A111403 -id:A111403 - cp's OEIS frontend

A220294 a(n) = 1 - 2^(2^n) + 2^(2^(n+1)).

3, 13, 241, 65281, 4294901761, 18446744069414584321, 340282366920938463444927863358058659841, 115792089237316195423570985008687907852929702298719625575994209400481361428481

Offset: 0

Michael Somos, Dec 10 2012

nonn

An infinite coprime sequence defined by recursion.

Cf. A001146, A002061, A002716, A087046, A111403, A220161, A255770, A255771, A255772.

[1 - 2^(2^n) + 2^(2^(n+1)): n in [0..10]]; // G. C. Greubel, Aug 10 2018

Table[4^(2^m) - 2^(2^m) + 1, {m, 0, 7}] (* Michael De Vlieger, Aug 02 2016 *)

A220294(n):=1 - 2^(2^n) + 2^(2^(n+1))$ makelist(A220294(n),n,0,10); /* Martin Ettl, Dec 10 2012 */

{a(n) = if( n<0, 0, 1 - 2^(2^n) + 2^(2^(n+1)))};

A220161(n+1) = a(n) * A220161(n).
a(n+1) = 1 + (a(n) - 1) * (A220161(n) - 1).
a(n) = A002716(2*n) = 1 + A087046(n+2) = 1 + A111403(n).
a(n) = A002061(A001146(n)). - Pontus von Brömssen, Aug 31 2021

A040996 Maximum number of distinct functions at the bottom of a Boolean (or Binary) Decision Diagram (or BDD) with negation by pointer complementation.

Original entry on oeis.org

1, 6, 120, 32640, 2147450880, 9223372034707292160, 170141183460469231722463931679029329920, 57896044618658097711785492504343953926464851149359812787997104700240680714240

Offset: 0

R. H. Hardin

nonn

At 0, the last variable, the only choice is (t, f) because the first entry is always uncomplemented and the 2nd must be different.
At level 1, the 2nd-to-last variable, the first entry is either t or a pointer to a following level (0) and the 2nd entry is either of these or its negation, except it may not equal the first entry.
At level n, the n-th-to-last variable, the first entry is either t or a pointer to one of the following levels' functions and the second entry is any of these or its negation, but not equal to the first entry.
From Luis H. Gallardo, Nov 18 2021: (Start)
Another description of a(n) follows: let TP(n) = t^(2^n-1)*(t+1)^(2^n-1) in the ring F_2[t]. Expand TP(n) as a sum of monomials c*t^k in F_2[t], with c equal 0 or 1. Lift TP(n) to LTP(n) in the ring Z[t], i.e., consider the coefficients c of TP(n) to be integers in LTP(n), instead of elements of F_2. Finally, substitute t by 2 in LTP(n). We get: a(n) = LTP(n).
Example: a(3) = subs(t=2, TP(3)) = 32640, where TP(3) = t^14 + t^13 + t^12 + t^10 + t^9 + t^8 + t^7 = t^7*(t+1)^7 in F_2[t]. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..12
Cezar Campeanu, Nelma Moreira, and Rogerio Reis, Expected Compression Ratio for DFCA: experimental average case analysis, Technical Report Series: DCC-2011-07, December 2011, Departamento de Ciencia de Computadores, Universidade do Porto.
Dagstuhl Seminar Design & Test, More about BDD's
Alan J. Hu, David L. Dill, Andreas J. Drexler and C. Han Yang, Higher-level specification and verification with BDDs, In: von Bochmann G., Probst D.K. (eds) Computer Aided Verification. CAV 1992. Lecture Notes in Computer Science (1993), vol 663. Springer, Berlin, Heidelberg.

Subsequence of A000217. Half of A111403.

[2^(2^n)*(2^(2^n)-1)/2: n in [0..10]]; // Vincenzo Librandi, Sep 30 2011

a(n) = subs(t=2,modp(expand(t^(2^n-1)*(t+1)^(2^n-1)),2)); # Luis H. Gallardo, Nov 18 2021

f[x_]:=Module[{c=2^(2^x)},(c(c-1))/2]; Array[f,10,0] (* Harvey P. Dale, Sep 29 2011 *)

a(n)=if(n<=0,n==0,2^(2^n)*(2^(2^n)-1)/2)

a(n) = (S(n-1) + 1) * (2*S(n-1) + 1) where S(n-1) = Sum_{k

a(n) is the (2^(2^n)-1)-th triangular number; i.e., a(n) = 2^(2^n)*(2^(2^n)-1)/2.

a(n) = A111403(n) / 2. - Tilman Piesk, Oct 04 2024

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cp's OEIS Frontend

A220294 a(n) = 1 - 2^(2^n) + 2^(2^(n+1)).

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