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 1-5 of 5 results.

A228024 Antiharmonic mean of the divisors of A228023(n) (the n-th primitive antiharmonic number).

Original entry on oeis.org

1, 13, 35, 85, 119, 185, 255, 313, 455, 481, 425, 455, 629, 841, 845, 741, 1183, 841, 1355, 1015, 1547, 1105, 1445, 2197, 1799, 2735, 3315, 2405, 3523, 2873, 2755, 3605, 4165, 4595, 4607, 5705, 5525, 5365, 7111, 7995, 5915, 7157, 6165, 8285, 5525, 9373, 10933, 11305, 7813
Offset: 1

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Comments

A228023 is the main entry for this sequence.

Crossrefs

Formula

a(n) = sigma_2(A228023(n))/sigma(A228023(n)).

A020487 Antiharmonic numbers: numbers k such that sigma_1(k) divides sigma_2(k).

Original entry on oeis.org

1, 4, 9, 16, 20, 25, 36, 49, 50, 64, 81, 100, 117, 121, 144, 169, 180, 196, 200, 225, 242, 256, 289, 324, 325, 361, 400, 441, 450, 468, 484, 500, 529, 576, 578, 605, 625, 650, 676, 729, 784, 800, 841, 900, 961, 968, 980, 1024, 1025, 1058, 1089, 1156, 1225, 1280, 1296
Offset: 1

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Comments

Numbers k such that antiharmonic mean of divisors of k is an integer. Antiharmonic mean of divisors of number m = Product (p_i^e_i) is A001157(m)/A000203(m) = Product ((p_i^(e_i+1)+1)/(p_i+1)). So a(n) = k, for some n, if A001157(k)/A000203(k) is an integer. - Jaroslav Krizek, Mar 09 2009
Squares are antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) - ... + 1 is an integer. The nonsquare antiharmonic numbers are A227771. They include the primitive antiharmonic numbers A228023, except for its first term. - Jonathan Sondow, Aug 02 2013
Sequence is infinite, see A227771. - Charles R Greathouse IV, Sep 02 2013
The term "antiharmonic" is also known as "contraharmonic". - Pahikkala Jussi, Dec 11 2013

Examples

			a(3) = 9 = 3^2; antiharmonic mean of divisors of 9 is (3^(2+1) + 1)/(3 + 1) = 7; 7 is an integer. - _Jaroslav Krizek_, Mar 09 2009
		

Crossrefs

Programs

  • Haskell
    a020487 n = a020487_list !! (n-1)
    a020487_list = filter (\x -> a001157 x `mod` a000203 x == 0) [1..]
    -- Reinhard Zumkeller, Jan 21 2014
    
  • Magma
    [n: n in [1..1300] | IsZero(DivisorSigma(2,n) mod DivisorSigma(1,n))]; // Bruno Berselli, Apr 10 2013
    
  • Mathematica
    Select[Range[2000], Divisible[DivisorSigma[2, #], DivisorSigma[1, #]]&] (* Jean-François Alcover, Nov 14 2017 *)
  • PARI
    is(n)=sigma(n,2)%sigma(n)==0 \\ Charles R Greathouse IV, Jul 02 2013
    
  • Python
    from sympy import divisor_sigma
    def ok(n): return divisor_sigma(n, 2)%divisor_sigma(n, 1) == 0
    print([k for k in range(1, 1300) if ok(k)]) # Michael S. Branicky, Feb 25 2024
    
  • Python
    # faster for producing initial segment of sequence
    from math import prod
    from sympy import factorint
    def ok(n):
        f = factorint(n)
        sigma1 = prod((p**(  e+1)-1)//(p-1)    for p, e in f.items())
        sigma2 = prod((p**(2*e+2)-1)//(p**2-1) for p, e in f.items())
        return sigma2%sigma1 ==  0
    print([k for k in range(1, 1300) if ok(k)]) # Michael S. Branicky, Feb 25 2024

A227771 Antiharmonic numbers that are not squares.

