Joshua Swanson - cp's OEIS frontend

User: Joshua Swanson

Joshua Swanson has authored 2 sequences.

A368881 a(n) = binomial(n+3, 4) + binomial(n+1, 3) + 1.

1, 2, 7, 20, 46, 91, 162, 267, 415, 616, 881, 1222, 1652, 2185, 2836, 3621, 4557, 5662, 6955, 8456, 10186, 12167, 14422, 16975, 19851, 23076, 26677, 30682, 35120, 40021, 45416, 51337, 57817, 64890, 72591, 80956, 90022, 99827, 110410, 121811, 134071

Offset: 0

Joshua Swanson, Jan 08 2024

The number of bigrassmannian permutations in the type B hyperoctahedral group of order 2^n*n!, i.e., those with a unique left and right type B descent or the identity. This can be characterized by avoiding 18 signed permutation patterns.

			For n=2, all eight 2 X 2 signed permutation matrices are bigrassmannian except the negative of the identity matrix, or equivalently the one with window notation [-1 -2], so a(2) = 7.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Joshua Swanson, Bigrassmannians and pattern avoidance notes.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A050407.

It appears that this is equal to {A005712}+1, also ({A212039}+2)/3 .

Table[Binomial[n + 3, 4] + Binomial[n + 1, 3] + 1, {n, 0, 20}]
LinearRecurrence[{5,-10,10,-5,1},{1,2,7,20,46},50] (* Harvey P. Dale, Jan 21 2025 *)

def A368881(n): return 1+(n*(n*(n*(n + 10) + 11) + 2))//24 # Chai Wah Wu, Jan 27 2024

a(n) = (1/24)*(n^4 + 10*n^3 + 11*n^2 + 2*n + 24).
G.f.: (x^4 - 5x^3 + 7x^2 - 3x + 1)/(1-x)^5.
E.g.f.: exp(x)*(24 + 24*x + 48*x^2 + 16*x^3 + x^4)/24. - Stefano Spezia, Jan 09 2024

A338621 Triangle read by rows: A(n, k) is the number of partitions of n with "aft" value k (see comments).

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 3, 2, 2, 4, 3, 2, 2, 4, 5, 2, 2, 2, 4, 6, 7, 1, 2, 2, 4, 6, 9, 6, 1, 2, 2, 4, 6, 10, 11, 7, 2, 2, 4, 6, 10, 13, 14, 5, 2, 2, 4, 6, 10, 14, 19, 15, 5, 2, 2, 4, 6, 10, 14, 21, 22, 17, 3, 2, 2, 4, 6, 10, 14, 22, 27, 29, 17, 2, 2, 2, 4, 6, 10, 14, 22, 29, 36, 33, 17

Offset: 0

Joshua Swanson, Nov 04 2020

The "aft" of an integer partition is the number of cells minus the larger of the number of parts or the largest part. For example, aft(4, 2, 2) = 8-4 = 4 = aft(3, 3, 1, 1).
Columns stabilize to twice the partition numbers: A(n, k) = 2p(n) = A139582(n) if n > 2k.
Row sums are partition numbers A000041.
Maximum value of k in row n is n - ceiling(sqrt(n)) = (n-1) - floor(sqrt(n-1)) = A028391(n-1).

			A(6, 2) = 4 since there are four partitions with 6 cells and aft 2, namely (4, 2), (2, 2, 1, 1), (4, 1, 1), (3, 1, 1, 1).
Triangle starts:
  1;
  1;
  2;
  2, 1;
  2, 2, 1;
  2, 2, 3;
  2, 2, 4, 3;
  2, 2, 4, 5,  2;
  2, 2, 4, 6,  7,  1;
  2, 2, 4, 6,  9,  6,  1;
  2, 2, 4, 6, 10, 11,  7;
  2, 2, 4, 6, 10, 13, 14,  5;
  2, 2, 4, 6, 10, 14, 19, 15,  5;
  2, 2, 4, 6, 10, 14, 21, 22, 17,  3;
  2, 2, 4, 6, 10, 14, 22, 27, 29, 17,  2;
  2, 2, 4, 6, 10, 14, 22, 29, 36, 33, 17,  1;
  2, 2, 4, 6, 10, 14, 22, 30, 41, 45, 39, 15,  1;
  2, 2, 4, 6, 10, 14, 22, 30, 43, 52, 57, 41, 14;
  2, 2, 4, 6, 10, 14, 22, 30, 44, 57, 69, 67, 47, 11;
  2, 2, 4, 6, 10, 14, 22, 30, 44, 59, 76, 85, 81, 46, 9; ...

S. C. Billey, M. Konvalinka, and J. P. Swanson, Asymptotic normality of the major index on standard tableaux, Adv. in Appl. Math. 113 (2020).

S. C. Billey, M. Konvalinka, and J. P. Swanson, Asymptotic normality of the major index on standard tableaux, arXiv:1905.00975 [math.CO], 2019.
FindStat - Combinatorial Statistic Finder, The aft of an integer partition

Cf. A000041, A139582.

CoefficientList[
SeriesCoefficient[
  1 + Sum[If[r == 0, 1, 2] q^(r + 1) Sum[
      q^(2 s) t^s QBinomial[2 s + r, s, q t], {s, 0, 30}], {r, 0,
     30}], {q, 0, 20}], t]

Row(n)={if(n==0, [1], my(v=vector(n)); forpart(p=n, v[1+n-max(#p, p[#p])]++); Vecrev(Polrev(v)))}
{ for(n=1, 15, print(Row(n))) } \\ Andrew Howroyd, Nov 04 2020

G.f.: Sum_{lambda} t^aft(lambda) * q^|lambda| = 1 + Sum_{r >= 0} c_r * q^(r+1) * Sum_{s >= 0} q^(2*s) * t^s * [2*s + r, s]_(q*t) where c_0 = 1, c_r = 2 for r >= 1, and [a, b]_q is a Gaussian binomial coefficient (see A022166).

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User: Joshua Swanson

A368881 a(n) = binomial(n+3, 4) + binomial(n+1, 3) + 1.

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