A143232 - OEIS

%I #13 Sep 13 2024 08:01:50

%S 2,0,2,0,2,1,2,0,2,1,3,1,3,1,2,0,2,1,3,2,3,2,4,2,4,2,3,2,3,1,2,0,2,1,

%T 3,2,4,2,4,3,5,3,5,4,5,4,5,3,5,4,5,4,5,3,5,3,4,2,4,2,3,1,2,0,2,1,3,2,

%U 4,3,4,3,5,4,6,4,6,5,7,5,7,6,7,6,7,5,7,6,8,6,8,7,8,7,8,6,8,7,8,7,8,6,8,6,7

%N Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2.

%C A004001 is the Hofstadter-Conway $10,000 Sequence and A004001(n) - n/2 is increasingly larger versions of the batrachion Blancmange function.

%H G. C. Greubel, <a href="/A143232/b143232.txt">Table of n, a(n) for n = 1..10000</a>

%H H. Rich, <a href="http://www.jsoftware.com/pipermail/programming/2008-August/011551.html">Hofstadter-Conway $10,000 Sequence</a>, Programming Archives, 2008.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Hofstadter-Conway10000-DollarSequence.html">Hofstadter-Conway $10,000 Sequence</a>.

%F a(n) = Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2 .

%F For practical purposes, a full Egyptian fraction algorithm is not necessary. Since the elements of A004001(n) - n/2 are either whole or their fractional part is .5, the sequence can be effected by a(n) = sefd(A004001(n) - n/2) with sefd(x) = x + 3 * (x - floor(x)) .

%e a(43) = 5 because A004001(43) = 25, so (A004001(43) - (43/2)) = 3.5 and the Egyptian fraction expansion of 3.5 is (1/1)+(1/1)+(1/1)+(1/2), so the denominators are 1,1,1,2 which sums to 5.

%t HC[n_]:= HC[n]= If[n<3, 1, HC[HC[n-1]] +HC[n-HC[n-1]]]; (*HC=A004001*)

%t f[x_]:= 4*x -3*Floor[x];

%t A143232[n_]:= f[HC[n] -n/2];

%t Table[A143232[n], {n,100}] (* _G. C. Greubel_, Sep 10 2024 *)

%o (J) a004001 =: ((] +&:$: -) $:@:<:)@.(2&<) M.

%o a143232 =: (+ 3 * 1&|)@:(a004001 - -:)"0

%o (Magma)

%o A004001:=[n le 2 select 1 else Self(Self(n-1))+ Self(n-Self(n-1)):n in [1..125]];

%o f:= func< x | 4*x - 3*Floor(x) >;

%o A143232:= func< n | f(A004001[n] - n/2) >;

%o [A143232(n): n in [1..100]]; // _G. C. Greubel_, Sep 10 2024

%o (SageMath)

%o @CachedFunction

%o def b(n): # b = A004001

%o     if n<3: return 1

%o     else: return b(b(n-1)) + b(n-b(n-1))

%o def f(x): return x + 3 * (x - floor(x))

%o def A143232(n): return f(b(n) - n/2)

%o [A143232(n) for n in range(1,101)] # _G. C. Greubel_, Sep 10 2024

%Y Cf. A004001.

%K nonn,easy

%O 1,1

%A Dan Bron (j (at) bron.us), Jul 31 2008