A375272 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.

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, 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, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2

Offset

1,6

Comments

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

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

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

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

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

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

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

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.

Example

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.
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.

Mathematica

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

Prog

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

Crossrefs

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

Sequence in context: A214715 A244145 A371734 * A086435 A266226 A376885

Adjacent sequences: A375269 A375270 A375271 * A375273 A375274 A375275

Keyword

nonn,easy,base

Author

Amiram Eldar, Aug 09 2024

Status

approved