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 1-4 of 4 results.

A361937 Numbers k with record values of the ratio A000005(k)/A246600(k) between the total number of divisors of k and the number of divisors d of k such that the bitwise OR of k and d is equal to k.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 336, 420, 840, 1680, 3360, 6720, 7560, 15120, 30240, 60480, 95760, 120960, 176400, 191520, 257040, 352800, 383040, 514080, 1028160, 1681680, 2056320, 2998800, 3112200, 5525520, 5997600, 6224400, 8353800, 12448800, 16216200, 24897600
Offset: 1

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Author

Amiram Eldar, Mar 31 2023

Keywords

Comments

This sequence is infinite since the ratio A000005(k)/A246600(k) is unbounded. For example, if k = 2^m then A000005(k)/A246600(k) = m+1.
All the terms except for 1 are in A355670.

Examples

			The ratios A000005(k)/A246600(k) for k = 1, 2, 3 and 4 are 1, 2, 1 and 3. The record values, 1, 2 and 3, occur at 1, 2 and 4, the first 3 terms of this sequence.
		

Crossrefs

Programs

  • Mathematica
    r[n_] := DivisorSigma[0,n]/DivisorSum[n, Boole[BitOr[#, n] == n] &];
    seq[kmax_] := Module[{rm = 0, k = 1, s = {}, r1}, Do[r1 = r[k]; If[r1 > rm, rm = r1; AppendTo[s, k]], {k, 1 , kmax}]; s]; seq[10^6]
  • PARI
    r(n) = numdiv(n)/sumdiv(n, d, bitor(d, n) == n);
    lista(kmax) = {my(rm = 0, r1); for(k = 1, kmax, r1 = r(k); if(r1 > rm, rm = r1; print1(k, ", "))); }

A361807 Numbers k with record values of the ratio A000005(k)/A049419(k) between the number of divisors of k and the number of exponential divisors of k.

Original entry on oeis.org

1, 2, 6, 30, 210, 2310, 30030, 480480, 510510, 8168160, 9699690, 155195040, 223092870, 3569485920, 6469693230, 103515091680, 200560490130, 3208967842080, 7420738134810, 118731810156960, 304250263527210, 4868004216435360, 13082761331670030, 209324181306720480
Offset: 1

Views

Author

Amiram Eldar, Mar 25 2023

Keywords

Comments

This sequence is infinite since the ratio A000005(k)/A049419(k) is unbounded. For example, for k = A002110(m) we have A000005(k)/A049419(k) = 2^m.
The corresponding record values are 1, 2, 4, 8, 16, 32, 64, 96, 128, ...

Examples

			The ratios A000005(k)/A049419(k) for k = 1, 2, 3, 4, 5 and 6 are 1, 2, 2, 3/2, 2 and 4. The record values, 1, 2 and 4, occur at 1, 2 and 6, the first 3 terms of this sequence.
		

Crossrefs

Subsequence of A025487.
Similar sequences: A307870, A335832.

Programs

  • Mathematica
    f[p_, e_] := (e+1)/DivisorSigma[0, e]; r[1] = 1; r[n_] := Times @@ f @@@ FactorInteger[n]; seq[kmax_] := Module[{rm = 0, s = {}, r1}, Do[r1 = r[k]; If[r1 > rm, rm = r1; AppendTo[s, k]], {k, 1 , kmax}]; s]; seq[10^6]
  • PARI
    r(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i, 2]+1)/numdiv(f[i, 2])); }
    lista(kmax) = {my(rm = 0, r1); for(k = 1, kmax, r1 = r(k); if(r1 > rm, rm = r1; print1(k, ", "))); }

A377562 Numbers that have twice as many infinitary divisors as noninfinitary divisors.

Original entry on oeis.org

4, 9, 12, 18, 20, 25, 28, 32, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 96, 98, 99, 108, 116, 117, 121, 124, 126, 132, 140, 147, 148, 150, 153, 156, 160, 164, 169, 171, 172, 175, 188, 198, 200, 204, 207, 212, 220, 224, 228, 234, 236, 242, 243, 244, 245
Offset: 1

Views

Author

Amiram Eldar, Nov 01 2024

Keywords

Comments

Numbers k such that A037445(k) = 2 * A348341(k).
Numbers k such that A037445(k)/A000005(k) = 2/3. For numbers k in A036537, all the divisors are infinitary divisors, so A037445(k)/A000005(k) = 1. For numbers k that are not in A036537, the largest possible ratio A037445(k)/A000005(k) is 2/3.
Numbers whose prime factorization has exactly one exponent of the form 3*2^k-1, with k >= 0, and the rest of the exponents, if there are any, are of the form 2^k-1, with k >= 1.
The asymptotic density of this sequence is d * Sum_{p prime} (Sum_{k>=0} 1/p^(3*2^k-1))/(1 + Sum_{k>=1} 1/p^(2^k-1)) = 0.23171917985739378623..., where d = A327839.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := 2^DigitCount[e, 2, 1]/(e + 1); q[1] = False; q[n_] := Times @@ f @@@ FactorInteger[n] == 2/3; Select[Range[250], q]
  • PARI
    is(n) = {my(f = factor(n)); vecprod(apply(x -> (1 << hammingweight(x)) / (x+1), f[, 2])) == 2/3;}

A377563 Numbers that have fewer infinitary divisors than noninfinitary divisors.

Original entry on oeis.org

16, 36, 48, 80, 81, 100, 112, 144, 162, 176, 180, 196, 208, 225, 240, 252, 256, 272, 288, 300, 304, 324, 336, 368, 396, 400, 405, 432, 441, 450, 464, 468, 484, 496, 512, 528, 560, 567, 576, 588, 592, 612, 624, 625, 648, 656, 676, 684, 688, 700, 720, 752, 768, 784, 800
Offset: 1

Views

Author

Amiram Eldar, Nov 01 2024

Keywords

Comments

Numbers whose prime factorization has at least one exponent that has at least two zeros in its binary representation (A158582), or at least two exponents that are not of the form 2^k-1, with k >= 1 (A062289).
The asymptotic density of this sequence is 1 - d * (1 + Sum_{p prime} (Sum_{k>=0} 1/p^(3*2^k-1))/(1 + Sum_{k>=1} 1/p^(2^k-1))) = 0.07306380398261191432..., where d = A327839.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := 2^DigitCount[e, 2, 1]/(e + 1); q[1] = False; q[n_] := Times @@ f @@@ FactorInteger[n] < 1/2; Select[Range[800], q]
  • PARI
    is(n) = {my(f = factor(n)); vecprod(apply(x -> (1 << hammingweight(x)) / (x+1), f[, 2])) < 1/2;}
Showing 1-4 of 4 results.