This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A197655 #73 Feb 27 2020 04:18:36
%S A197655 1,6,1,63,126,1,364,4374,1092,1,1365,85120,127680,5460,1,3906,984375,
%T A197655 6000000,1968750,19530,1,9331,7562646,157828125,210437500,18906615,
%U A197655 55986,1,19608,42824236,2628749256,11029593750,4381248760,128472708,137256,1
%N A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*6 and containing (k+1)*6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 degrees which are positively or negatively oriented.
%C A197655 Definition of a meander:
%C A197655 A binary curve C is a triple (m, S, dir) such that
%C A197655 (a) S is a list with values in {L,R} which starts with an L,
%C A197655 (b) dir is a list of m different values, each value of S being allocated a value of dir,
%C A197655 (c) consecutive Ls increment the index of dir,
%C A197655 (d) consecutive Rs decrement the index of dir,
%C A197655 (e) the integer m>0 divides the length of S and
%C A197655 (f) C is a meander if each value of dir occurs length(S)/m times.
%C A197655 For this sequence, m = 6.
%C A197655 The values in the triangle are proved by brute force for 0 <= n <= 5. The formulas are not yet proved in general.
%C A197655 The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595, A197653 and A197654. For n > 0, the first column seems to be A053700. The diagonal right hand is A000012. Row sums are in A198258.
%C A197655 The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 19. - _Susanne Wienand_, Jul 04 2015
%H A197655 Susanne Wienand, <a href="/A197655/b197655.txt">Table of n, a(n) for n = 0..209</a>
%H A197655 Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/Meander">Meanders and walks on the circle</a>.
%F A197655 recursive formula (conjectured):
%F A197655 T(n,k) = T(6,n,k)
%F A197655 T(6,n,k) = T(1,n,k)^6 + T(1,n,k)*T(5,n,n-1-k), 0 <= k < n
%F A197655 T(6,n,n) = 1 k = n
%F A197655 T(5,n,k) = T(1,n,k)^5 + T(1,n,k)*T(4,n,n-1-k), 0 <= k < n
%F A197655 T(5,n,n) = 1 k = n
%F A197655 T(4,n,k) = T(1,n,k)^4 + T(1,n,k)*T(3,n,n-1-k), 0 <= k < n
%F A197655 T(4,n,n) = 1 k = n
%F A197655 T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
%F A197655 T(3,n,n) = 1 k = n
%F A197655 T(2,n,k) = T(1,n,k)^2 + T(1,n,k)*T(1,n,n-1-k), 0 <= k < n
%F A197655 T(2,n,n) = 1 k = n
%F A197655 T(5,n,k) = A197654
%F A197655 T(4,n,k) = A197653
%F A197655 T(3,n,k) = A194595
%F A197655 T(2,n,k) = A103371
%F A197655 T(1,n,k) = A007318 (Pascal's Triangle)
%F A197655 closed formula (conjectured): T(n,n) = 1, k = n
%F A197655 T(n,k) = A + B + C + D + E + F, k < n
%F A197655 A = (C(n,k))^6
%F A197655 B = (C(n,k))^5 * C(n,n-1-k)
%F A197655 C = (C(n,k))^4 *(C(n,n-1-k))^2
%F A197655 D = (C(n,k))^3 *(C(n,n-1-k))^3
%F A197655 E = (C(n,k))^2 *(C(n,n-1-k))^4
%F A197655 F = C(n,k) *(C(n,n-1-k))^5
%F A197655 [Susanne Wienand]
%F A197655 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,5). (Cf. A103371, A194595, A197653). [_Peter Luschny_, Oct 21 2011]
%F A197655 T(n,k) = A198065(n+1,k+1)C(n,k)^6/(k+1)^5. [_Peter Luschny_, Oct 29 2011]
%F A197655 T(n,k) = h(n,k)*binomial(n,k)^6, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 6. [_Peter Luschny_, Nov 24 2011]
%e A197655 For n = 4 and k = 2, T(n,k) = 127680
%e A197655 Example for recursive formula:
%e A197655 T(1,4,2) = 6
%e A197655 T(5,4,4-1-2) = T(5,4,1) = 13504
%e A197655 T(6,4,2) = 6^6 + 6*13504 = 127680
%e A197655 Example for closed formula:
%e A197655 T(4,2) = A + B + C + D + E + F
%e A197655 A = 6^6 = 46656
%e A197655 B = 6^5 * 4 = 31104
%e A197655 C = 6^4 * 4^2 = 20736
%e A197655 D = 6^3 * 4^3 = 13824
%e A197655 E = 6^2 * 4^4 = 9216
%e A197655 F = 6 * 4^5 = 6144
%e A197655 T(4,2) = 127680
%e A197655 Some examples of list S and allocated values of dir if n = 4 and k = 2:
%e A197655 Length(S) = (4+1)*6 = 30 and S contains (2+1)*6 = 18 Ls.
%e A197655 S: L,L,L,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
%e A197655 dir: 1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5,0,0,5,4,3,2,1,0,5,4,3,2,1
%e A197655 S: L,L,L,L,L,L,L,L,L,L,R,L,L,R,L,R,R,R,L,R,R,L,R,R,R,L,L,R,R,L
%e A197655 dir: 1,2,3,4,5,0,1,2,3,4,4,4,5,5,5,5,4,3,3,3,2,2,2,1,0,0,1,1,0,0
%e A197655 S: L,L,L,L,R,L,L,R,L,L,R,L,R,R,L,L,L,L,L,R,R,L,L,L,R,R,R,R,L,R
%e A197655 dir: 1,2,3,4,4,4,5,5,5,0,0,0,0,5,5,0,1,2,3,3,2,2,3,4,4,3,2,1,1,1
%e A197655 Each value of dir occurs 30/6 = 5 times.
%p A197655 A197655 := (n,k) -> (1+n)*(1+3*k+3*k^2-n-3*k*n+n^2)*(1+k+k^2+n-k*n+n^2)* binomial(n,k)^6/(1+k)^5; seq(print(seq(A197655(n, k), k=0..n)), n=0..7); # _Peter Luschny_, Oct 21 2011
%t A197655 T[n_, k_] := (1 + n)(1 + 3k + 3k^2 - n - 3k*n + n^2)(1 + k + k^2 + n - k*n + n^2) Binomial[n, k]^6/(1 + k)^5;
%t A197655 Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 30 2018, after _Peter Luschny_ *)
%o A197655 (C#) static int[] A197655_row(int r) { return GenBinomial(r, 6); } // The function GenBinomial(r, s) is defined in A194595.
%o A197655 // This C#-program causes numerical overflow for results
%o A197655 // larger than 2147483647. - _Susanne Wienand_, Jul 04 2015
%o A197655 (Sage)
%o A197655 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 A197655 def A197655(n,k) : return S(5,n,k)
%o A197655 for n in (0..5) : print([A197655(n,k) for k in (0..n)]) # _Peter Luschny_, Oct 24 2011
%o A197655 (PARI)
%o A197655 A197655(n,k) = {if(n==1+2*k,6,(1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n))*binomial(n,k)^6} \\ _Peter Luschny_, Nov 24 2011
%Y A197655 Cf. A000012, A007318, A053700, A103371, A194595, A197653, A197654, A198258.
%K A197655 nonn,tabl
%O A197655 0,2
%A A197655 _Susanne Wienand_, Oct 19 2011