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Harry Richman: I only wanted credit for the last comment, since the others are trivial. But I will do so in the future if that is the convention.

Andrew Howroyd: 1/2 * A000984(2^n) is considered ambiguous (see 5th bullet of https://oeis.org/wiki/Style_Sheet#Spelling_and_notation). Rather write (1/2) * A000984(2^n) or A000984(2^n)/2, assuming that is what you mean.

Harry Richman: Ok that makes sense about ambiguous notation, thanks for pointing this out!

Andrew Howroyd: We usually want you to take credit for trivial formulas too. (triviality is too much of a slippery slope - anything added to Formulas section should be signed - the more routine it is, the more people understand that the addition of a formula really doesn't really carry a lot of deeper significance. It's up to the reader to decide if a formula is likely original work or just fact filling - sometimes obvious, often not. In this sense 'credit' is almost impossible to establish. In contrast, in a mathematical paper authors go to great length to attribute all sources of information, but a paper takes a lot of effort to write and research whereas here the emphasis is on efficiently collecting the facts and ease of research)

From Harry Richman, May 18 2023: (Start)

log log a(n) ~ (log 2) * (n + 1) + log log 2 + O(n / 2^n). - _Harry Richman_, May 18 2023.). (End)

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Andrew Howroyd: When adding more than one formula, you need to use a block comment - otherwise it is impossible to know where your contribution starts. See 2nd bullet of https://oeis.org/wiki/Style_Sheet#Signing_your_name_when_you_contribute_to_an_existing_sequence

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a(n) = A001790(2^n). - _Harry Richman_, May 18 2023.).

a(n) = 1/2 * A000984(2^n).

a(n) = 1/2 * (2^n + 1) * A000108(2^n).

log log a(n) ~ (log 2) * (n + 1) + log log 2 + O(n / 2^n). - Harry Richman, May 18 2023.

Cf. A000051, A000108, A000984, A001790, A061548.

a(n) = A001790(2^n) _). - _Harry Richman_, May 18 2023.

a(n) = A001790(2^n) Harry Richman, May 18 2023.

Cf. A000051, A001790, A061548.

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a(n) = binomial(2^(n+1), 2^n)/2 = binomial(2^(n+1) - 1, 2^n) = binomial(2^(n+1) - 1, 2^n - 1).

Peter Luschny: Neil, please stop these repetitions.