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.

A005820 3-perfect (triply perfect, tri-perfect, triperfect or sous-double) numbers: numbers such that the sum of the divisors of n is 3n.

Original entry on oeis.org

120, 672, 523776, 459818240, 1476304896, 51001180160
Offset: 1

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Author

Keywords

Comments

These six terms are believed to comprise all 3-perfect numbers. - cf. the MathWorld link. - Daniel Forgues, May 11 2010
If there exists an odd perfect number m (a famous open problem) then 2m would be 3-perfect, since sigma(2m) = sigma(2)*sigma(m) = 3*2m. - Jens Kruse Andersen, Jul 30 2014
According to the previous comment from Jens Kruse Andersen, proving that this sequence is complete would imply that there are no odd perfect numbers. - Farideh Firoozbakht, Sep 09 2014
If 2 were prepended to this sequence, then it would be the sequence of integers k such that numerator(sigma(k)/k) = A017665(k) = 3. - Michel Marcus, Nov 22 2015
From Antti Karttunen, Mar 20 2021, Sep 18 2021, (Start):
Obviously, any odd triperfect numbers k, if they exist, have to be squares for the condition sigma(k) = 3*k to hold, as sigma(k) is odd only for k square or twice a square. The square root would then need to be a term of A097023, because in that case sigma(2*k) = 9*k. (See illustration in A347391).
Conversely to Jens Kruse Andersen's comment above, any 3-perfect number of the form 4k+2 would be twice an odd perfect number. See comment in A347870.
(End)

Examples

			120 = 2^3*3*5;  sigma(120) = (2^4-1)/1*(3^2-1)/2*(5^2-1)/4 = (15)*(4)*(6) = (3*5)*(2^2)*(2*3) = 2^3*3^2*5 = (3) * (2^3*3*5) = 3 * 120. - _Daniel Forgues_, May 09 2010
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 120, p. 42, Ellipses, Paris 2008.
  • R. K. Guy, Unsolved Problems in Number Theory, B2.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. Stewart, L'univers des nombres, "Les nombres multiparfaits", Chap.15, pp 82-5, Belin/Pour la Science, Paris 2000.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 142.
  • David Wells, "The Penguin Book of Curious and Interesting Numbers," Penguin Books, London, 1986, pages 135, 159 and 185.

Crossrefs

Subsequence of the following sequences: A007691, A069085, A153501, A216780, A292365, A336458, A336461, A336745, and if there are no odd terms, then also of A334410.
Positions of 120's in A094759, 119's in A326200.

Programs

Formula

a(n) = 2*A326051(n). [provided no odd triperfect numbers exist] - Antti Karttunen, Jun 13 2019

Extensions

Wells gives the 6th term as 31001180160, but this is an error.
Edited by Farideh Firoozbakht and N. J. A. Sloane, Sep 09 2014 to remove some incorrect statements.

A095301 Numbers n such that there is some k < n with n*sigma(k) = k*sigma(n).

Original entry on oeis.org

28, 140, 200, 224, 234, 270, 308, 364, 476, 496, 532, 600, 644, 672, 700, 812, 819, 868, 936, 1036, 1148, 1170, 1204, 1316, 1400, 1484, 1488, 1540, 1638, 1652, 1708, 1800, 1820, 1876, 1988, 2016, 2044, 2200, 2212, 2324, 2380, 2464, 2480, 2492, 2574, 2600
Offset: 1

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Author

Klaus Brockhaus, Jun 01 2004

Keywords

Comments

Original name: Numbers n such that A094759(n) < n.
Agrees with A050973 without duplicates.
Also numbers n such that the value sigma(n)/n has already been reached before n. If n belongs to the sequence then A214701(n) = A214701(n-1). - Michel Marcus, Aug 19 2012

Examples

			A094759(28) = 6 < 28, hence 28 is in the sequence.
		

References

  • B. Spearman and K. S. Williams, Handbook of Estimates in the Theory of Numbers, Carleton Math. Lecture Note Series No. 14, 1975; see p. 3.2, Eq. (3.9).

Crossrefs

Programs

  • PARI
    for(n=1,2600,s=sigma(n);k=1;while(n*sigma(k)!=k*s,k++);if(k
    				
  • PARI
    allab = []; nb = 0; for (i=1, n, ab = sigma(i)/i; already = 0; if (length(allab) > 0, for (j=1, length(allab), if (ab == allab[j], already = 1; break););); if (already == 1,  nb++; print1(i, ","), allab = concat(allab, ab););)
    /* Michel Marcus, Aug 19 2012 */

Extensions

New name from Charles R Greathouse IV, Jul 21 2015

A094757 Least positive k <= n such that n*pi(k) = k*pi(n), where pi(n) is the number of primes <= n (A000720).

