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%I A197654 #96 Mar 07 2020 06:47:19
%S A197654 1,5,1,31,62,1,121,1215,363,1,341,13504,20256,1364,1,781,96875,500000,
%T A197654 193750,3905,1,1555,501066,7321875,9762500,1252665,9330,1,2801,
%U A197654 2033647,72656661,262609375,121094435,6100941,19607,1
%N A197654 Triangle by rows T(n,k), showing the number of meanders with length 5(n+1) and containing 5(k+1) L's and 5(n-k) R's, where L's and R's denote arcs of equal length and a central angle of 72 degrees which are positively or negatively oriented.
%C A197654 Definition of a meander:
%C A197654 A binary curve C is a triple (m, S, dir) such that:
%C A197654 (a) S is a list with values in {L,R} which starts with an L,
%C A197654 (b) dir is a list of m different values, each value of S being allocated a value of dir,
%C A197654 (c) consecutive L's increment the index of dir,
%C A197654 (d) consecutive R's decrement the index of dir,
%C A197654 (e) the integer m>0 divides the length of S.
%C A197654 Then C is a meander if each value of dir occurs length(S)/m times.
%C A197654 Let T(m,n,k) = number of meanders (m, S, dir) in which S contains m(k+1) L's and m(n-k) R's, so that length(S) = m(n+1).
%C A197654 For this sequence, m = 5, T(n,k) = T(5,n,k).
%C A197654 The values in the triangle were proved by brute force for 0 <= n <= 6. The formulas have not yet been proved in general.
%C A197654 The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595 and A197653. The first column seems to be A053699. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A152031 and to start with the second number of A152031. Row sums are in A198257.
%C A197654 The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 29. - _Susanne Wienand_, Jul 01 2015
%H A197654 Alois P. Heinz, <a href="/A197654/b197654.txt">Rows n = 0..140, flattened</a>
%H A197654 Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/Meander">Meanders and walks on the circle</a>.
%H A197654 Susanne Wienand, <a href="https://oeis.org/wiki/User:Susanne_Wienand#Example_of_a_meander_counted_by_A197654">Example of a meander counted by A197654</a>
%F A197654 Recursive formula (conjectured):
%F A197654 T(n,k) = T(5,n,k) = T(1,n,k)^5 + T(1,n,k)*T(4,n,n-1-k), 0 <= k < n
%F A197654 T(5,n,n) = 1 k = n
%F A197654 T(4,n,k) = T(1,n,k)^4 + T(1,n,k) * T(3,n,n-1-k), 0 <= k < n
%F A197654 T(4,n,n) = 1 k = n
%F A197654 T(3,n,k) = T(1,n,k)^3 + T(1,n,k) * T(2,n,n-1-k), 0 <= k < n
%F A197654 T(3,n,n) = 1 k = n
%F A197654 T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k), 0<= k < n
%F A197654 T(2,n,n) = 1 k = n
%F A197654 T(4,n,k) = A197653
%F A197654 T(3,n,k) = A194595
%F A197654 T(2,n,k) = A103371
%F A197654 T(1,n,k) = A007318 (Pascal's Triangle)
%F A197654 Closed formula (conjectured): T(n,n) = 1, k = n
%F A197654 T(n,k) = A + B + C + D + E, k < n
%F A197654 A = (C(n,k))^5
%F A197654 B = (C(n,k))^4 * C(n,n-1-k)
%F A197654 C = (C(n,k))^3 *(C(n,n-1-k))^2
%F A197654 D = (C(n,k))^2 *(C(n,n-1-k))^3
%F A197654 E = C(n,k) *(C(n,n-1-k))^4 [Susanne Wienand]
%F A197654 Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,4). [_Peter Luschny_, Oct 20 2011]
%F A197654 T(n,k) = A198064(n+1,k+1)C(n,k)^5/(k+1)^4. [_Peter Luschny_, Oct 29 2011]
%F A197654 T(n,k) = h(n,k)*binomial(n,k)^5, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 5. [_Peter Luschny_, Nov 24 2011]
%e A197654 For n = 5 and k = 2, T(n,k) = 500000
%e A197654 Example for recursive formula:
%e A197654 T(1,5,2) = 10
%e A197654 T(4,5,5-1-2) = T(4,5,2) = 40000
%e A197654 T(5,5,2) = 10^5 + 10*40000 = 500000
%e A197654 Example for closed formula:
%e A197654 T(5,2) = A + B + C + D + E
%e A197654 A = 10^5
%e A197654 B = 10^4 * 10
%e A197654 C = 10^3 * 10^2
%e A197654 D = 10^2 * 10^3
%e A197654 E = 10 * 10^4
%e A197654 T(5,2) = 5 * 10^5 = 500000
%e A197654 Some examples of list S and allocated values of dir if n = 5 and k = 2:
%e A197654 Length(S) = (5+1)*5 = 30 and S contains (2+1)*5 = 15 Ls.
