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A307451 Sum of binary weights of two consecutive Fibonacci numbers minus the binary weight of the following Fibonacci number.

%I A307451 #25 Apr 14 2019 19:07:25
%S A307451 0,1,0,1,3,0,1,4,0,3,7,1,1,7,2,5,11,6,1,7,6,3,13,11,3,4,9,10,15,9,10,
%T A307451 7,10,14,11,9,16,12,6,11,19,15,12,19,12,23,22,7,12,19,18,17,18,11,17,
%U A307451 27,13,20,28,17,9,22,29,18,26,18,30,18,15,24,20,20,28,28,24,24,18,21,28
%N A307451 Sum of binary weights of two consecutive Fibonacci numbers minus the binary weight of the following Fibonacci number.
%C A307451 The binary weight of a positive Fibonacci number is at least 1 (and at least 2 for positive Fibonacci numbers other than 1, 2, 8) but not more than the sum of the binary weights of the previous two Fibonacci numbers.
%C A307451 Therefore a(n) is at least 0, at most n - 1.
%C A307451 Number of carries in base-2 addition of A000045(n-2)+A000045(n-1)=A000045(n). - _Robert Israel_, Apr 14 2019
%H A307451 Robert Israel, <a href="/A307451/b307451.txt">Table of n, a(n) for n = 2..10000</a>
%H A307451 Dario Carrasquel, <a href="https://dariocarrasquel.com/2016/08/26/5-ways-to-solve-fibonacci-in-scala-tail-recursion-memoization-the-pisano-period-more/">5 ways to solve Fibonacci in Scala - Tail Recursion, Memoization, The Pisano Period & More</a>, August 26, 2016.
%F A307451 a(n) = (A011373(n - 2) + A011373(n - 1)) - A011373(n).
%e A307451 Fibonacci(8) = 21 = 10101 in binary.
%e A307451 Fibonacci(9) = 34 = 100010 in binary.
%e A307451 Fibonacci(10) = 55 = 110111 in binary, which has five 1s. We see that 10101 has three 1s and 100010 just two. Thus a(10) = 0.
%p A307451 B:= map(t -> convert(convert(combinat:-fibonacci(t),base,2),`+`), [$0..100]):
%p A307451 B[1..-3]-B[2..-2]-B[3..-1]; # _Robert Israel_, Apr 14 2019
%t A307451 Table[(DigitCount[Fibonacci[n - 2], 2, 1] + DigitCount[Fibonacci[n - 1], 2, 1]) - DigitCount[Fibonacci[n], 2, 1], {n, 2, 100}]
%o A307451 (Scala) def fibonacci(n: BigInt): BigInt = {
%o A307451   val zero = BigInt(0)
%o A307451   def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
%o A307451     case `zero` => a
%o A307451     case _ => fibTail(n - 1, b, a + b)
%o A307451   }
%o A307451   fibTail(n, 0, 1)
%o A307451 } // Based on "Case 3: Tail Recursion" from Carrasquel (2016) link
%o A307451 (2 to 100).map(n => (fibonacci(n - 2).bitCount + fibonacci(n - 1).bitCount) - fibonacci(n).bitCount)
%o A307451 (PARI) f(n) = hammingweight(fibonacci(n)); \\ A011373
%o A307451 a(n) = f(n-1) + f(n-2) - f(n); \\ _Michel Marcus_, Apr 14 2019
%Y A307451 Cf. A000045, A000120, A011373.
%K A307451 nonn,easy,base
%O A307451 2,5
%A A307451 _Alonso del Arte_, Apr 08 2019