Original entry on oeis.org

1, 2, 3, 2, 5, 2, 7, 2, 9, 10, 11, 12, 13, 14, 10, 16, 17, 18, 19, 10, 21, 22, 23, 16, 25, 26, 27, 28, 29, 27, 31, 32, 27, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 40, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 56, 64, 65, 66, 67, 68, 69, 70, 71, 72
Offset: 1

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Author

Amarnath Murthy, May 30 2004

Keywords

Comments

Conjecture: For every n there exists a k different from n (possibly k > n) such that n*pi(k) = k*pi(n).
From David A. Corneth, Nov 15 2019: (Start)
If n*pi(k) = k*pi(n) then n/pi(n) = k/pi(k). So to find terms, one can make a list of pairs (k/pi(k), k) and sort them.
Then if for two such pairs (m/pi(m), m) and (k/pi(k), k), m > k have the same first element, i.e., m/pi(m) = k/pi(k) then a(m) = k for the least k with that ratio.
Amarnath Murthy's conjecture above is false. For n = 3 we have pi(n)/n = 2/3. For no other k we have pi(k)/k = 2/3. Therefore the conjecture is false. (End)

Examples

			a(15) = 10 as 15*pi(10) = 15*4 = 60 = 10*pi(15) = 10*6.
For k in {2, 4, 6, 8} we have pi(k)/k = 1/2 and for no k < 2 this holds. So for all these values a(k) = 2. - _David A. Corneth_, Nov 15 2019
		

Crossrefs

Cf. A095299 for n such that a(n) < n.

Programs

  • Mathematica
    Table[SelectFirst[Range[n], n PrimePi[#] == # PrimePi[n] &], {n, 72}] (* Michael De Vlieger, Dec 14 2019 *)
  • PARI
    {m=72;pi=vector(m,n,omega(n!));for(n=1,m,k=1;while(n*pi[k]!=k*pi[n],k++);print1(k,","))}
    
  • PARI
    first(n) = {n = nextprime(n); my(v = vector(n), t = -1, q = 1, res = vector(n), m); v[1] = [0, 1]; v[2] = [1/2, 2]; forprime(p = 2, n, t++; for(c = q, p - 1, v[c] = [t/c, c]; ); q = p ); v[n] = [t/n,n]; v = vecsort(v); res[1] = 1; for(i = 2, #v, if(v[i-1][1] != v[i][1], m = v[i][2]; ); res[v[i][2]] = m ); res } \\ David A. Corneth, Nov 15 2019

Extensions

Edited and extended by Klaus Brockhaus, Jun 01 2004

A094758 Least k <= n such that n*tau(k) = k*tau(n), where tau(n) is the number of divisors of n (A000005).

Original entry on oeis.org

1, 1, 3, 4, 5, 3, 7, 8, 9, 5, 11, 8, 13, 7, 15, 16, 17, 9, 19, 20, 21, 11, 23, 9, 25, 13, 27, 28, 29, 15, 31, 32, 33, 17, 35, 36, 37, 19, 39, 40, 41, 21, 43, 44, 45, 23, 47, 48, 49, 25, 51, 52, 53, 27, 55, 56, 57, 29, 59, 40, 61, 31, 63, 64, 65, 33, 67, 68, 69, 35, 71, 72, 73, 37
Offset: 1

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Author

Amarnath Murthy, May 30 2004

Keywords

Examples

			6*tau(3) = 6*2 = 3*4 = 3*tau(6), hence a(6) = 3.
		

Crossrefs

Cf. A095300 for n such that a(n) < n.

Programs

  • Maple
    A094758 := proc(n)
        for k from 1 to n do
            if n*numtheory[tau](k) = k*numtheory[tau](n) then
                return k;
            end if;
        end do:
    end proc: # R. J. Mathar, Nov 15 2019
  • Mathematica
    a[n_] := Module[{k = 1, r = DivisorSigma[0, n]/n}, While[DivisorSigma[0, k]/k != r, k++]; k]; Array[a, 100] (* Amiram Eldar, Aug 19 2019 *)
  • PARI
    for(n=1,75,s=numdiv(n);k=1;while(n*numdiv(k)!=k*s,k++);print1(k,","));

Extensions

Edited and extended by Klaus Brockhaus, Jun 01 2004

A326200 Lexicographically earliest sequence such that a(i) = a(j) => sigma(i)/i = sigma(j)/j for all i, j.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 6, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90
Offset: 1

Views

Author

Antti Karttunen, Jun 13 2019

Keywords

Comments

Restricted growth sequence transform of the abundancy index of n.
For all i, j:
a(i) = a(j) <=> A094759(i) = A094759(j),
a(i) = a(j) => A017665(i) = A017665(j),
a(i) = a(j) => A017666(i) = A017666(j).

Crossrefs

Cf. A000396 (positions of 6's), A005820 (positions of 119's).

Programs

  • PARI
    up_to = 105664; \\ (In the same equivalence class as 78, 364 and 6448).
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    v326200 = rgs_transform(vector(up_to,n,sigma(n)/n));
    A326200(n) = v326200[n];
Showing 1-5 of 5 results.