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.

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A036878 a(n) = p^(p-1) where p = prime(n).

Original entry on oeis.org

2, 9, 625, 117649, 25937424601, 23298085122481, 48661191875666868481, 104127350297911241532841, 907846434775996175406740561329, 88540901833145211536614766025207452637361, 550618520345910837374536871905139185678862401
Offset: 1

Views

Author

Simon Colton (simonco(AT)cs.york.ac.uk)

Keywords

Comments

Also the least refactorable number (A033950) that has the n-th prime as its least prime factor. - Robert G. Wilson v, Jun 28 2006

Examples

			5^(5-1) = 5^4 = 625.
		

Crossrefs

These integers are refactorable -- i.e., the number of divisors divides the number itself, cf. A033950.
Subset of A062981. Subsequence of A000169.
Subsequence of A111134 and A246655.

Programs

A036896 Odd refactorable numbers.

Original entry on oeis.org

1, 9, 225, 441, 625, 1089, 1521, 2025, 2601, 3249, 4761, 5625, 6561, 7569, 8649, 12321, 15129, 16641, 19881, 25281, 31329, 33489, 35721, 40401, 45369, 47961, 50625, 56169, 62001, 71289, 84681, 91809, 95481, 99225, 103041, 106929, 114921
Offset: 1

Views

Author

Simon Colton (simonco(AT)cs.york.ac.uk)

Keywords

Comments

Odd refactorable numbers are always squares.
All terms = 1 (mod 8). [Zak Seidov, May 25 2010]

Examples

			9 is refactorable because tau(9)=3 and 3 divides 9.
		

Crossrefs

Subsequence of A033950 and of A016754.

Programs

  • Mathematica
    Do[If[IntegerQ[n/DivisorSigma[0, n]], Print[n]], {n, 1, 100000, 2}]
    Select[Range[1,1001,2]^2,Divisible[#,DivisorSigma[0,#]]&] (* Harvey P. Dale, Jan 22 2012 *)
  • PARI
    is(n)=n%2&&issquare(n)&&n%numdiv(n)==0 \\ Charles R Greathouse IV, Apr 23 2013
    
  • PARI
    list(lim)=my(v=List(),f);forstep(n=1,sqrtint(lim\1),2,f=factor(n)[,2];if(n^2%prod(i=1,#f,2*f[i]+1)==0,listput(v,n^2))); Vec(v) \\ Charles R Greathouse IV, Apr 23 2013
    
  • Python
    from itertools import count, islice
    from math import prod
    from sympy import factorint
    def A036896_gen(): # generator of terms
        for n in count(1,2):
            if not (m:=n**2)%prod(e<<1|1 for e in factorint(n).values()): yield m
    A036896_list = list(islice(A036896_gen(),40)) # Chai Wah Wu, Oct 04 2024

A051278 Numbers n such that n = k/d(k) has a unique solution, where d(k) = number of divisors of k.

Original entry on oeis.org

4, 6, 9, 10, 12, 14, 15, 20, 21, 22, 26, 32, 33, 34, 35, 36, 38, 39, 42, 46, 50, 51, 55, 57, 58, 60, 62, 65, 66, 69, 70, 74, 75, 77, 78, 82, 85, 86, 87, 90, 91, 93, 94, 95, 96, 98, 100, 102, 106, 108, 110, 111, 114, 115, 118, 119, 122, 123, 126, 128, 129, 130
Offset: 1

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Author

Keywords

Comments

Because d(k) <= 2*sqrt(k), it suffices to check k from 1 to 4*n^2. - Nathaniel Johnston, May 04 2011
A051521(a(n)) = 1. - Reinhard Zumkeller, Dec 28 2011

Examples

			36 is the unique number k with k/d(k)=4.
		

Crossrefs

Programs

  • Haskell
    a051278 n = a051278_list !! (n-1)
    a051278_list = filter ((== 1) . a051521) [1..]
    -- Reinhard Zumkeller, Dec 28 2011
  • Maple
    with(numtheory): A051278 := proc(n) local ct,k: ct:=0: for k from 1 to 4*n^2 do if(n=k/tau(k))then ct:=ct+1: fi: od: if(ct=1)then return n: else return NULL: fi: end: seq(A051278(n),n=1..40);
  • Mathematica
    cnt[n_] := Count[Table[n == k/DivisorSigma[0, k], {k, 1, 4*n^2}], True]; Select[Range[130], cnt[#] == 1 &]  (* Jean-François Alcover, Oct 22 2012 *)

A051279 Numbers n such that n = k/d(k) has exactly 2 solutions, where d(k) = number of divisors of k.

