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.
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
Examples
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;
Links
- G. C. Greubel, Rows n = 0..100 of the irregular triangle, flattened
Programs
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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 *) -
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
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.
Extensions
More terms and edited by G. C. Greubel, Apr 29 2021