A298647 Number of Dyck paths of semilength n having maximal degree of asymmetry, namely n-1 for n>2 and 0 otherwise.
1, 1, 2, 2, 2, 8, 16, 52, 134, 432, 1248, 4104, 12648, 42464, 136512, 466568, 1545566, 5361632, 18165184, 63804952, 219997832, 780895392, 2730730176, 9780049008, 34598622616, 124873507904, 446068180608, 1620786592416, 5837657948832, 21336295622016, 77395590570240
Offset: 0
Keywords
Examples
a(1) = 1, counting UD; a(2) = 2 since both UDUD and UUDD have maximal degree of asymmetry 0; a(5) = 8 counting UDUUUUDDDD, UDUDUUUDDD, UDUUDDUUDD, UDUDUUDUDD, and their reflections in a vertical axis.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1676
Programs
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Maple
b:= proc(x, y, v) option remember; expand( `if`(min(y, v, x-max(y, v))<0, 0, `if`(x=0, 1, (l-> add(add( `if`(y=v+(j-i)/2, 1, z)*b(x-1, y+i, v+j), i=l), j=l))([-1, 1])))) end: a:= n-> lcoeff(add(b(n, j$2), j=0..n)): seq(a(n), n=0..40); # second Maple program: a:= proc(n) option remember; `if`(n<8, [1$2, 2$3, 8, 16, 52][n+1], (256*(n-4)*(n-5)*(n-8)*(147*n^5-1708*n^4+7165*n^3-12896*n^2+8882*n -1362)*a(n-6)-64*(294*n^7-6139*n^6+52088*n^5-227713*n^4+534530*n^3 -630478*n^2+295718*n-24240)*a(n-5)-16*(3675*n^8-96943*n^7+1072857 *n^6-6448749*n^5+22718880*n^4-46984656*n^3+53484228*n^2-28042052*n +3963360)*a(n-4)-24*(1470*n^7-30345*n^6+256611*n^5-1138707*n^4 +2807363*n^3-3706740*n^2+2239444*n-369120)*a(n-3)+4*(n-2)*(1470 *n^7-27958*n^6+211755*n^5-810763*n^4+1642737*n^3-1686751*n^2+790686 *n-150480)*a(n-2)+2*(n-1)*(882*n^6-14357*n^5+89078*n^4-262023*n^3 +370096*n^2-233344*n+64320)*a(n-1))/((147*n^5-2443*n^4+15467*n^3 -46109*n^2+63736*n-32160)*(n-1)*n*(n+1))) end: seq(a(n), n=0..40);
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Mathematica
b[x_, y_, v_] := b[x, y, v] = Expand[ If[Min[y, v, x - Max[y, v]]<0, 0, If[x==0, 1, Function[l, Sum[Sum[ If[y == v+(j-i)/2, 1, z] b[x-1, y+i, v+j], {i, l}], {j, l}]][{-1, 1}]]]]; a[n_] := With[{p = Sum[b[n, j, j], {j, 0, n}]}, Coefficient[p, z, Exponent[p, z]]]; a /@ Range[0, 40] (* Jean-François Alcover, Dec 30 2020, after Alois P. Heinz *)
Formula
a(n) = A298645(n,n-1) for n > 2.
a(n) ~ 2^(2*n + 3) / (Pi * n^3). - Vaclav Kotesovec, Mar 06 2018
Comments