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 1-7 of 7 results.

A038548 Number of divisors of n that are at most sqrt(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 2, 3, 1, 4, 1, 3, 2, 2, 2, 5, 1, 2, 2, 4, 1, 4, 1, 3, 3, 2, 1, 5, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 6, 1, 2, 3, 4, 2, 4, 1, 3, 2, 4, 1, 6, 1, 2, 3, 3, 2, 4, 1, 5, 3, 2, 1, 6, 2, 2, 2, 4, 1, 6, 2, 3, 2, 2, 2, 6, 1, 3, 3, 5, 1, 4, 1, 4, 4
Offset: 1

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Author

Keywords

Comments

Number of ways to arrange n identical objects in a rectangle, modulo rotation.
Number of unordered solutions of x*y = n. - Colin Mallows, Jan 26 2002
Number of ways to write n-1 as n-1 = x*y + x + y, 0 <= x <= y <= n. - Benoit Cloitre, Jun 23 2002
Also number of values for x where x+2n and x-2n are both squares (e.g., if n=9, then 18+18 and 18-18 are both squares, as are 82+18 and 82-18 so a(9)=2); this is because a(n) is the number of solutions to n=k(k+r) in which case if x=r^2+2n then x+2n=(r+2k)^2 and x-2n=r^2 (cf. A061408). - Henry Bottomley, May 03 2001
Also number of sums of sequences of consecutive odd numbers or consecutive even numbers including sequences of length 1 (e.g., 12 = 5+7 or 2+4+6 or 12 so a(12)=3). - Naohiro Nomoto, Feb 26 2002
Number of partitions whose consecutive parts differ by exactly two.
a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24=2^3*3 and 375=3*5^3 both have prime signature (3,1). - Christian G. Bower, Jun 06 2005
Also number of partitions of n such that if k is the largest part, then each of the parts 1,2,...,k-1 occurs exactly twice. Example: a(12)=3 because we have [3,3,2,2,1,1],[2,2,2,2,2,1,1] and [1,1,1,1,1,1,1,1,1,1,1,1]. - Emeric Deutsch, Mar 07 2006
a(n) is also the number of nonnegative integer solutions of the Diophantine equation 4*x^2 - y^2 = 16*n. For example, a(24)=4 because there are 4 solutions: (x,y) = (10,4), (11,10), (14,20), (25,46). - N-E. Fahssi, Feb 27 2008
a(n) is the number of even divisors of 2*n that are <= sqrt(2*n). - Joerg Arndt, Mar 04 2010
First differences of A094820. - John W. Layman, Feb 21 2012
a(n) = #{k: A027750(n,k) <= A000196(n)}; a(A008578(n)) = 1; a(A002808(n)) > 1. - Reinhard Zumkeller, Dec 26 2012
Row lengths of the tables in A161906 and A161908. - Reinhard Zumkeller, Mar 08 2013
Number of positive integers in the sequence defined by x_0 = n, x_(k+1) = (k+1)*(x_k-2)/(k+2) or equivalently by x_k = n/(k+1) - k. - Luc Rousseau, Mar 03 2018
Expanding the first comment: Number of rectangles with area n and integer side lengths, modulo rotation. Also number of 2D grids of n congruent squares, in a rectangle, modulo rotation (cf. A000005 for rectangles instead of squares; cf. A034836 for the 3D case). - Manfred Boergens, Jun 08 2021
Number of divisors of n that have an even number of prime divisors (counted with multiplicity), or in other words, number of terms of A028260 that divide n. - Antti Karttunen, Apr 17 2022

Examples

			a(4) = 2 since 4 = 2 * 2 = 4 * 1. Also A034178(4*4) = 2 since 16 = 4^2 - 0^2 = 5^2 - 3^2. - _Michael Somos_, May 11 2011
x + x^2 + x^3 + 2*x^4 + x^5 + 2*x^6 + x^7 + 2*x^8 + 2*x^9 + 2*x^10 + x^11 + ...
		

References

  • George E. Andrews and Kimmo Eriksson, Integer Partitions, Cambridge Univ. Press, 2004, page 18, exer. 21, 22.

Crossrefs

Different from A068108. Records give A038549, A004778, A086921.
Cf. A066839, A033676, row sums of A303300.
Inverse Möbius transform of A065043.
Cf. A244664 (Dgf at s=2), A244665 (Dgf at s=3).

