A318464 -id:A318464 - cp's OEIS frontend

A318465 The number of Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the Zeckendorf expansion (A014417) of each s(i) contains only terms that are in the Zeckendorf expansion of r(i).

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 2, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4, 4, 2, 8, 4, 4, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 8

Offset: 1

Antti Karttunen, Aug 30 2018

Zeckendorf-infinitary divisors are analogous to infinitary divisors (A077609) with Zeckendorf expansion instead of binary expansion. - Amiram Eldar, Jan 09 2020

			a(16) = 4 since 16 = 2^4 and the Zeckendorf expansion of 4 is 101, i.e., its Zeckendorf representation is a set with 2 terms: {1, 3}. There are 4 possible exponents of 2: 0, 1, 3 and 4, corresponding to the subsets {}, {1}, {3} and {1, 3}. Thus 16 has 4 Zeckendorf-infinitary divisors: 2^0 = 1, 2^1 = 2, 2^3 = 8, and 2^4 = 16.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A007895, A318464, A318469.
Cf. A014417, A037445, A038148, A074848, A077609.
Cf. also A318472, A318474.

fb[n_] := Block[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; Fibonacci[1 + Position[Reverse@fr, ?(# == 1 &)]]]; f[p, e_] := 2^Length@fb[e]; a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n])); Array[a, 100] (* Amiram Eldar, Jan 09 2020 after Robert G. Wilson v at A014417 *)

A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
A007895(n) = { my(s=0); while(n>0, s++; n -= fibonacci(1+A072649(n))); (s); }
A318465(n) = factorback(apply(e -> 2^A007895(e),factor(n)[,2]));

Multiplicative with a(p^e) = 2^A007895(e), where A007895(n) gives the number of terms in the Zeckendorf representation of n.

a(n) = 2^A318464(n).

Name edited and interpretation in terms of divisors added by Amiram Eldar, Jan 09 2020

A376885 The number of factors of n of the form p^(k!) counted with multiplicity, where p is a prime and k >= 1, when the factorization is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Original entry on oeis.org

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, 3, 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, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 2, 2, 4, 1, 2, 2, 2, 1, 3, 1, 3, 3

Offset: 1

Amiram Eldar, Oct 08 2024

Let n = Product p^e be the canonical prime factorization of n. The factorization of n that is based on the factorial-base representation of the exponents is done by factoring each prime power p^e into prime powers, p^e = Product_{k} (p^(k!))^d_k where e = Sum_{k>=1} d_k * k! is the factorial-base representation of e. So, the factors in this factorization are prime powers with exponents that are factorial numbers. Each factor in the factorization, p^(k!), can have a multiplicity d in the range [1, k], so this factorization of n is a product of numbers of the form (p^(k!))^d.
The number of factors counted with multiplicity (the sum of the multiplicities d in (p^(k!))^d) is given by this sequence (analogous to A001222 for the canonical prime factorization).
The number of distinct factors p^(k!) of n is A376886(n) (analogous to A001221).
The number of divisors of n that can be constructed from partial sets of these factors (with multiplicities that are not larger than those in n) is A376887(n) (analogous to A000005), and their sum is A376888(n) (analogous to A000203).

			For n = 8 = 2^3, the representation of 3 in factorial base is 11, i.e., 3 = 1! + 2!, so 8 = (2^(1!))^1 * (2^(2!))^1 and a(8) = 1 + 1 = 2.
For n = 16 = 2^4, the representation of 4 in factorial base is 20, i.e., 4 = 2 * 2!, so 16 = (2^(2!))^2 and a(16) = 2.

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

Cf. A000142, A007623, A034968, A077761.
Cf. A000005, A000203, A001221, A001222, A376886, A376887, A376888.
Similar sequences: A064547, A318464.

fdigsum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s += r; m++]; s]; f[p_, e_] := fdigsum[e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s;}
a(n) = {my(e = factor(n)[, 2]); sum(i = 1, #e, fdigsum(e[i]));}

Additive with a(p^e) = A034968(e).

Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 0.18682321026088865388..., where f(x) = -x + (1-x) * Sum_{k>=1} A034968(k)* x^k.

A376886 The number of distinct factors of n of the form p^(k!), where p is a prime and k >= 1, when the factorization is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 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, 2, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 2, 2, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 3, 3

Offset: 1

Amiram Eldar, Oct 08 2024

See A376885 for details about this factorization.
First differs from A371090 at n = 2^18 = 262144.
Differs from A064547 at n = 64, 128, 192, 256, 320, 384, 448, 512, ... .
Differs from A058061 at n = 128, 384, 512, 640, 896, ... .

			For n = 8 = 2^3, the representation of 3 in factorial base is 11, i.e., 3 = 1! + 2!, so 8 = (2^(1!))^1 * (2^(2!))^1 and a(8) = 1 + 1 = 2.
For n = 16 = 2^4, the representation of 4 in factorial base is 20, i.e., 4 = 2 * 2!, so 16 = (2^(2!))^2 and a(16) = 1.