Original entry on oeis.org

1, 2, 5, 7, 8, 11, 13, 16, 17, 19, 23, 24, 28, 29, 31, 37, 41, 43, 44, 47, 48, 52, 53, 56, 59, 61, 67, 68, 71, 73, 76, 79, 80, 81, 83, 84, 88, 89, 92, 97, 101, 103, 104, 107, 109, 113, 116, 120, 124, 127, 131, 132, 136, 137, 139, 148, 149, 151, 152, 154, 156
Offset: 1

Views

Author

Keywords

Comments

Because d(k) <= 2*sqrt(k), it suffices to check k from 1 to 4*n^2. - Nathaniel Johnston, May 04 2011
A051521(a(n)) = 2. - Reinhard Zumkeller, Dec 28 2011

Examples

			There are exactly 2 numbers k, 40 and 60, with k/d(k)=5.
		

Crossrefs

Programs

  • Haskell
    a051279 n = a051279_list !! (n-1)
    a051279_list = filter ((== 2) . a051521) [1..]
    -- Reinhard Zumkeller, Dec 28 2011
  • Maple
    with(numtheory): A051279 := proc(n) local ct, k: ct:=0: for k from 1 to 4*n^2 do if(n=k/tau(k))then ct:=ct+1: fi: od: if(ct=2)then return n: else return NULL: fi: end: seq(A051279(n), n=1..40); # Nathaniel Johnston, May 04 2011
  • Mathematica
    A051279 = Reap[Do[ct = 0; For[k = 1, k <= 4*n^2, k++, If[n == k/DivisorSigma[0, k], ct++]]; If[ct == 2, Print[n]; Sow[n]], {n, 1, 160}]][[2, 1]](* Jean-François Alcover, Apr 16 2012, after Nathaniel Johnston *)

A036898 List of pairs of consecutive refactorable numbers.

Original entry on oeis.org

1, 2, 8, 9, 1520, 1521, 50624, 50625, 62000, 62001, 103040, 103041, 199808, 199809, 221840, 221841, 269360, 269361, 463760, 463761, 690560, 690561, 848240, 848241, 986048, 986049, 1252160, 1252161, 1418480, 1418481, 2169728, 2169729, 2692880
Offset: 1

Views

Author

Simon Colton (simonco(AT)cs.york.ac.uk)

Keywords

Comments

Zelinsky (2002, Theorem 59, p. 15) proved that if k > 1, k and k+1 are both refactorable numbers, then k is even. As a result, a(n) == n-1 (mod 2) for n >= 3. See also A114617. - Jianing Song, Apr 01 2021

Examples

			8 is refactorable because tau(8)=4 and 4 divides 8.
9 is refactorable because tau(9)=3 and 3 divides 9.
		

Crossrefs

Programs

  • Mathematica
    SequencePosition[Table[If[Divisible[n,DivisorSigma[0,n]],1,0],{n,27*10^5}],{1,1}]//Flatten (* Harvey P. Dale, Dec 07 2021 *)
  • PARI
    isrefac(n) = ! (n % numdiv(n));
    lista(nn) = {for (n = 1, nn, if (isrefac(n) && isrefac(n+1), print1(n, ", ", n+1, ", ")););} \\ Michel Marcus, Aug 31 2013

A048166 Numbers k that are divisible by the number of unitary divisors of k (A034444).

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 64, 68, 72, 76, 80, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 136, 144, 148, 152, 160, 164, 168, 172, 176, 184, 188, 192, 196, 200, 208, 212, 216, 224, 232, 236, 240, 244, 248, 256, 264, 268
Offset: 1

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Author

Keywords

Examples

			a(81) = 392 = 2^3*7^2 has 4 unitary divisors, {1, 392, 8, 49}, and 4 divides 392.
		

Crossrefs

Programs

  • Mathematica
    okQ[n_] := Divisible[n, DivisorSum[n, Boole[CoprimeQ[#, n/#]]&]]; Select[ Range[300], okQ] (* Jean-François Alcover, Dec 05 2015 *)
    Select[Range[270], Divisible[#, 2^PrimeNu[#]] &] (* Amiram Eldar, Jul 16 2019 *)
  • PARI
    isok(n) = !(n % sumdiv(n, d, gcd(d, n/d)==1)); \\ Michel Marcus, Feb 25 2014
    
  • PARI
    isok(n) = !(n % 2^omega(n)); \\ Amiram Eldar, Jul 16 2019

Formula

Binomial transform of [1, 1, 1, 1, -3, 5, -7, 9, -11, 13, ...]. Binomial transform of this sequence is A048481. - Gary W. Adamson, Oct 23 2007

A049439 Numbers k such that the number of odd divisors of k is an odd divisor of k.

