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%I #8 Dec 01 2023 18:50:46
%S 1,4,19,103,634,4393,33893,288158,2674849,26888251,290614732,
%T 3356438587,41203019361,535141595208,7324289215167,105271669493307,
%U 1584113665608394,24890073684310405,407378999173905545,6930779764599424550,122334506551009552893,2236412875771806004767
%N a(n) = Sum_{k=0..n} A011971(n, k) * 2^(n - k).
%C The Peirce/Aitken polynomials evaluated at 1/2 and the result normalized with 2^n.
%o (Python)
%o from functools import cache
%o @cache
%o def b(n: int) -> list[int]:
%o if n == 0: return [1]
%o row = [b(n - 1)[n - 1]] + b(n - 1)
%o for k in range(1, n + 1): row[k] += row[k - 1]
%o return row
%o def a(n): return sum(b(n)[k] * 2 ** (n - k) for k in range(n + 1))
%o print([a(n) for n in range(22)])
%Y Cf. A011971, A367809.
%K nonn
%O 0,2
%A _Peter Luschny_, Dec 01 2023
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