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

Cf. A007623, A058061, A060130, A077761, A371090.

Similar sequences: A064547, A318464, A376885.

fdignum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, s++]; m++]; s]; f[p_, e_] := fdignum[e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

fdignum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, if(r > 0, s ++); m++); s;}
a(n) = {my(e = factor(n)[, 2]); sum(i = 1, #e, fdignum(e[i]));}

Additive with a(p^e) = A060130(e).

Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 0.12589120926760155013..., where f(x) = -x + (1-x) * Sum_{k>=1} A060130(k) * x^k.

A318469 Multiplicative with a(p^e) = A019565(A003714(e)).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 10, 2, 6, 2, 6, 4, 4, 2, 10, 3, 4, 5, 6, 2, 8, 2, 7, 4, 4, 4, 9, 2, 4, 4, 10, 2, 8, 2, 6, 6, 4, 2, 20, 3, 6, 4, 6, 2, 10, 4, 10, 4, 4, 2, 12, 2, 4, 6, 14, 4, 8, 2, 6, 4, 8, 2, 15, 2, 4, 6, 6, 4, 8, 2, 20, 10, 4, 2, 12, 4, 4, 4, 10, 2, 12, 4, 6, 4, 4, 4, 14, 2, 6, 6, 9, 2, 8, 2, 10, 8

Offset: 1

Antti Karttunen, Aug 30 2018

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A003714, A019565, A072649, A318464, A318465.

Cf. also A293442, A293443, A300834, A318470.

A003714(n) = { my(s=0,w); while(n>2, w = A072649(n); s += 2^(w-1); n -= fibonacci(w+1)); (s+n); }
A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
A318469(n) = factorback(apply(e -> A019565(A003714(e)),factor(n)[,2]));

For all n >= 1, A001222(a(n)) = A318464(n).

A375271 Partial products of A375270.

Original entry on oeis.org

1, 2, 6, 30, 210, 1680, 18480, 240240, 4084080, 77597520, 1784742960, 48188059920, 1397453737680, 43321065868080, 1602879437118960, 65718056921877360, 2825876447640726480, 132816193039114144560, 7039258231073049661680, 415316235633309930039120, 25334290373631905732386320

Offset: 1

Amiram Eldar, Aug 09 2024

nonn

Numbers with a record number of Zeckendorf-infinitary divisors (A318465). Also, indices of records in A318464.

a(n) is the least number k such that A318464(k) = n-1 and A318465(k) = 2^(n-1).

			A375270 begins with 1, 2, 3, 5, ..., so, a(1) = 1, a(2) = 1 * 2 = 2, a(3) = 1 * 2 * 3 = 6, a(4) = 1 * 2 * 3 * 5 = 30.

Amiram Eldar, Table of n, a(n) for n = 1..353

Cf. A037992 (analogous with "Fermi-Dirac primes", A050376), A318464, A318465, A375270.

Subsequence of A025487.

fib[lim_] := Module[{s = {}, f = 1, k = 2}, While[f <= lim, AppendTo[s, f]; k += 2; f = Fibonacci[k]]; s];
seq[max_] := Module[{s = {}, p = 2, e = 1, f = {}}, While[e > 0, e = Floor[Log[p, max]]; If[f == {}, f = fib[e], f = Select[f, # <= e &]]; s = Join[s, p^f]; p = NextPrime[p]]; FoldList[Times, 1, Sort[s]]]; seq[100]

fib(lim) = {my(s = List(), f = 1, k = 2); while(f <= lim, listput(s, f); k += 2; f = fibonacci(k)); Vec(s);}
lista(pmax) = {my(s = [1], p = 2, e = 1, f = [], r = 1); while(e > 0, e = logint(pmax, p); if(#f == 0, f = fib(e), f = select(x -> x <= e, f)); s = concat(s, apply(x -> p^x, f)); p = nextprime(p+1)); s = vecsort(s); for(i = 1, #s, r *= s[i]; print1(r, ", "))}

a(n) = Product_{k=1..n} A375270(k).

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.

Original entry on oeis.org

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

Amiram Eldar, Aug 09 2024

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).

			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.

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

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

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]

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]));

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.

cp's OEIS Frontend

A318465 The number of Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the Zeckendorf expansion (A014417) of each s(i) contains only terms that are in the Zeckendorf expansion of r(i).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A376885 The number of factors of n of the form p^(k!) counted with multiplicity, where p is a prime and k >= 1, when the factorization is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A376886 The number of distinct factors of n of the form p^(k!), where p is a prime and k >= 1, when the factorization is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A318469 Multiplicative with a(p^e) = A019565(A003714(e)).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A375271 Partial products of A375270.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula