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.

Showing 1-7 of 7 results.

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.

A377019 Numbers whose prime factorization has exponents that are all factorial numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25, 26, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78
Offset: 1

Views

Author

Amiram Eldar, Oct 13 2024

Keywords

Comments

First differs from its subsequence A004709 and from A344742 at n = 55: a(55) = 64 = 2^6 is not a term of A004709 and A344742.
Numbers k such that A376885(k) = A001221(k).
The asymptotic density of this sequence is Product_{p prime} (1 - 1/p^3 + (1 - 1/p) * (Sum_{k>=3} 1/p^(k!))) = 0.84018238588352905855... .

Crossrefs

Subsequence of A377020.
Subsequences: A005117, A004709.

Programs

  • Mathematica
    factorialQ[n_] := factorialQ[n] = Module[{m = n, k = 2}, While[Divisible[m, k], m /= k; k++]; m == 1]; q[n_] := AllTrue[FactorInteger[n][[;;, 2]], factorialQ]; Select[Range[100], q]
  • PARI
    isf(n) = {my(k = 2); while(!(n % k), n /= k; k++); n == 1;}
    is(k) = {my(e = factor(k)[, 2]); for(i = 1, #e, if(!isf(e[i]), return(0))); 1;}

A376888 The sum of divisors of n that are products of factors of the form p^(k!) with multiplicities not larger than their multiplicity in n, where p is a prime and k >= 1, when the factorization of n is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Original entry on oeis.org

1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 21, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 63, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 84, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 65, 84, 144
Offset: 1

Views

Author

Amiram Eldar, Oct 08 2024

Keywords

Comments

See A376885 for details about this factorization.
First differs from A188999 at n = 16.
The number of these divisors is given by A376887(n).

Examples

			For n = 12 = 2^2 * 3^1, the representation of 2 in factorial base is 10, i.e., 2 = 2!, so 12 = (2^(2!))^1 * (3^(1!))^1 and a(12) is the sum of the 4 divisors 1 + 3 + 4 + 12 = 20.
		

Crossrefs

Programs

  • Mathematica
    ff[q_, s_] := (q^(s + 1) - 1)/(q - 1); f[p_, e_] := Module[{k = e, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, AppendTo[s, {p^(m - 1)!, r}];]; m++]; Times @@ ff @@@ s]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    fdigits(n) = {my(k = n, m = 2, r, s = []); while([k, r] = divrem(k, m); k != 0 || r != 0, s = concat(s, r); m++); s;}
    a(n) = {my(f = factor(n), p = f[, 1], e = f[, 2], d); prod(i = 1, #p, prod(j = 1, #d=fdigits(e[i]), (p[i]^(j!*(d[j]+1)) - 1)/(p[i]^j! - 1)));}

Formula

Multiplicative: if e = Sum_{k>=1} d_k * k! (factorial base representation), then a(p^e) = Product_{k>=1} (p^(k!*{d_k+1}) - 1)/(p^(k!) - 1).

A377021 Numbers whose prime factorization has exponents that are all sums of distinct factorials (A059590, where 0! and 1! are not considered distinct).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71
Offset: 1

Views

Author

Amiram Eldar, Oct 13 2024

Keywords

Comments

First differs from its subsequence A046100 at n = 61: a(61) = 64 is not a term of A046100.
Numbers k such that A376885(k) = A376886(k).
Numbers that are "squarefree" when they are factorized into factors of the form p^(k!), where p is a prime and k >= 1, a factorization that is done using the factorial-base representation of the exponents in the prime factorization (see A376885 for more details). Each factor p^(k!) has a multiplicity 1.
The asymptotic density of this sequence is Product_{p prime} (1 - 1/p^2 + (1 - 1/p) * (Sum_{k>=2} 1/p^A059590(k))) = 0.93973112474919498992... .

Crossrefs

Programs

  • Mathematica
    expQ[n_] := expQ[n] = Module[{k = n, m = 2, r, s = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 1, s = 0; Break[]]; m++]; s == 1]; q[n_] := AllTrue[FactorInteger[n][[;;, 2]], expQ]; Select[Range[100], q]
  • PARI
    isexp(n) = {my(k = n, m = 2, r); while([k, r] = divrem(k, m); k != 0 || r != 0, if(r > 1, return(0)); m++); 1;}
    is(k) = {my(e = factor(k)[, 2]); for(i = 1, #e, if(!isexp(e[i]), return(0))); 1;}

A377022 Numbers whose prime factorization has exponents that have no digit 1 in their factorial-base representation (A255411).

Original entry on oeis.org

1, 16, 81, 625, 1296, 2401, 4096, 10000, 14641, 28561, 38416, 50625, 65536, 83521, 130321, 194481, 234256, 262144, 279841, 331776, 456976, 531441, 707281, 810000, 923521, 1185921, 1336336, 1500625, 1874161, 2085136, 2313441, 2560000, 2825761, 3111696, 3418801
Offset: 1

Views

Author

Amiram Eldar, Oct 13 2024

Keywords

Comments

Numbers that are "powerful" when they are factorized into factors of the form p^(k!), where p is a prime and k >= 1, a factorization that is done using the factorial-base representation of the exponents in the prime factorization (see A376885 for more details). Each factor p^(k!) has a multiplicity that is larger than 1.

Crossrefs

Analogous to A001694.
Subsequence of A036967.

Programs

  • Mathematica
    expQ[n_] := expQ[n] = Module[{k = n, m = 2, r, s = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r == 1, s = 0; Break[]]; m++]; s == 1]; seq[lim_] := Module[{p = 2, s = {1}, emax, es}, While[(emax = Floor[Log[p, lim]]) > 3, es = Select[Range[0, emax], expQ]; s = Union[s, Select[Union[Flatten[Outer[Times, s, p^es]]], # <= lim &]]; p = NextPrime[p]]; s]; seq[4*10^6]
  • PARI
    isexp(n) = {my(k = n, m = 2, r); while([k, r] = divrem(k, m); k != 0 || r != 0, if(r == 1, return(0)); m++); 1;}
    is(k) = {my(e = factor(k)[, 2]); for(i = 1, #e, if(!isexp(e[i]), return(0))); 1;}

Formula

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + Sum_{k>=1} 1/p^A255411(k)) = 1.07819745085315583226... .

A376887 The number of divisors of n that are products of factors of the form p^(k!) with multiplicities not larger than their multiplicity in n, where p is a prime and k >= 1, when the factorization of n is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 6, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 2, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Oct 08 2024

Keywords

Comments

See A376885 for details about this factorization.
If n = Product p_i^e_i is the canonical prime factorization of n, then the divisors that are counted by this function are d = Product p_i^s_i, where all the digits of s_i in factorial base are not larger than the corresponding digits of e_i.
The sum of these divisors is given by A376888(n).

Examples

			For n = 12 = 2^2 * 3^1, the representation of 2 in factorial base is 10, i.e., 2 = 2!, so 12 = (2^(2!))^1 * (3^(1!))^1 and a(12) = (1+1) * (1+1) = 4, corresponding to the 4 divisors 1, 3, 4 and 12.
		

Crossrefs

Programs

  • Mathematica
    fdigprod[n_] := Module[{k = n, m = 2, r, s = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s *= (r+1); m++]; s]; f[p_, e_] := fdigprod[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    fdigprod(n) = {my(k = n, m = 2, r, s = 1); while([k, r] = divrem(k, m); k != 0 || r != 0, s *= (r+1); m++); s;}
    a(n) = {my(e = factor(n)[, 2]); prod(i = 1, #e, fdigprod(e[i]));}

Formula

Multiplicative with a(p^e) = A227154(e).

A385378 The maximum possible number of distinct factors in the factorization of n into prime powers (A246655).

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

Views

Author

Amiram Eldar, Jun 27 2025

Keywords

Comments

Differs from A376885 and A384422 at n = 32, 64, 96, 128, 160, 192, ... .
Differs from A086435 at n = 36, 100, 144, 180, 196, 225, ... .
Differs from A375272 at n = 128, 384, 640, 896, 1024, 1152, ... .
a(n) depends only on the prime signature of n (A118914).
The indices of records in this sequence are the partial products of the sequence of powers of primes (A000961), i.e., the terms in A024923.
The least index n such that a(n) = k, for k = 0, 1, 2, ..., is A024923(k+1).

Examples

			      n | a(n) | factorization
  ------+------+--------------------------------
      2 |  1   | 2
      6 |  2   | 2 * 3
     24 |  3   | 2 * 3 * 2^2
    120 |  4   | 2 * 3 * 2^2 * 5
    840 |  5   | 2 * 3 * 2^2 * 5 * 7
   6720 |  6   | 2 * 3 * 2^2 * 5 * 7 * 2^3
  60480 |  7   | 2 * 3 * 2^2 * 5 * 7 * 2^3 * 3^2
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Floor[(Sqrt[8*e + 1] - 1)/2]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecsum(apply(x -> (sqrtint(8*x+1)-1)\2 , factor(n)[, 2]));

Formula

Additive with a(p^e) = A003056(e).
a(n) >= A001221(n), with equality if and only if n is cubefree (A004709).
a(n) >= 1 for n >= 2, with equality if and only if n is a prime or a square of a prime (A000430).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761), C = Sum_{k>=2} P(k*(k+1)/2) = 0.19285739770001405035..., and P is the prime zeta function.
Showing 1-7 of 7 results.