Original entry on oeis.org

20, 50, 117, 180, 200, 242, 325, 450, 468, 500, 578, 605, 650, 800, 968, 980, 1025, 1058, 1280, 1300, 1445, 1476, 1620, 1682, 1700, 1800, 1872, 2178, 2312, 2340, 2420, 2450, 2600, 2645, 2925, 3200, 3362, 3380, 3757, 3872, 4050, 4100, 4205, 4232, 4352, 4418, 4500, 4693, 5200
Offset: 1

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Jonathan Sondow, Aug 02 2013

Keywords

Comments

Given prime factorization m = product (p_i^e_i), the antiharmonic (or contraharmonic) mean of the divisors of m is sigma_2(m)/sigma_1(m) = product (p_i^(e_i+1)+1)/(p_i+1). If this is an integer, then m is called antiharmonic.
All squares are trivially antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) - ... + 1 is an integer. Sequence gives the nontrivial antiharmonic numbers.
The antiharmonic means of their divisors are A227986.
Sequence is infinite, since if n is in the sequence and gcd(n, k) = 1 then nk^2 is also in the sequence. - Charles R Greathouse IV, Aug 02 2013
Removing such terms nk^2 leaves the primitive antiharmonic numbers A228023. - Jonathan Sondow, Aug 04 2013

Examples

			sigma_2(20)/sigma_1(20) = (1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2)/(1 + 2 + 4 + 5 + 10 + 20) = 546/42 = 13 is an integer, 20 is not a square, and no smaller number has these properties, so a(1) = 20.
		

References

  • R. Guy, Unsolved Problems in Number Theory, B2 (see harmonic number).

Crossrefs

Programs

  • PARI
    is(n)=if(issquare(n),return(0)); my(f=factor(n)); denominator(prod(i=1,#f~,(f[i,1]^(f[i,2]+1)+1)/(f[i,1]+1)))==1 \\ Charles R Greathouse IV, Aug 02 2013

Formula

A001157(a(n))/A000203(a(n)) = A227986(n).

A228036 (10^n)-th primitive antiharmonic number.

Original entry on oeis.org

1, 605, 51005, 5837732, 599407380, 60462121402
Offset: 0

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We conjecture that lim_{n->oo} a(n)/100^n = lim_{n->oo} A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.10, 5.84, 5.99, 6.05 for n = 1..5, as well as by the values of A228023(n)/n^2.

Crossrefs

Formula

a(n) = A228023(10^n).

Extensions

a(5) from Charles R Greathouse IV, Sep 03 2013

A335389 Numbers k such that k and k+1 are both antiharmonic numbers (A020487).

Original entry on oeis.org

49, 324, 1024, 1444, 1681, 2600, 9800, 265225, 332928, 379456, 421200, 1940449, 4198400, 4293184, 4739328, 8346320, 11309768, 27050400, 65918161, 203694425, 384199200, 418488849, 546717924, 2239277041, 2687489280, 4866742025, 5783450400, 6933892900, 7725003664
Offset: 1

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Amiram Eldar, Jun 04 2020

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Comments

Terms of this sequence k such that k and k+1 are both nonsquares (A227771) are 203694425, 4866742025, ...
Can two consecutive numbers be both primitive antiharmonic numbers (A228023)? Numbers k such that k and k+2 are both primitive antiharmonic numbers exist - the first two are 38246258 and 344321280.

Examples

			49 is a term since both 49 and 50 are antiharmonic: sigma_2(49)/sigma(49) = 43 and sigma_2(50)/sigma(50) = 35 are both integers.
		

Crossrefs

Programs

  • Mathematica
    antihQ[n_] := Divisible[DivisorSigma[2, n], DivisorSigma[1, n]]; seq = {}; q1 = antihQ[1];  Do[q2 = antihQ[n]; If[q1 && q2, AppendTo[seq, n - 1]]; q1 = q2, {n, 2, 2 * 10^6}]; seq
Showing 1-5 of 5 results.