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.

A374784 Denominator of the mean unitary abundancy index of the unitary divisors of n.

Original entry on oeis.org

1, 4, 6, 8, 10, 24, 14, 16, 18, 8, 22, 16, 26, 56, 60, 32, 34, 72, 38, 80, 4, 88, 46, 96, 50, 104, 54, 112, 58, 48, 62, 64, 132, 136, 28, 16, 74, 152, 52, 160, 82, 16, 86, 176, 180, 184, 94, 64, 98, 40, 204, 208, 106, 216, 20, 224, 76, 232, 118, 160, 122, 248
Offset: 1

Views

Author

Amiram Eldar, Jul 20 2024

Keywords

Examples

			For n = 4, 4 has 2 unitary divisors, 1 and 4. Their unitary abundancy indices are usigma(1)/1 = 1 and usigma(4)/4 = 5/4, and their mean unitary abundancy index is (1 + 5/4)/2 = 9/8. Therefore a(4) = denominator(9/8) = 8.
		

Crossrefs

Cf. A034444, A034448 (usigma), A077610, A374783 (numerators).
Similar sequences: A374777/A374778, A374786/A374787.

Programs

  • Mathematica
    f[p_, e_] := 1 + 1/(2*p^e); a[1] = 1; a[n_] := Denominator[Times @@ f @@@ FactorInteger[n]]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); denominator(prod(i = 1, #f~, 1 + 1/(2*f[i,1]^f[i,2])));}

A374786 Numerator of the mean infinitary abundancy index of the infinitary divisors of n.

Original entry on oeis.org

1, 5, 7, 9, 11, 35, 15, 45, 19, 11, 23, 21, 27, 75, 77, 33, 35, 95, 39, 99, 5, 115, 47, 105, 51, 135, 133, 135, 59, 77, 63, 165, 161, 175, 33, 19, 75, 195, 63, 99, 83, 25, 87, 207, 209, 235, 95, 77, 99, 51, 245, 243, 107, 665, 23, 675, 91, 295, 119, 231, 123, 315
Offset: 1

Views

Author

Amiram Eldar, Jul 20 2024

Keywords

Comments

The infinitary abundancy index of a number k is A049417(k)/k.
The record values of a(n)/A374787(n) are attained at the terms of A037992.
The least number k such that a(k)/A374787(k) is larger than 2, 3, 4, ..., is A037992(6) = 7560, A037992(33) = 1370819010042780920891599455129161859473627856000, ... .

Examples

			For n = 4, 4 has 2 infinitary divisors, 1 and 4. Their infinitary abundancy indices are isigma(1)/1 = 1 and isigma(4)/4 = 5/4, and their mean infinitary abundancy index is (1 + 5/4)/2 = 9/8. Therefore a(4) = numerator(9/8) = 9.
		

Crossrefs

Similar sequences: A374777/A374778, A374783/A374784.

Programs

  • Mathematica
    f[p_, e_] := p^(2^(-1 + Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)])); a[1] = 1; a[n] := Numerator[Times @@ (1 + 1/(2*Flatten@ (f @@@ FactorInteger[n])))]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), b); numerator(prod(i = 1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], 1 + 1/(2*f[i, 1]^(2^(#b-k))), 1))));}

Formula

Let f(n) = a(n)/A374787(n). Then:
f(n) = (Sum_{d infinitary divisor of n} isigma(d)/d) / id(n), where isigma(n) is the sum of infinitary divisors of n (A049417), and id(n) is their number (A037445).
f(n) is multiplicative with f(p^e) = Product{k>=1, e_k=1} (1 + 1/(2*p^(2^(k+1)))), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
f(n) = (Sum_{d infinitary divisor of n} d*id(d)) / (n*id(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} f(k) = Product_{P} (1 + 1/(2*P*(P+1))) = 1.21407233718434377029..., where P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376). For comparison, the asymptotic mean of the infinitary abundancy index over all the positive integers is 1.461436... = 2 * A327574.
Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).
f(n) <= A374777(n)/A374778(n) with equality if and only if n is squarefree (A005117).
f(n) >= A374783(n)/A374784(n) with equality if and only if n is in A138302.

A374787 Denominator of the mean infinitary abundancy index of the infinitary divisors of n.

Original entry on oeis.org

1, 4, 6, 8, 10, 24, 14, 32, 18, 8, 22, 16, 26, 56, 60, 32, 34, 72, 38, 80, 4, 88, 46, 64, 50, 104, 108, 112, 58, 48, 62, 128, 132, 136, 28, 16, 74, 152, 52, 64, 82, 16, 86, 176, 180, 184, 94, 64, 98, 40, 204, 208, 106, 432, 20, 448, 76, 232, 118, 160, 122, 248
Offset: 1

Views

Author

Amiram Eldar, Jul 20 2024

Keywords

Examples

			For n = 4, 4 has 2 infinitary divisors, 1 and 4. Their infinitary abundancy indices are isigma(1)/1 = 1 and isigma(4)/4 = 5/4, and their mean infinitary abundancy index is (1 + 5/4)/2 = 9/8. Therefore a(4) = denominator(9/8) = 8.
		

Crossrefs

Cf. A037445, A049417 (isigma), A077609, A374786 (numerators).
Similar sequences: A374777/A374778, A374783/A374784.

Programs

  • Mathematica
    f[p_, e_] := p^(2^(-1 + Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)])); a[1] = 1; a[n] := Denominator[Times @@ (1 + 1/(2*Flatten@ (f @@@ FactorInteger[n])))]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), b); denominator(prod(i = 1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], 1 + 1/(2*f[i, 1]^(2^(#b-k))), 1))));}

A374785 Numbers whose unitary divisors have a mean unitary abundancy index that is larger than 2.

Original entry on oeis.org

223092870, 281291010, 300690390, 6469693230, 6915878970, 8254436190, 8720021310, 9146807670, 9592993410, 10407767370, 10485364890, 10555815270, 11125544430, 11532931410, 11797675890, 11823922110, 12095513430, 12328305990, 12598876290, 12929686770, 13162479330
Offset: 1

Views

Author

Amiram Eldar, Jul 20 2024

Keywords

Comments

Numbers k such that A374783(k)/A374784(k) > 2.
The least odd term is A070826(43) = 5.154... * 10^74, and the least term that is coprime to 6 is Product_{k=3..219} prime(k) = 1.0459... * 10^571.
The least nonsquarefree (A013929) term is a(613) = 148802944290 = 2 * 3 * 5 * 7 * 11 * 13 * 17 *19 * 23^2 * 29.
All the terms are nonpowerful numbers (A052485). For powerful numbers (A001694) k, A374783/(k)/A374784(k) < Product_{p prime} (1 + 1/(2*p)) = 1.242534... (A366586).

Examples

			223092870 is a term since A374783(223092870)/A374784(223092870) = 666225/330752 = 2.014... > 2.
		

Crossrefs

Subsequence of A052485.
Similar sequences: A245214, A374788.

Programs

  • Mathematica
    f[p_, e_] := 1 + 1/(2*p^e); r[1] = 1; r[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[4*10^8], s[#] > 2 &]
  • PARI
    is(n) = {my(f = factor(n)); prod(i = 1, #f~, 1 + 1/(2*f[i,1]^f[i,2])) > 2;}

Formula

A001221(a(n)) >= 9.
Showing 1-4 of 4 results.