A375272 - OEIS

%I #10 Aug 11 2024 10:35:00

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

%T 2,2,1,2,2,3,1,3,1,2,2,2,1,3,1,2,2,2,1,3,2,3,2,2,1,3,1,2,2,3,2,3,1,2,

%U 2,3,1,3,1,2,2,2,2,3,1,3,2,2,1,3,2,2,2

%N The number of factors of n of the form p^Fibonacci(k), where p is a prime and k >= 2, when the factorization is uniquely done using the dual Zeckendorf representation of the exponents in the prime factorization of n.

%C First differs from A086435 at n = 36. Differs from A266226 at n = 1, 36, ... .

%C The number of dual-Zeckendorf-infinitary divisors of n (defined in A331109) that are prime powers (A246655).

%C a(n) depends only on the prime signature of n.

%C Analogous to A064547 (binary representation) and A318464 (Zeckendorf representation).

%H Amiram Eldar, <a href="/A375272/b375272.txt">Table of n, a(n) for n = 1..10000</a>

%F Additive with a(p^e) = A112310(e).

%F a(n) = log_2(A331109(n)).

%F Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761), C = Sum_{k>=2} (A112310(k)-A112310(k-1)) * P(k) = 0.18790467121403662496..., and P(s) is the prime zeta function.

%e For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.

%e For n = 256 = 2^8, the dual Zeckendorf representation of 8 is 1011, i.e., 8 = Fibonacci(2) + Fibonacci(3) + Fibonacci(5) = 1 + 2 + 5. Therefore 256 = 2^(1+2+5) = 2^1 * 2^2 * 2^5, and a(256) = 3.

%t toDualZeck[n_] := Module[{s = 0, v = 0, i = 0, f}, While[s < n, s += Fibonacci[i + 2]; v += 2^i; i++]; i--; While[i >= 0, f = Fibonacci[i + 2]; If[s - f >= n, s -= f; v -= 2^i]; i--]; v]; (* A003754, after _Rémy Sigrist_'s PARI code in A112309 *)

%t f[p_, e_] := DigitCount[toDualZeck[e], 2, 1]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) todualzeck(n) = {my (s=0, v=0); for (i=0, oo, if (s>=n, forstep (j=i-1, 0, -1, if (s-fibonacci(2+j)>=n, s-=fibonacci(2+j); v-=2^j;);); return (v);); s+=fibonacci(2+i); v+=2^i;);} \\ A003754, _Rémy Sigrist_'s code in A112309

%o a(n) = vecsum(apply(x -> hammingweight(todualzeck(x)), factor(n)[, 2]));

%Y Cf. A000045, A064547, A077761, A086435, A112309, A112310, A246655, A266226, A318464, A331109.

%K nonn,easy,base

%O 1,6

%A _Amiram Eldar_, Aug 09 2024