cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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

Views

Author

Amiram Eldar, Oct 08 2024

Keywords

Comments

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).

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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]
  • PARI
    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]));}

Formula

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.

A069903 Number of distinct prime factors of n-th triangular number.

Original entry on oeis.org

0, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 3, 3, 2, 2, 2, 3, 4, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 4, 3, 2, 3, 4, 3, 3, 3, 3, 4, 3, 2, 3, 3, 2, 3, 4, 3, 2, 3, 4, 4, 3, 2, 4, 4, 2, 3, 3, 3, 4, 3, 3, 4, 4, 3, 3, 3, 2, 3, 4, 4, 4, 3, 3, 3, 2, 2, 4, 5, 3, 3, 4, 3, 3, 4
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 10 2002

Keywords

Examples

			A000217(11) = 11*(11+1)/2 = 66 = 2*3*11, therefore a(11) = 3.
		

Crossrefs

Programs

Formula

a(n) = A001221(A000217(n)).
Sum_{k=1..n} a(k) = 2 * n * (log(log(n)) + B - 1/4) + O(n/log(n)), where B is Mertens's constant (A077761). - Amiram Eldar, Sep 21 2024

A238949 Degree of divisor lattice D(n).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 4, 1, 2, 2, 3, 1, 3, 1, 3, 3, 2, 1, 3, 2, 3, 2, 3, 1, 3, 2, 3, 2, 2, 1, 4, 1, 2, 3, 2, 2, 3, 1, 3, 2, 3, 1, 4, 1, 2, 3, 3, 2, 3, 1, 3, 2, 2, 1, 4, 2, 2, 2, 3, 1, 4, 2, 3, 2, 2, 2, 3, 1, 3, 3, 4, 1, 3, 1, 3, 3
Offset: 1

Views

Author

Sung-Hyuk Cha, Mar 07 2014

Keywords

Crossrefs

Programs

Formula

a(n) = A001221(n) + A056170(n) as given in the Cha, DuCasse, Quintas reference. - Geoffrey Critzer, Mar 02 2015
Additive with a(p^e) = 1+A057427(e-1). - Antti Karttunen, Jul 23 2017
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} 1/p^2 (A085548). - Amiram Eldar, Feb 13 2024

Extensions

More terms from Antti Karttunen, Jul 23 2017

A318464 Additive with a(p^e) = A007895(e), where A007895(n) gives the number of terms in Zeckendorf representation of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 3
Offset: 1

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Author

Antti Karttunen, Aug 30 2018

Keywords

Comments

From Amiram Eldar, Aug 09 2024: (Start)
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 Zeckendorf representation of the exponents in the prime factorization of n.
Equivalently, the number of Zeckendorf-infinitary divisors of n (defined in A318465) that are prime powers (A246655). (End)

Crossrefs

Programs

  • Mathematica
    z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; a[n_] := Total[z /@ FactorInteger[n][[;; , 2]]]; a[1] = 0; Array[a, 100] (* Amiram Eldar, May 15 2023 *)
  • PARI
    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); }
    A318464(n) = vecsum(apply(e -> A007895(e),factor(n)[,2]));

Formula

a(n) = A007814(A318465(n)).
a(n) = A001222(A318469(n)).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{k>=2} (A007895(k)-A007895(k-1)) * P(k) = 0.05631817952062180045..., where P(s) is the prime zeta function. - Amiram Eldar, Oct 09 2023

A376365 The number of distinct prime factors of the cubefree numbers.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 2, 1, 2, 2, 1, 3, 1, 2, 2, 2, 1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 3, 1, 2, 2, 3, 1, 1, 2, 2, 2, 2, 3, 1, 2, 1, 3, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 3, 2, 1, 1, 3, 2, 1, 3, 2, 2, 2, 2, 2, 1, 2, 2, 2
Offset: 1

Views

Author

Amiram Eldar, Sep 21 2024

Keywords

Crossrefs

Programs

  • Mathematica
    f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # < 3 &], Length[e], Nothing]]; Array[f, 150]
  • PARI
    lista(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] > 2, is = 0; break)); if(is, print1(#e, ", ")));}

Formula

a(n) = A001221(A004709(n)).
Sum_{A004709(k) <= x} a(k) = (6/Pi^2) * x * (log(log(x)) + B - C) + O(x/log(x)), where B is Mertens's constant (A077761) and C = Sum_{p prime} (p-1)/(p*(p^3-1)) = 0.10770743252352371604... (Das et al., 2024).

A008481 If n = Product (p_j^k_j) then a(n) = Sum partition(k_j).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 5, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 3, 1, 7, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 6, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 4, 1, 2, 3, 11, 2, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 2, 3, 1, 6, 5, 2, 1, 4, 2, 2, 2
Offset: 1

Views

Author

Keywords

Comments

a(n) is a function of the prime signature of n (cf. A025487). - Matthew Vandermast, Jun 24 2012

Crossrefs

Differs from A318473 for the first time at n=32, where a(32)=7, while A318473(32)=8.

