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Number of ways to represent n using the binary operator a * b = 2^a + b.
1

%I #38 Dec 19 2021 09:39:34

%S 1,1,2,3,7,12,23,41,81,149,282,522,987,1843,3463,6473,12160,22773,

%T 42719,80025,150074,281258,527320,988334,1852849,3473061,6510681,

%U 12204139,22877649,42884585,80389797,150692973,282481747,529522496,992614937,1860696007

%N Number of ways to represent n using the binary operator a * b = 2^a + b.

%D D. E. Knuth, personal communication.

%H D. E. Knuth, <a href="/A000630/a000630.pdf">Letter to N. J. A. Sloane, Jan. 1995</a>

%F Sum_{n>=0} a(n)*q^n = 1/(1 - Sum_{n>=0} a(n)*q^(2^n) ).

%F As n increases, a(n+1)/a(n) approaches the value x = 1.874542... satisfying 1 = Sum_{j>=0} a(j)/x^(2^j). [_David W. Wilson_]

%e E.g. 4 = 1+1+1+1 = 2^1 + 1+1 = 2^1 + 2^1 = 2^2 = 2^1+1 = 1+2^1 + 1 = 1+1+2^1.

%p N:= 100: # to get all terms up to a(N)

%p F:= add(a[i]*q^i,i=0..N) - 1/(1 - add(a[i]*q^(2^i),i=0..ilog2(N))):

%p S:= series(F, q, N+1);

%p sol:= {}:

%p for i from 0 to N do

%p eq:= subs(sol, coeff(S,q,i));

%p sol:= sol union {a[i]=solve(eq,a[i])};

%p od:

%p subs(sol, [seq(a[i],i=0..N)]); # _Robert Israel_, Dec 12 2014

%K nonn,easy,nice

%O 0,3

%A _N. J. A. Sloane_

%E More terms from _David W. Wilson_