OFFSET
0,5
COMMENTS
Generating rule: Start with a single '1' in row 0; let S(n) denote the initial [2^(n+1) - (n+1)] terms of the partial sums of row n; generate row n+1 by concatenating the following: S(n), 2^[n*(n-1)/2] repeated (n-1) times and the terms of S(n) when read in reverse order.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..1990
Jordan Stoyanov, Christophe Vignat, Non-Conventional Limits of Random Sequences Related to Partitions of Integers, arXiv:1901.04029 [math.PR], 2019. [See page 2 and appendix for additional properties of this sequence.]
C. Vignat, T. Wakhare, Finite generating functions for the sum of digits sequence, arXiv:1708.06479 [math.NT], 2017.
Tanay Wakhare, Christophe Vignat, Settling some sum suppositions, arXiv:1805.10569 [math.NT], 2018.
Tanay Wakhare, Christophe Vignat, Settling some sum suppositions, Acta Math. Hungar. (2018).
FORMULA
Row sums are 2^(n*(n+1)/2) for n>=0.
EXAMPLE
Triangle begins:
1;
1,1;
1,2, 2, 2,1;
1,3,5,7,8, 8,8, 8,7,5,3,1;
1,4,9,16,24,32,40,48,55,60,63,64, 64,64,64, 64,63,60,55,48,40,32,24,16,9,4,1; ...
Illustrate the row g.f.s by:
(1+x)^2*(1+x^2) = g.f. of row 2: [1,2,2,2,1];
(1+x)^3*(1+x^2)^2*(1+x^4) = g.f. of row 3: [1,3,5,7,8,8,8,8,7,5,3,1];
(1+x)^4*(1+x^2)^3*(1+x^4)^2*(1+x^8) = g.f. of row 4.
MATHEMATICA
Rest@ Flatten@ Array[{1}~Join~CoefficientList[Series[Product[(1 + x^(2^i))^(# - i), {i, 0, # - 1}], {x, 0, 2^(# + 1) - (# + 1)}], x] &, 5] (* Michael De Vlieger, Aug 21 2018 *)
PROG
(PARI) {T(n, k)=local(A=[1]); if(n==0, 1, for(i=0, n-1, A=concat(Vec((Polrev(A)+O(x^(#A+i)))/(1-x)), Vec(O(x^(#A))+Pol(Vec(Ser(A)/(1-x)))))); A[k+1])}
for(n=0, 6, for(k=0, 2^(n+1)-(n+2), print1(T(n, k), ", ")); print(""))
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Paul D. Hanna, Jul 19 2007
STATUS
approved