

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
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OFFSET

0,5


COMMENTS

Row sums (and central terms) form A028361: Product_{i=0..n1} (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(n1). In this setup I think that the limit curve f(x) satisfies f(0)=0, f(1x)=f(x), f(1/2)=1, f'(x)=2f(2x) for x<=1/2. Is this equation solvable?  Martin Fuller, Aug 31 2007
Comment 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(1x), 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Richard Kenyon, Infinite scaled convolutions, Preprint, 1992 (apparently unpublished)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.


FORMULA

T(n, 2^(n1)) = A028361(n1) for n>=1.
T(n, 2^(n2)) = A028362(n1) for n>=2.
Sum_{k=0..2^n1} (k+1)*T(n,k) = A028362(n+1) for n>=0.
G.f. of row n: Product_{j=0..n1} (1  x^(2^j+1))/(1x).  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[0]:=(1x^2)/(1x):
lprint(seriestolist(series(p[0], x, 2)));
for n from 1 to 4 do
p[n]:=p[n1]*(1x^(2^n+1))/(1x);
lprint(seriestolist(series(p[n], x, 2^(n+1))));
od: # N. J. A. Sloane, Nov 13 2018


PROG

(PARI) T(n, k)=local(A=[1], B=[1]); if(n==0, 1, for(i=0, n1, B=Vec(Ser(A)/(1x)); A=concat(B, Vec(Pol(B)+O(x^#B)))); A[k+1])
for(n=0, 6, for(k=0, 2^n1, print1(T(n, k), ", ")); print())
(PARI) T(n, k)=polcoeff(prod(j=0, n1, (1x^(2^j+1))/(1x)), k)
for(n=0, 6, for(k=0, 2^n1, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Aug 09 2009


CROSSREFS

Cf. A131792 (main diagonal); A028361, A028362.
Sequence in context: A272689 A274887 A008302 * A010358 A155865 A156133
Adjacent sequences: A131788 A131789 A131790 * A131792 A131793 A131794


KEYWORD

nonn,tabf,look


AUTHOR

Paul D. Hanna, Jul 15 2007


STATUS

approved