Programs

Formula

a(n) = ceiling(d(n)/2), where d(n) = number of divisors of n (A000005).
a(2k) = A034178(2k) + A001227(k). a(2k+1) = A034178(2k+1). - Naohiro Nomoto, Feb 26 2002
G.f.: Sum_{k>=1} x^(k^2)/(1-x^k). - Jon Perry, Sep 10 2004
Dirichlet g.f.: (zeta(s)^2 + zeta(2*s))/2. - Christian G. Bower, Jun 06 2005 [corrected by Vaclav Kotesovec, Aug 19 2019]
a(n) = (A000005(n) + A010052(n))/2. - Omar E. Pol, Jun 23 2009
a(n) = A034178(4*n). - Michael Somos, May 11 2011
2*a(n) = A161841(n). - R. J. Mathar, Mar 07 2021
a(n) = A000005(n) - A056924(n) = A056924(n) + A010052(n) = Sum_{d|n} A065043(d). - Antti Karttunen, Apr 17 2022
Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma - 1/2)*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 27 2022

A004778 Where records occur in A038548.

Original entry on oeis.org

1, 4, 12, 24, 36, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 7560, 10080, 15120, 20160, 25200, 27720, 45360, 50400, 55440, 83160, 110880, 166320, 221760, 277200, 332640, 498960, 554400, 665280, 720720, 1081080, 1441440, 2162160
Offset: 1

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Author

Chittaranjan Pardeshi, Sep 19 2003

Keywords

Comments

Records are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, ...
This sequence coincides with the highly composite numbers (A002182) excluding 2, 6, and 48. These excluded numbers occur directly after the only squares of A002182, namely 1, 4, and 36. - Jaycob Coleman, Mar 08 2015

Crossrefs

Programs

  • Mathematica
    s={}; dm = 0; Do[d =Ceiling[DivisorSigma[0,n]/2]; If[d > dm, dm = d; AppendTo[s, n]], {n, 10^5}]; s (* Amiram Eldar, Jul 07 2019 *)
  • PARI
    lista(nn) = {rec = 0; for (n=1, nn, nb = ceil(numdiv(n)/2); if (nb > rec, print1(n, ", "); rec = nb;););} \\ Michel Marcus, Sep 01 2013

A086921 Least number with at least n divisors that are at most its square root.

Original entry on oeis.org

1, 4, 12, 24, 36, 60, 120, 120, 180, 240, 360, 360, 720, 720, 720, 840, 1260, 1260, 1680, 1680, 2520, 2520, 2520, 2520, 5040, 5040, 5040, 5040, 5040, 5040, 7560, 7560, 10080, 10080, 10080, 10080, 15120, 15120, 15120, 15120, 20160, 20160, 25200, 25200
Offset: 1

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Author

N. J. A. Sloane, Sep 19 2003

Keywords

Comments

Smallest number m such that can write m = p*q (with p <= q) in at least n ways.

Crossrefs

Formula

a(n) = A061799(2n-1). - David Wasserman, Apr 12 2005

Extensions

More terms from Vladeta Jovovic and Ray Chandler, Sep 20 2003

A227068 Least number with exactly n divisors less than its square root.

Original entry on oeis.org

2, 6, 12, 24, 48, 60, 144, 120, 180, 240, 3072, 360, 900, 960, 720, 840, 5184, 1260, 36864, 1680, 2880, 3600, 12582912, 2520, 6480, 61440, 6300, 6720, 805306368, 5040, 14400, 7560, 46080, 983040, 25920, 10080, 32400, 746496, 184320, 15120
Offset: 1

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Author

T. D. Noe, Jul 11 2013

Keywords

Comments

This is similar to A038549, which counts divisors of n <= sqrt(n). Note that an upper bound on a(n) is 3*2^(n-1), which is attained at n = 2, 3, 4, 5, 11, 23, and 29 -- the number 4 and the primes in A005384 (Sophie Germain primes, p and 2p+1 are prime).
A056924(a(n)) = n and A056924(m) <> n for m < a(n). - Reinhard Zumkeller, Jul 12 2013

Examples

			The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Only 6 of these are < sqrt(60). And 60 is the first such number.
		

