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A173122
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.
5
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, 2, 1024, 256, 64, 16, 4, 2048, 512, 128, 32, 8, 2, 4096, 1024, 256, 64, 16, 4, 8192, 2048, 512, 128, 32, 8, 16384, 4096, 1024, 256, 64, 16, 32768, 8192, 2048, 512, 128, 32, 65536, 16384, 4096, 1024, 256, 64
OFFSET
0,2
FORMULA
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.
EXAMPLE
Irregular triangle begins as:
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, 2;
1024, 256, 64, 16, 4;
MATHEMATICA
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[n_]:= CoefficientList[Series[Sum[t[n, k, q], {k, 0, n}], {q, 0, n}], q];
Table[T[n], {n, 0, 12}]//Flatten (* modified by G. C. Greubel, Apr 29 2021 *)
PROG
(Sage)
@CachedFunction
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) )
def s(n, x): return sum( t(n, k, x) for k in (0..n) )
flatten([taylor(s(n, x), x, 0, n).list() for n in (0..12)]) # G. C. Greubel, Apr 29 2021
CROSSREFS
KEYWORD
nonn,tabf,easy,less
AUTHOR
Roger L. Bagula, Feb 10 2010
EXTENSIONS
More terms and edited by G. C. Greubel, Apr 29 2021
STATUS
approved