A339358 Maximum number of copies of a 1234567 permutation pattern in an alternating (or zig-zag) permutation of length n + 11.
32, 64, 320, 576, 1696, 2816, 6400, 9984, 19392, 28800, 50304, 71808, 116160, 160512, 244992, 329472, 480480, 631488, 887744, 1144000, 1560416, 1976832, 2629120, 3281408, 4271488, 5261568, 6723840, 8186112, 10294656, 12403200, 15379968, 18356736, 22480800, 26604864
Offset: 1
Keywords
Examples
a(1) = 32. The alternating permutation of length 1+11=12 with the maximum number of copies of 1234567 is 132547698(11)(10)(12). The 32 copies are 12468(10)(12), 12469(10)(12), 12478(10)(12), 12479(10)(12), 12568(10)(12), 12569(10)(12), 12578(10)(12), 12579(10)(12), 13468(10)(12), 13469(10)(12), 13478(10)(12), 13479(10)(12), 13568(10)(12), 13569(10)(12), 13578(10)(12), 13579(10)(12), 12468(11)(12), 12469(11)(12), 12478(11)(12), 12479(11)(12), 12568(11)(12), 12569(11)(12), 12578(11)(12), 12579(11)(12), 13468(11)(12), 13469(11)(12), 13478(11)(12), 13479(11)(12), 13568(11)(12), 13569(11)(12), 13578(11)(12), and 13579(11)(12).
Links
- Lara Pudwell, From permutation patterns to the periodic table, Notices of the American Mathematical Society. 67.7 (2020), 994-1001.
Crossrefs
Cf. A168380.
Programs
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Maple
A339358 := proc(n) nhalf := ceil(n/2) ; if type(n,'even') then 128*binomial(nhalf+6,7)-64*binomial(nhalf+5,6) ; else 128*binomial(nhalf+4,7)+128*binomial(nhalf+4,6)+32*binomial(nhalf+4,5) ; end if; end proc: seq(A339358(n),n=1..40) ; # R. J. Mathar, Jan 11 2024
Formula
a(2n) = 64*A050486(n-1) = 128*C(n+6,7) - 64*C(n+5,6).
a(2n-1) = 128*C(n+4,7) + 128*C(n+4,6) + 32*C(n+4,5).
D-finite with recurrence (-n+1)*a(n) +2*a(n-1) +16*a(n-2) +2*a(n-3) +(n+7)*a(n-4)=0. - R. J. Mathar, Jan 11 2024
Comments