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A131791
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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.
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3
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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
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OFFSET
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0,5
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COMMENTS
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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
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)
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REFERENCES
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Richard Kenyon, Infinite scaled convolutions, Preprint, 1992 (apparently unpublished)
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LINKS
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FORMULA
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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
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EXAMPLE
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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].
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MAPLE
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p[-1]:=1:
lprint(seriestolist(series(p[-1], x, 0)));
p[0]:=(1-x^2)/(1-x):
lprint(seriestolist(series(p[0], 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))));
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MATHEMATICA
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T[n_, k_] := SeriesCoefficient[Product[(1-x^(2^j+1))/(1-x), {j, 0, n-1}], {x, 0, k}];
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PROG
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(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])
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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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