%e A197654 S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R,R,R,R,R,R,R,R
%e A197654 dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,0,4,3,2,1,0,4,3,2,1,0,4,3,2,1
%e A197654 S: L,L,L,L,L,L,L,R,R,L,L,R,R,R,L,R,R,R,L,R,L,L,L,R,R,L,R,R,R,R
%e A197654 dir: 1,2,3,4,0,1,2,2,1,1,2,2,1,0,0,0,4,3,3,3,3,4,0,0,4,4,4,3,2,1
%e A197654 S: L,L,L,L,L,R,L,L,L,L,L,R,L,R,R,R,R,R,R,R,R,R,R,L,L,L,L,R,R,R
%e A197654 dir: 1,2,3,4,0,0,0,1,2,3,4,4,4,4,3,2,1,0,4,3,2,1,0,0,1,2,3,3,2,1
%e A197654 Each value of dir occurs 30/5 = 6 times.
%e A197654 The triangle begins:
%e A197654 1,
%e A197654 5, 1,
%e A197654 31, 62, 1,
%e A197654 121, 1215, 363, 1,
%e A197654 341, 13504, 20256, 1364, 1,
%e A197654 781, 96875, 500000, 193750, 3905, 1,
%e A197654 ...
%p A197654 A197654 := (n,k)->(k^4+2*k^3*(1-n)+2*k^2*(2+n+2*n^2)+k*(3+n-n^2-3*n^3)+ n^4+n^3+n^2+n+1)*binomial(n,k)^5/(1+k)^4;
%p A197654 seq(print(seq(A197654(n, k), k=0..n)), n=0..7); # _Peter Luschny_, Oct 20 2011
%t A197654 T[n_, k_] := (k^4 + 2*k^3*(1-n) + 2*k^2*(2+n+2*n^2) + k*(3+n-n^2-3*n^3) + n^4+n^3+n^2+n+1)*Binomial[n, k]^5/(1+k)^4; Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Feb 20 2017, after _Peter Luschny_ *)
%o A197654 (C#) static int[] A197654_row(int r) { return GenBinomial(r, 5); } // The function GenBinomial(r, s) is defined in A194595.
%o A197654 // This C#-program causes numerical overflow for results larger than 2147483647. - _Susanne Wienand_, Jul 01 2015
%o A197654 (Sage)
%o A197654 def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
%o A197654 def A197654(n,k) : return S(4,n,k)
%o A197654 for n in (0..5) : print([A197654(n,k) for k in (0..n)]) # Peter Luschny, Oct 24 2011
%o A197654 (PARI) A197654(n,k) = {if(n ==1+2*k,5,(1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n))*binomial(n,k)^5} \\ _Peter Luschny_, Nov 24 2011
%Y A197654 Cf. A000012, A007318, A053699, A103371, A152031, A194595, A197653, A197655, A198257.
%K A197654 nonn,tabl
%O A197654 0,2
%A A197654 _Susanne Wienand_, Oct 19 2011