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 A131791 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. 3
 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, 4, 1, 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, 1, 6, 20, 49, 99, 175, 280, 415 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Row sums (and central terms) form A028361: Product_{i=0..n-1} (2^i + 1). 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 From N. J. A. Sloane, Nov 13 2018: (Start) Kenyon (1992) defines p_n(x) (n >= 0) to be the polynomial p_n(x) = (1+x)*(1+x+x^2)*(1+x+x^2+x^3+x^4)*...*(1+x+...+x^(2^n)). 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 (f(c)+o(1))*p_n(1)/2^(n+1), where f : R -> R is a nonzero C^1 function satisfying (i) support(f) is a subset of [0,1], (ii) f(x) = f(1-x), and (iii) f'(x) = 4*f(2*x) for 0 <= x <= 1/2. 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". This provides a fairly complete answer to Martin Fuller's question above. (End) REFERENCES Richard Kenyon, Infinite scaled convolutions, Preprint, 1992 (apparently unpublished) LINKS Paul D. Hanna, Rows 0 to 11 of the triangle, flattened. Julien Clément, Antoine Genitrini, Binary Decision Diagrams: from Tree Compaction to Sampling, arXiv:1907.06743 [cs.DS], 2019. FORMULA T(n, 2^(n-1)) = A028361(n-1) for n>=1. T(n, 2^(n-2)) = A028362(n-1) for n>=2. Sum_{k=0..2^n-1} (k+1)*T(n,k) = A028362(n+1) for n>=0. G.f. of row n: Product_{j=0..n-1} (1 - x^(2^j+1))/(1-x). - Paul D. Hanna, Aug 09 2009 EXAMPLE Triangle begins: 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, 4, 1; 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; ... ILLUSTRATION OF GENERATING METHOD. 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]. MAPLE p[-1]:=1: lprint(seriestolist(series(p[-1], x, 0))); p:=(1-x^2)/(1-x): lprint(seriestolist(series(p, x, 2))); for n from 1 to 4 do p[n]:=p[n-1]*(1-x^(2^n+1))/(1-x); lprint(seriestolist(series(p[n], x, 2^(n+1)))); od: # N. J. A. Sloane, Nov 13 2018 MATHEMATICA T[n_, k_] := SeriesCoefficient[Product[(1-x^(2^j+1))/(1-x), {j, 0, n-1}], {x, 0, k}]; Table[T[n, k], {n, 0, 6}, {k, 0, 2^n-1}] // Flatten (* Jean-François Alcover, Oct 01 2019 *) PROG (PARI) T(n, k)=local(A=, B=); 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]) for(n=0, 6, for(k=0, 2^n-1, print1(T(n, k), ", ")); print()) (PARI) T(n, k)=polcoeff(prod(j=0, n-1, (1-x^(2^j+1))/(1-x)), k) for(n=0, 6, for(k=0, 2^n-1, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Aug 09 2009 CROSSREFS Cf. A131792 (main diagonal); A028361, A028362. Sequence in context: A272689 A274887 A008302 * A308497 A010358 A155865 Adjacent sequences:  A131788 A131789 A131790 * A131792 A131793 A131794 KEYWORD nonn,tabf,look AUTHOR Paul D. Hanna, Jul 15 2007 STATUS approved

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Last modified April 3 00:35 EDT 2020. Contains 333195 sequences. (Running on oeis4.)