1, 7, 16, 16, 7, 1, 1, 15, 61, 115, 115, 61, 15, 1, 1, 31, 206, 626, 1056, 1056, 626, 206, 31, 1, 1, 63, 659, 2989, 7554, 11774, 11774, 7554, 2989, 659, 63, 1, 1, 127, 2052, 13308, 47349, 105099, 154624, 154624, 105099, 47349, 13308, 2052, 127, 1, 1, 255, 6297, 56935, 274677
Offset: 1
Triangle of T(n,k), k=-n..n-1 begins:
1, 1
1, 3, 3, 1
1, 7, 16, 16, 7, 1
1, 15, 61, 115, 115, 61, 15, 1
1, 31, 206, 626, 1056, 1056, 626, 206, 31, 1
1, 63, 659, 2989, 7554, 11774, 11774, 7554, 2989, 659, 63, 1
The sequence of polynomials starts:
p(0, x) = 1
p(1, x) = x + 1
p(2, x) = x^3 + 3*x^2 + 3*x + 1
p(3, x) = x^5 + 7*x^4 + 16*x^3 + 16*x^2 + 7*x + 1
p(4, x) = x^7 + 15*x^6 + 61*x^5 + 115*x^4 + 115*x^3 + 61*x^2 + 15*x + 1
- Peter Doyle, Myths about card shuffling, talk given at DIMACS Workshop on Puzzling Mathematics and Mathematical Puzzles: a Gathering in Honor of Peter Winkler's 60th Birthday, Rutgers University, Jun 08, 2007
- Paul D. Hanna, Table of n, a(n) for n = 1..2070 for rows 1..45 of flattened triangle
- R. M. Adin, F. Brenti and Y. Roichman, Descent numbers and major indices for the hyperoctahedral group, Adv. Appl. Math. 27 (2001), 210-224.
- R. Stanley, Eulerian partitions of a unit hypercube, in Higher Combinatorics (M. Aigner, ed.), Reidel, Dordrecht/Boston, 1977, p. 49.
See also
A008292 (Eulerian numbers).
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gf := (1 - x)/(exp((x + 1)*z) - x): ser := series(gf, z, 12):
for n from 0 to 6 do p := simplify(n!*(x - 1)^n*coeff(ser,z, n));
print(PolynomialTools:-CoefficientList(p, x)) od: # Peter Luschny, Jun 24 2019
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T[n_, k_] /; -n <= k <= n-1 := T[n, k] = (n-k) T[n-1, k-1] + T[n-1, k] + (n+k+1) T[n-1, k+1]; T[1, -1] = T[1, 0] = 1; T[, ] = 0;
Table[T[n, k], {n, 1, 8}, {k, -n, n-1}] (* Jean-François Alcover, Jun 27 2019 *)
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{T(n,k) = polcoeff( n!*polcoeff( -1 + (1-y) / (1 - y*exp( (1-y^2)*x +x*O(x^n) )) ,n,x),k,y)}
for(n=1,10,for(k=1,2*n,print1(T(n,k),", "));print("")) \\ Paul D. Hanna, Jan 03 2017
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