Original entry on oeis.org

1, 2, 4, 8, 9, 16, 18, 32, 36, 64, 72, 128, 144, 225, 256, 288, 441, 450, 512, 576, 625, 882, 900, 1024, 1089, 1152, 1250, 1521, 1764, 1800, 2025, 2048, 2178, 2304, 2500, 2601, 3042, 3249, 3528, 3600, 4050, 4096, 4356, 4608, 4761, 5000, 5202, 5625, 6084
Offset: 1

Views

Author

Simon Colton (simonco(AT)cs.york.ac.uk)

Keywords

Comments

Invented by the HR concept formation program.
Sequence consists of all numbers of the form A000079(k)*A036896(m). - Matthew Vandermast, Nov 14 2010

Examples

			There are 3 odd divisors of 18, namely 1,3 and 9 and 3 itself is an odd divisor of 18.
		

Crossrefs

Contains A000079 and A036896.
Subsequence of A028982. Includes A120349, A120358, A120359, A120361, A181795. See also A181794.

Programs

  • Haskell
    a049439 n = a049439_list !! (n-1)
    a049439_list = filter (\x -> ((length $ oddDivs x) `elem` oddDivs x)) [1..]
       where oddDivs n = [d | d <- [1,3..n], mod n d == 0]
    -- Reinhard Zumkeller, Aug 17 2011
    
  • Mathematica
    ok[n_] := (d = Length @ Select[Divisors[n], OddQ] ;
      IntegerQ[n/d] && OddQ[d]); Select[Range[6100], ok]
    (* Jean-François Alcover, Apr 22 2011 *)
    odQ[n_]:=Module[{ods=Select[Divisors[n],OddQ]},MemberQ[ods,Length[ ods]]]; Select[Range[7000],odQ] (* Harvey P. Dale, Dec 18 2011 *)
    Select[Range[6000], OddQ[(d = DivisorSigma[0, #/2^IntegerExponent[#, 2]])] && Divisible[#, d] &] (* Amiram Eldar, Jun 12 2022 *)
  • PARI
    is(n)=my(d=numdiv(n>>valuation(n,2))); d%2 && n%d==0 \\ Charles R Greathouse IV, Feb 07 2017

Formula

a(n) = A000079(k)*A016754(m) for appropriate k, m. - Reinhard Zumkeller, Jun 05 2008

Extensions

Example corrected by Harvey P. Dale, Jul 14 2011

A051280 Numbers n such that n = k/d(k) has exactly 3 solutions, where d(k) = number of divisors of k.

Original entry on oeis.org

3, 25, 40, 49, 54, 121, 125, 135, 140, 169, 189, 216, 220, 250, 260, 289, 297, 340, 351, 361, 375, 380, 400, 459, 460, 500, 513, 529, 580, 620, 621, 675, 729, 740, 770, 783, 820, 837, 841, 860, 875, 882, 910, 940, 961, 999, 1060, 1107, 1152, 1161, 1180, 1188, 1190
Offset: 1

Views

Author

Keywords

Comments

Many terms are of the form a(k) * p^m/(m+1), where p is coprime to the three solutions for k. The sequence of "primitive" terms (i.e. not expressible this way) begins 3, 40, 54, 125, 135, 216, 250.... Are there any such numbers that admit a fourth solution? - Charlie Neder, Feb 13 2019

Examples

			There are exactly 3 numbers k, 9, 18 and 24, with k/d(k) = 3.
		

Crossrefs

Programs

  • Mathematica
    (Select[Table[k / Length @ Divisors[k], {k, 1, 200000}], IntegerQ] // Sort // Split // Select[#, Length[#] == 3 &] &)[[All, 1]][[1 ;; 53]] (* Jean-François Alcover, Apr 22 2011 *)

A057265 Even refactorable numbers (i.e., the number of divisors is itself a divisor and it is also even).

Original entry on oeis.org

2, 8, 12, 18, 24, 40, 56, 60, 72, 80, 84, 88, 96, 104, 108, 128, 132, 136, 152, 156, 180, 184, 204, 228, 232, 240, 248, 252, 276, 288, 296, 328, 344, 348, 360, 372, 376, 384, 396, 424, 444, 448, 450, 468, 472, 480, 488, 492
Offset: 1

Views

Author

Simon Colton (simonco(AT)cs.york.ac.uk), Aug 21 2000

Keywords

Comments

Invented by the HR mathematical theory formation program.

