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%I #38 Oct 02 2019 05:19:06
%S 1,1,1,1,2,2,1,1,3,5,6,6,5,3,1,1,4,9,15,21,26,29,30,30,29,26,21,15,9,
%T 4,1,1,5,14,29,50,76,105,135,165,194,220,241,256,265,269,270,270,269,
%U 265,256,241,220,194,165,135,105,76,50,29,14,5,1,1,6,20,49,99,175,280,415
%N Triangle read by rows of 2^n terms for n>=0: let S(n) denote the initial 2^n terms of the partial sums of row n; starting with a single '1' in row 0, generate row n+1 by concatenating S(n) with the terms of S(n) when read in reverse order.
%C Row sums (and central terms) form A028361: Product_{i=0..n-1} (2^i + 1).
%C I'm interested in the graph of S(n). It appears to tend to a limit curve if scaled appropriately, e.g., scaled to fit a [0,1] box by f_n(x) = T(n,[x*2^n])/A028361(n-1). In this setup I think that the limit curve f(x) satisfies f(0)=0, f(1-x)=f(x), f(1/2)=1, f'(x)=2f(2x) for x<=1/2. Is this equation solvable? - _Martin Fuller_, Aug 31 2007
%C From _N. J. A. Sloane_, Nov 13 2018: (Start)
%C Kenyon (1992) defines p_n(x) (n >= 0) to be the polynomial
%C p_n(x) = (1+x)*(1+x+x^2)*(1+x+x^2+x^3+x^4)*...*(1+x+...+x^(2^n)).
%C He shows among many other things that the coefficient of x^(floor(c*2^(n+1))) in p_n(x), for c in [0,1], is given by
%C (f(c)+o(1))*p_n(1)/2^(n+1),
%C where f : R -> R is a nonzero C^1 function satisfying
%C (i) support(f) is a subset of [0,1],
%C (ii) f(x) = f(1-x), and
%C (iii) f'(x) = 4*f(2*x) for 0 <= x <= 1/2.
%C These three properties define f uniquely up to multiplication by a scalar. Also f is C^oo, is nowhere analytic on [0,1], and is a "bump function", since its graph looks like a "bump".
%C This provides a fairly complete answer to _Martin Fuller_'s question above. (End)
%D Richard Kenyon, Infinite scaled convolutions, Preprint, 1992 (apparently unpublished)
%H Paul D. Hanna, <a href="/A131791/b131791.txt">Rows 0 to 11 of the triangle, flattened.</a>
%H Julien Clément, Antoine Genitrini, <a href="https://arxiv.org/abs/1907.06743">Binary Decision Diagrams: from Tree Compaction to Sampling</a>, arXiv:1907.06743 [cs.DS], 2019.
%F T(n, 2^(n-1)) = A028361(n-1) for n>=1.
%F T(n, 2^(n-2)) = A028362(n-1) for n>=2.
%F Sum_{k=0..2^n-1} (k+1)*T(n,k) = A028362(n+1) for n>=0.
%F G.f. of row n: Product_{j=0..n-1} (1 - x^(2^j+1))/(1-x). - _Paul D. Hanna_, Aug 09 2009
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 2, 1;
%e 1, 3, 5, 6, 6, 5, 3, 1;
%e 1, 4, 9, 15, 21, 26, 29, 30, 30, 29, 26, 21, 15, 9, 4, 1;
%e 1, 5, 14, 29, 50, 76, 105, 135, 165, 194, 220, 241, 256, 265, 269, 270, 270, 269, 265, 256, 241, 220, 194, 165, 135, 105, 76, 50, 29, 14, 5, 1; ...
%e ILLUSTRATION OF GENERATING METHOD.
%e From row 2: [1,2,2,1], take the partial sums: [1,3,5,6] and concatenate to this the terms in reverse order: [6,5,3,1] to obtain row 3: [1,3,5,6, 6,5,3,1].
%p p[-1]:=1:
%p lprint(seriestolist(series(p[-1],x,0)));
%p p[0]:=(1-x^2)/(1-x):
%p lprint(seriestolist(series(p[0],x,2)));
%p for n from 1 to 4 do
%p p[n]:=p[n-1]*(1-x^(2^n+1))/(1-x);
%p lprint(seriestolist(series(p[n],x,2^(n+1))));
%p od: # _N. J. A. Sloane_, Nov 13 2018
%t T[n_, k_] := SeriesCoefficient[Product[(1-x^(2^j+1))/(1-x), {j, 0, n-1}], {x, 0, k}];
%t Table[T[n, k], {n, 0, 6}, {k, 0, 2^n-1}] // Flatten (* _Jean-François Alcover_, Oct 01 2019 *)
%o (PARI) T(n,k)=local(A=[1],B=[1]);if(n==0,1,for(i=0,n-1, B=Vec(Ser(A)/(1-x));A=concat(B,Vec(Pol(B)+O(x^#B))));A[k+1])
%o for(n=0,6,for(k=0,2^n-1,print1(T(n,k),", "));print())
%o (PARI) T(n,k)=polcoeff(prod(j=0,n-1,(1-x^(2^j+1))/(1-x)),k)
%o for(n=0,6,for(k=0,2^n-1,print1(T(n,k),", "));print()) \\ _Paul D. Hanna_, Aug 09 2009
%Y Cf. A131792 (main diagonal); A028361, A028362.
%K nonn,tabf,look
%O 0,5
%A _Paul D. Hanna_, Jul 15 2007
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