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A260005 a(n) = f(2,n,2), where f is the Sudan function defined in A260002.

%I A260005 #48 Nov 14 2024 23:21:49
%S A260005 19,10228,15569256417,5742397643169488579854258,
%T A260005 36681813266165915713665394441869800619098139628586701684547
%N A260005 a(n) = f(2,n,2), where f is the Sudan function defined in A260002.
%C A260005 Naturally a subsequence of A260004.
%C A260005 See A260002-A260003 for the evaluation of the Sudan function.
%C A260005 Using f(2,n,2) = f(1, f(2,n,1), f(2,n,1)+2) = 2^(f(2,n,1)+2)*(f(2,n,1)+2)-f(2,n,1)-4 and f(2,n,1) = f(1, n, n+1) = 2^(n+1)*(n+2)-(n+3) we have:
%C A260005 a(n)=f(2,n,2)
%C A260005     =f(1, 2^(n+1)*(n+2)-(n+3), 2^(n+1)*(n+2)-(n+3)+2)
%C A260005     =2^(2^(n+1)*(n+2)-(n+3)+2)*(2^(n+1)*(n+2)-(n+3)+2)-2^(n+1)*(n+2)+(n+3)-4
%C A260005     =2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1).
%H A260005 Natan Arie' Consigli, <a href="/A260005/b260005.txt">Table of n, a(n) for n = 0..6</a>
%H A260005 Wikipedia, <a href="https://en.wikipedia.org/wiki/Sudan_function">Sudan function</a> (see 3rd line of "Values of F2(x, y)" table).
%F A260005 a(n) = 2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1).
%e A260005 a(1) = f(2,1,2) = f(1,f(2,1,1),f(2,1,1)+2) = f(1,8,10) = 2^10*(8+2)-10-2 = 10228.
%t A260005 Table[2^(2^(n + 1) (n + 2) - (n + 1)) (2^(n + 1) (n + 2) - (n + 1)) - 2^(n + 1) (n + 2) + (n - 1), {n, 0, 5}] (* _Vincenzo Librandi_, Jul 27 2015 *)
%o A260005 (Magma) [2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1):n in [0..5]]; // _Vincenzo Librandi_, Jul 27 2015
%o A260005 (PARI) a(n) = 2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1);
%o A260005 vector(10, n, a(n-1)) \\ _Altug Alkan_, Oct 01 2015
%Y A260005 Cf. A048493 (f(2,n,1)), A260002, A260003, A260004, A260006.
%K A260005 nonn
%O A260005 0,1
%A A260005 _Natan Arie Consigli_, Jul 23 2015