Crossrefs

Programs

  • C
    /* See Tek link. */
  • Haskell
    import Data.List (elemIndex); import Data.Maybe (fromJust)
    a227068 = (+ 1) . fromJust . (`elemIndex` a056924_list)
    -- Reinhard Zumkeller, Jul 12 2013
    
  • Mathematica
    nn = 22; t = Table[0, {nn}]; found = 0; n = 0; While[found < nn, n++; c = Length[Select[Divisors[n], # < Sqrt[n] &]]; If[c > 0 && c <= nn && t[[c]] == 0, t[[c]] = n; found++]]; t
    Map[Function[k, FirstPosition[#, k]], Range@ 22] &@ Table[Count[Divisors@ n, m_ /; m < Sqrt@ n], {n, 10^5}] // Flatten (* Michael De Vlieger, May 13 2016, Version 10 *)

Extensions

a(29) from Paul Tek, Jul 13 2013

A072671 Least m which can be written as i*j+i+j in n different ways: A072670(m)=n.

Original entry on oeis.org

0, 3, 11, 23, 35, 59, 191, 119, 179, 239, 575, 359, 1295, 899, 719, 839, 9215, 1259, 786431, 1679, 2879, 15359, 3599, 2519, 6479, 61439, 6299, 6719, 2359295, 5039, 3221225471, 7559, 46079, 983039, 25919, 10079, 206158430207, 32399, 184319, 15119, 44099, 20159, 5308415, 107519, 25199, 2985983, 9663676415, 27719, 233279, 45359
Offset: 0

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Author

Reinhard Zumkeller, Jun 30 2002

Keywords

Comments

a(n) = A038549(n+1) - 1.

Crossrefs

A334853 a(n) is the least number with exactly n odd divisors that are <= sqrt(n).

Original entry on oeis.org

1, 9, 30, 90, 225, 315, 630, 945, 1575, 2835, 4410, 3465, 8190, 6930, 14175, 10395, 27720, 17325, 35910, 31185, 66150, 62370, 99225, 45045, 198450, 90090, 121275, 103950, 207900, 155925, 415800, 135135, 311850, 368550, 353430, 225225, 1158300, 706860
Offset: 1

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Author

Ilya Gutkovskiy, May 13 2020

Keywords

Comments

Index of first occurrence of n in A069288.

Crossrefs

A363981 Integers k such that the smallest integer with k factor pairs has an odd number of divisors.

Original entry on oeis.org

1, 2, 5, 11, 13, 14, 17, 23, 29, 38, 41, 43, 46, 47, 53, 58, 59, 61, 67, 68, 71, 73, 74, 83, 86, 89, 94, 95, 101, 103, 107, 109, 111, 113, 116, 118, 122, 123, 127, 131, 137, 138, 143, 149, 151, 158, 163, 167, 172, 173, 178, 179, 181, 188, 191, 193, 194, 197
Offset: 1

Views

Author

Henry Nonnemaker, Jul 02 2023

Keywords

Comments

A factor pair of an integer k is an unordered pair of positive integers (a,b) such that a*b=k.
A038549(n) = min(A005179(2n-1), A005179(2n)). This sequence contains values of k where A005179(2k-1) is smaller.
Also values k such that A038549(k) is a perfect square.
I do not know if this sequence is infinite or finite, however I have checked integers up to 20000 and continued to find values at a similar density.

Examples

			The smallest number with 5 factor pairs is 36: (1,36), (2,18), (3,12), (4,9), (6,6). 36 has an odd number of divisors, 9. Thus, 5 is a term.
		

Crossrefs

Programs

  • PARI
    f(n) = min(A005179(2*n-1), A005179(2*n)); \\ A038549
    isok(k) = issquare(f(k)); \\ Michel Marcus, Jul 07 2023
  • Python
    from sympy.utilities.iterables import multiset_partitions
    from sympy.ntheory import factorint, prime
    import math
    def smallestNumWithNDivisors(n):
        partitionsOfPrimeFactors = multiset_partitions(factorint(n, multiple=True))
        candidates = []
        for partition in partitionsOfPrimeFactors:
            factorization = []
            for subset in partition:
                factorization.append(math.prod(subset))
            factorization.sort()
            factorization.reverse()
            candidate = 1
            for j in range(0, len(factorization)):
                candidate *= prime(j+1)**(factorization[j]-1)
            candidates.append(candidate)
        return min(candidates)
    for k in range(1,200):
        if smallestNumWithNDivisors(2*k-1)
    				
Showing 1-7 of 7 results.