A294175
a(n) = 2^(n-1) + ((1+(-1)^n)/4)*binomial(n, n/2) - binomial(n, floor(n/2)).
Original entry on oeis.org
0, 0, 1, 1, 5, 6, 22, 29, 93, 130, 386, 562, 1586, 2380, 6476, 9949, 26333, 41226, 106762, 169766, 431910, 695860, 1744436, 2842226, 7036530, 11576916, 28354132, 47050564, 114159428, 190876696, 459312152, 773201629, 1846943453, 3128164186, 7423131482
Offset: 0
For example, for n=5, a(5)=6 and the 6 subsets are {2}, {4}, {2,4}, {1,2,4}, {2,3,4}, {2,4,5}.
- Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO] (2023).
The following relate to compositions of n + 1 with alternating sum k < 0.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse:
A344612).
A345197 counts compositions by length and alternating sum.
Cf.
A000070,
A001700,
A007318,
A025047,
A032443,
A034871,
A106356,
A114121,
A126869,
A163493,
A344743,
A345908,
A289871,
A359066,
A359067.
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f:= gfun:-rectoproc({(8+8*n)*a(n)+(4*n+16)*a(1+n)+(-20-6*n)*a(n+2)+(-5-n)*a(n+3)+(5+n)*a(n+4), a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 1}, a(n), remember):
map(f, [$0..40]); # Robert Israel, Feb 12 2018
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f[n_] := 2^(n - 1) + ((1 + (-1)^n)/4) Binomial[n, n/2] - Binomial[n, Floor[n/2]]; Array[f, 38, 0] (* Robert G. Wilson v, Feb 10 2018 *)
Table[Length[Select[Tuples[{0,1},{n+1}],First[#]==1&&Count[#,0]>Count[#,1]&]],{n,0,10}] (* Gus Wiseman, Jul 22 2021 *)
A359067
a(2*n) = Sum_{k=0..n-1} binomial(2*n,k) binomial(2*n-1-k, n-1-k). a(2*n+1) = (Sum_{k=0..n} binomial(2*n+1,k) binomial(2*n-k, n-k)) - binomial(2*n-1, n).
Original entry on oeis.org
0, 1, 4, 7, 28, 49, 199, 351, 1436, 2561, 10499, 18943, 77617, 141569, 579149, 1066495, 4354780, 8085505, 32954635, 61616127, 250713893, 471556097, 1915928117, 3621830655, 14696701553, 27902803969, 113099318869, 215530668031, 872780984131, 1668644405249, 6751457741849
Offset: 1
For n = 3, the a(3) = 4 admissible pinnacle sets in S_3^D are {}, {1}, {2}, {3}.
- Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO] (2023).
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a := n -> if irem(n - 1, 2) = 1 then binomial(n, n/2 - 1)*hypergeom([n/2 + 1, -n/2 + 1], [n/2 + 2], -1) else binomial(n + 1, n/2 + 1/2)*hypergeom([n/2 + 1/2, -n/2 + 1/2], [n/2 + 3/2], -1)/2 - binomial(n - 2, n/2 - 1/2) fi:
seq(simplify(a(n)), n = 3..31); # Peter Luschny, Jan 03 2023
A363582
Number of admissible mesa sets among Stirling permutations of order n.
Original entry on oeis.org
1, 2, 3, 6, 12, 22, 44, 88, 169, 338, 676, 1322, 2644, 5288, 10433, 20866, 41732, 82736, 165472, 330944, 658012, 1316024, 2632048, 5242778, 10485556, 20971112, 41822049, 83644098, 167288196, 333885702, 667771404, 1335542808, 2667053601, 5334107202, 10668214404
Offset: 1
For n = 4, the a(4) = 6 admissible pinnacle sets for Stirling permutations of order 4 are {}, {2}, {3}, {4}, {2,4}, and {3,4}.
- Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, "Mesas of Stirling permutations," preprint.
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a:= proc(n) option remember; `if`(n<4, n, (2*n*(2*n-3)*
a(n-1)+27*(n-4)*(n-2)*(a(n-3)/2-a(n-4)))/(n*(2*n-3)))
end:
seq(a(n), n=1..45); # Alois P. Heinz, Jun 13 2023
Showing 1-3 of 3 results.
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