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From Robert Israel, Jan 16 2018: (Start)
a(n) = Product_{0 <= i < j <= n-1} (2^j - 2^i) = 2^(n*(n-1)*(n-2)/6) * Product_{1<=k<=n-1} (2^k-1)^(n-k). - _Robert Israel_, Jan 16 2018
a(n) = 2^(n*(n-1)*(n-2)/6) * Product_{1<=k<=n-1} (2^k-1)^(n-k). (End)
a(n) = Product_{k=0..n-2} ( 2^(k+1)^2 * QPochhammer(2^(-k-1); 2; k+1) ). - G. C. Greubel, Aug 31 2023
# First program
# Second program
(* First program *)
f[j_] := 2^(j - 1); z = 15;
v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]
Table[v[n], {n, 1, z}] (* A203303 *)
Table[v[n + 1]/v[n], {n, 1, z - 1}] (* A002884 *)
Table[v[n] *v[n + 2]/(2*v[n + 1]^2), {n, 1, z - 1}] (* A171499 *)
Table[FactorInteger[v[n]], {n, 1, z - 1}]
(* Second program *)
Table[Product[2^(k+1) -2^j, {k, 0, n-2}, {j, 0, k}], {n, 15}] (* G. C. Greubel, Aug 31 2023 *)
(Magma) [1] cat [(&*[(&*[2^(k+1) -2^j: j in [0..k]]): k in [0..n-2]]): n in [2..15]]; // G. C. Greubel, Aug 31 2023
(SageMath) [product(product(2^(k+1) -2^j for j in range(k+1)) for k in range(n-1)) for n in range(1, 16)] # G. C. Greubel, Aug 31 2023
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a(n) ~ 1/A335011 * 2^(n*(n-1)*(2*n-1)/6) * QPochhammer(1/2)^n. - Vaclav Kotesovec, May 19 2020
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f:= n -> 2^(n*(n-1)*(n-2)/6)*mul((2^k-1)^(n-k), k=1..n-1):
seq(f(n), n=1..12); # Robert Israel, Jan 16 2018
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