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.

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

Offset: 0

Author

Paul D. Hanna, Jul 15 2007

Keywords

• nonn
• tabf
• look

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)

Examples

			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].
		

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• Richard Kenyon, Infinite scaled convolutions, Preprint, 1992 (apparently unpublished)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.

Crossrefs

Cf. A131792 (main diagonal); A028361, A028362.

Programs

Maple

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))));
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 *)

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

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