cp's OEIS Frontend

A118340 Pendular triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + T(n-1,k), for n>=k>0, with T(n,0) = 1 and T(n,n) = 0^n.

%I A118340 #30 Feb 18 2021 00:28:43
%S A118340 1,1,0,1,1,0,1,2,1,0,1,3,4,1,0,1,4,9,5,1,0,1,5,15,20,6,1,0,1,6,22,48,
%T A118340 28,7,1,0,1,7,30,85,113,37,8,1,0,1,8,39,132,282,169,47,9,1,0,1,9,49,
%U A118340 190,519,688,237,58,10,1,0,1,10,60,260,837,1762,1074,318,70,11,1,0
%N A118340 Pendular triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + T(n-1,k), for n>=k>0, with T(n,0) = 1 and T(n,n) = 0^n.
%C A118340 Definitions. A pendular triangle is a triangle in which row n is generated from the pendular sums of row n-1. Pendular sums of a row are partial sums taken in back-and-forth order, starting with the leftmost term, jumping to the rightmost term, back to the leftmost unused term, then forward to the rightmost unused term, etc.
%C A118340 In each pass, the partial sum is placed in the new row directly under the term most recently used in the sum. Continue in this way until all the terms of the prior row have been used and then complete the new row by appending a zero at the end. Pendular sums are so named because the process resembles a swinging pendulum that slows down on each pass until it eventually comes to rest in the center.
%C A118340 In the simplest case, pendular triangles obey the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + p*T(n-1,k), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n, for some fixed number p.
%C A118340 In which case the g.f. G=G(x) of the central terms satisfies: G = 1 - p*x*G + p*x*G^2 + x*G^3.
%C A118340 More generally, a pendular triangle is defined by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + Sum_{j>=1} p(j)*T(n-1,k-1+j), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n.
%C A118340 Remarkably, the g.f. G=G(x) of the central terms satisfies: G = 1 + x*G^3 + Sum_{j>=1} p(j)*x^j*[G^(2*j) - G^(2*j-1)].
%C A118340 Further, the g.f. of the m-th lower semi-diagonal equals G(x)^(m+1) for m>=0, where the m-th semi-diagonal consists of those terms located at m rows below the central terms.
%C A118340 For variants of pendular triangles, the main diagonal may be nonzero, but then the g.f.s of the semi-diagonals are more complex.
%H A118340 G. C. Greubel, <a href="/A118340/b118340.txt">Rows n = 0..100 of the triangle, flattened</a>(Rows n = 0..20 from Paul D. Hanna)
%F A118340 T(2*n+m,n) = [A108447^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A108447; compare semi-diagonals to the diagonals of convolution triangle A118343.
%e A118340 Row 6 equals the pendular sums of row 5:
%e A118340   [1,  4,  9,  5,  1,  0], where the sums proceed as follows:
%e A118340   [1, __, __, __, __, __], T(6,0) = T(5,0) = 1;
%e A118340   [1, __, __, __, __,  1], T(6,5) = T(6,0) + T(5,5) = 1 + 0 = 1;
%e A118340   [1,  5, __, __, __,  1], T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
%e A118340   [1,  5, __, __,  6,  1], T(6,4) = T(6,1) + T(5,4) = 5 + 1 = 6;
%e A118340   [1,  5, 15, __,  6,  1], T(6,2) = T(6,4) + T(5,2) = 6 + 9 = 15;
%e A118340   [1,  5, 15, 20,  6,  1], T(6,3) = T(6,2) + T(5,3) = 15 + 5 = 20;
%e A118340   [1,  5, 15, 20,  6,  1, 0] finally, append a zero to obtain row 6.
%e A118340 Triangle begins:
%e A118340   1;
%e A118340   1,  0;
%e A118340   1,  1,  0;
%e A118340   1,  2,  1,   0;
%e A118340   1,  3,  4,   1,    0;
%e A118340   1,  4,  9,   5,    1,    0;
%e A118340   1,  5, 15,  20,    6,    1,    0;
%e A118340   1,  6, 22,  48,   28,    7,    1,    0;
%e A118340   1,  7, 30,  85,  113,   37,    8,    1,   0;
%e A118340   1,  8, 39, 132,  282,  169,   47,    9,   1,  0;
%e A118340   1,  9, 49, 190,  519,  688,  237,   58,  10,  1,  0;
%e A118340   1, 10, 60, 260,  837, 1762, 1074,  318,  70, 11,  1, 0;
%e A118340   1, 11, 72, 343, 1250, 3330, 4404, 1568, 413, 83, 12, 1, 0; ...
%e A118340 Central terms are T(2*n,n) = A108447(n);
%e A118340 semi-diagonals form successive self-convolutions of the central terms:
%e A118340 T(2*n+1,n) = A118341(n) = [A108447^2](n),
%e A118340 T(2*n+2,n) = A118342(n) = [A108447^3](n).
%t A118340 T[n_, k_, p_]:= T[n,k,p] = If[n<k || k<0, 0, If[k==0, 1, If[k==n, 0, If[n<=2*k, T[n,n-k-1,p] + p*T[n-1,k,p], T[n,n-k,p] + T[n-1,k,p] ]]]];
%t A118340 Table[T[n,k,1], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Feb 17 2021 *)
%o A118340 (PARI) {T(n,k) = if(n<k || k<0,0, if(k==0,1, if(n==k,0, if(n>2*k, T(n-1,k) + T(n,n-k), T(n-1,k) + T(n,n-1-k)))))}
%o A118340 for(n=0,12, for(k=0,n, print1(T(n,k),", "));print(""))
%o A118340 (Sage)
%o A118340 @CachedFunction
%o A118340 def T(n, k, p):
%o A118340     if (k<0 or n<k): return 0
%o A118340     elif (k==0): return 1
%o A118340     elif (k==n): return 0
%o A118340     elif (n>2*k): return T(n,n-k,p) + T(n-1,k,p)
%o A118340     else: return T(n, n-k-1, p) + p*T(n-1, k, p)
%o A118340 flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Feb 17 2021
%o A118340 (Magma)
%o A118340 function T(n,k,p)
%o A118340   if k lt 0 or n lt k then return 0;
%o A118340   elif k eq 0 then return 1;
%o A118340   elif k eq n then return 0;
%o A118340   elif n gt 2*k then return T(n,n-k,p) + T(n-1,k,p);
%o A118340   else return T(n,n-k-1,p) + p*T(n-1,k,p);
%o A118340   end if;
%o A118340   return T;
%o A118340 end function;
%o A118340 [T(n,k,1): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Feb 17 2021
%Y A118340 Cf. A108447 (central terms), A118341, A118343.
%Y A118340 variants: A118344 (Catalan), A118362 (ternary), A118350, A118355.
%Y A118340 Cf. A167763 (p=0), this sequence (p=1), A118345 (p=2), A118350 (p=3).
%K A118340 nonn,tabl
%O A118340 0,8
%A A118340 _Paul D. Hanna_, Apr 25 2006