Programs

  • Maple
    a:= n-> add(combinat[numbpart](i[2]), i=ifactors(n)[2]):
    seq(a(n), n=1..100);  # Alois P. Heinz, Aug 30 2018
  • Mathematica
    Prepend[ Array[ Plus @@ (PartitionsP /@ Last[ Transpose[ FactorInteger[ # ] ] ])&, 100, 2 ], 0 ]
    (* Second program: *)
    Array[Total[PartitionsP /@ FactorInteger[#][[All, -1]] - Boole[# == 1]] &, 87] (* Michael De Vlieger, Sep 02 2018 *)
  • PARI
    A008481(n) = vecsum(apply(e -> numbpart(e),factor(n)[,2])); \\ Antti Karttunen, Aug 30 2018

Formula

From Antti Karttunen, Aug 30 2018: (Start)
Additive with a(p^e) = A000041(e).
a(n) = A007814(A318312(n)). (End)
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) = 1.03089282973521424158..., where f(x) = -1 + (1-x) * Product_{k>=1} (1 + x^k)/(1 - x^(2*k)). - Amiram Eldar, Sep 29 2023

Extensions

Term a(1) corrected from 1 to 0 (for an empty sum) by Antti Karttunen, Aug 30 2018

A329354 a(n) = Sum_{d|n} d*omega(d).

Original entry on oeis.org

0, 2, 3, 6, 5, 17, 7, 14, 12, 27, 11, 45, 13, 37, 38, 30, 17, 62, 19, 71, 52, 57, 23, 101, 30, 67, 39, 97, 29, 162, 31, 62, 80, 87, 82, 162, 37, 97, 94, 159, 41, 220, 43, 149, 137, 117, 47, 213, 56, 152, 122, 175, 53, 197, 126, 217, 136, 147, 59, 410, 61, 157, 187, 126, 148, 336, 67, 227, 164, 342, 71, 362, 73, 187, 213, 253, 172, 394, 79
Offset: 1

Views

Author

Antti Karttunen, Nov 15 2019

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[d*PrimeNu[d], {d, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Aug 18 2021 *)
  • PARI
    A329354(n) = sumdiv(n,d,omega(d)*d);

Formula

a(n) = Sum_{d|n} d*A001221(d).
a(n) = A180253(n) - A323599(n).
a(n) = A328260(n) + A329375(n).
a(n) = Sum_{d|n} (n/d) * sopf(d). - Wesley Ivan Hurt, May 24 2021
Dirichlet g.f.: primezeta(s-1) * zeta(s-1) * zeta(s). - Ilya Gutkovskiy, Aug 18 2021
Conjecture: Sum_{k=1..n} a(k) ~ Pi^2 * n^2 * (log(log(n)) + A077761) / 12. - Vaclav Kotesovec, Mar 03 2023

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

Views

Author

Amiram Eldar, Oct 08 2024

Keywords

Comments

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, ... .

Examples

			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.
		

Crossrefs

Similar sequences: A064547, A318464, A376885.

Programs

  • Mathematica
    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]
  • PARI
    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]));}

Formula

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.

A092523 Number of distinct prime factors of n-th odd number.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 3, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 3, 1, 1, 2, 2
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 07 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Table[PrimeNu[n],{n,1,211,2}] (* Harvey P. Dale, May 25 2015 *)
  • PARI
    a(n) = omega(2*n-1); \\ Amiram Eldar, Sep 21 2024

Formula

a(n) = A001221(2*n-1).
Sum_{k=1..n} a(k) = n * (log(log(n)) + B - 1/2) + O(n/log(n)), where B is Mertens's constant (A077761). - Amiram Eldar, Sep 21 2024

A125029 a(n) = number of exponents in the prime factorization of n that are noncomposite.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 0, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 0, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 1, 0, 2, 1, 3, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 3
Offset: 1

Views

Author

Leroy Quet, Nov 16 2006

Keywords

Examples

			a(720) = 2, since the prime factorization of 720 is 2^4 * 3^2 * 5^1 and two of the exponents in this factorization are noncomposites (the exponents 2 and 1).
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Length @ Select[Last /@ FactorInteger[n], # == 1 || PrimeQ[ # ] &]; a[1] = 0; Table[a[n], {n, 110}] (* Ray Chandler, Nov 19 2006 *)
  • PARI
    A125029(n) = vecsum(apply(e -> if((1==e)||isprime(e),1,0), factorint(n)[, 2])); \\ Antti Karttunen, Jul 07 2017

Formula

From Amiram Eldar, Sep 30 2023: (Start)
Additive with a(p^e) = A080339(e).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = P(3) - Sum_{p prime >= 3} (P(p) - P(p+1)) = 0.05377157198303445809..., where P(s) is the prime zeta function. (End)

Extensions

Extended by Ray Chandler, Nov 19 2006
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