Examples

			18 is refactorable because tau(18) = 6 and 6 divides 18 and 18 is even.
		

References

  • S. Colton, Unpublished PhD Thesis, University of Edinburgh, 2000

Crossrefs

Programs

  • Magma
    [k:k in [2..500 by 2]| IsIntegral(k/d) and IsEven(d) where d is #Divisors(k)]; // Marius A. Burtea, Jan 13 2020
  • Mathematica
    rfnQ[n_]:=Module[{ds=DivisorSigma[0,n]},Divisible[n,ds] && EvenQ[ds]];Select[Range[2,500,2],rfnQ]  (* Harvey P. Dale, Mar 14 2011 *)

Extensions

Corrected (erroneous term 36 removed) by Harvey P. Dale, Mar 14 2011

A073904 Smallest multiple k*n of n having n divisors.

Original entry on oeis.org

1, 2, 9, 8, 625, 12, 117649, 24, 36, 80, 25937424601, 60, 23298085122481, 448, 2025, 384, 48661191875666868481, 180, 104127350297911241532841, 240, 35721, 11264, 907846434775996175406740561329, 360, 10000, 53248, 26244, 1344
Offset: 1

Views

Author

Amarnath Murthy, Aug 18 2002

Keywords

Comments

Smallest refactorable number, m, such that m=k*n has n divisors. - Robert G. Wilson v, Oct 31 2005

Examples

			Smallest multiple a(n)=k*n; a(1)=1*1, a(2)=1*2, a(3)=3*3, a(4)=2*4, a(5)=125*5, a(6)=2*6, ... having d(k*n)=n divisors; d(1)=1, d(2)=2, d(3^2)=3, d(2^3)=4, d(5^4)=5, d(2^2*3)=3*2=6, ...
		

Crossrefs

Cf. A033950 (refactorable numbers, also known as tau numbers).
Cf. A110821 (SuperRefactorable numbers).

Programs

  • Mathematica
    f[n_] := Block[{k = 1}, If[ PrimeQ[n], n^(n - 1), While[d = DivisorSigma[0, k*n]; d != n, k++ ]; k*n]]; Table[ f[n], {n, 28}] (* Robert G. Wilson v *)

Formula

If p is a prime then a(p) = p^(p-1). If n = p^2 then a(n) = 2^(p-1)*p^(p-1).
a(p^r) = (2*3*5*...*p_r)^(p-1) for r < p <= p_r. a(p^r) = (2*3*...*p_(r-1))^(p-1)*p^(p-1) for p > p_r. Else a(p^r) = ...? for r >= p. Problem a(2^r) = ...? Cf. A005179(p^n)=(2*3*...*p_n)^(p-1) for p_n < 2^p. - Thomas Ordowski, Aug 20 2005
a(p^r) = (2*3...*p_(r-1)*p)^(p-1) for p > p_r; else a(p^r) = (2*3...*p...*p_m)^(p-1)*p^(p^k-p) for p <= p_r and p_m < 2^p, where m=r-k+1 for smallest k such that p^k > r, so k=floor(log(r)/log(p))+1 and p > log(p_m)/log(2). Examples: If k=1 then a(p^r) = (2*3*...*p_r)^(p-1) for r < p <= p_r. If p=2 then a(2^r) = (2*3*...*p_m)*2^(2^k-2) for r < 5. For instance, let r=4 so k=3, m=2 and a(2^4)=384. - Thomas Ordowski, Aug 22 2005
If p is a prime and n=p^r then a(p^r) = (s_1*s_2*...*s_r)^(p-1) where (s_r) is a permutation of the (ascending sequence) numbers of the form q^(p^j) for every prime q and j>=0; permutation such that s_(p^j)=p^(p^j) and shifted remainder. For example, if p=3 then (s_r): 3, 2, 3^3, 5, 7, 2^3, 11, 13, 3^9, 17, 19, ... so a(3^r) = (3*2*27*5*...*s_r)^2. - Thomas Ordowski, Aug 29 2005
If n=2^r then a(2^r) is the product of the first r members of the A109429 sequence. - Thomas Ordowski, Aug 29 2005
a(n) = n * A076931(n). - Thomas Ordowski, Oct 07 2005
a(4) = 8; a(2*prime(n)) = A299795(n), for n>1. - Bernard Schott, Nov 06 2022

Extensions

a(12) corrected by Thomas Ordowski, Aug 18 2005
Further corrections from Thomas Ordowski, Oct 07 2005
a(21), a(27) & a(28) from Robert G. Wilson v, Oct 31 2005
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