%I #9 Apr 29 2021 03:50:43
%S 1,2,4,1,8,2,16,4,32,8,2,64,16,4,128,32,8,2,256,64,16,4,512,128,32,8,
%T 2,1024,256,64,16,4,2048,512,128,32,8,2,4096,1024,256,64,16,4,8192,
%U 2048,512,128,32,8,16384,4096,1024,256,64,16,32768,8192,2048,512,128,32,65536,16384,4096,1024,256,64
%N Irregular triangle T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with t(n,0,q) = t(n,n,q) = 1, read by rows.
%H G. C. Greubel, <a href="/A173122/b173122.txt">Rows n = 0..100 of the irregular triangle, flattened</a>
%F T(n) = coefficients of Sum_{k=0..n} t(n,k,q) for powers of q, where t(n,k,q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with t(n,0,q) = t(n,n,q) = 1.
%e Irregular triangle begins as:
%e 1;
%e 2;
%e 4, 1;
%e 8, 2;
%e 16, 4;
%e 32, 8, 2;
%e 64, 16, 4;
%e 128, 32, 8, 2;
%e 256, 64, 16, 4;
%e 512, 128, 32, 8, 2;
%e 1024, 256, 64, 16, 4;
%t t[n_, k_, q_]:= If[k==0 || k==n, 1, q*Boole[n==2] + Sum[q^j*Binomial[n-2*j, k-j] *Boole[n>2*j], {j, 0, 5}]];
%t T[n_]:= CoefficientList[Series[Sum[t[n,k,q], {k,0,n}], {q,0,n}], q];
%t Table[T[n], {n, 0, 12}]//Flatten (* modified by _G. C. Greubel_, Apr 29 2021 *)
%o (Sage)
%o @CachedFunction
%o def t(n, k, x): return 1 if (k==0 or k==n) else x*bool(n==2) + sum( x^j*binomial(n-2*j, k-j)*bool(n>2*j) for j in (0..5) )
%o def s(n,x): return sum( t(n,k,x) for k in (0..n) )
%o flatten([taylor(s(n,x), x, 0, n).list() for n in (0..12)]) # _G. C. Greubel_, Apr 29 2021
%Y Cf. A007318, A072405, A173117, A173118, A173119, A173120.
%K nonn,tabf,easy,less
%O 0,2
%A _Roger L. Bagula_, Feb 10 2010
%E More terms and edited by _G. C. Greubel_, Apr 29 2021