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 1-3 of 3 results.

A046951 a(n) is the number of squares dividing n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 2, 2, 4, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2
Offset: 1

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Author

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

Keywords

Comments

Rediscovered by the HR automatic theory formation program.
a(n) depends only on prime signature of n (cf. A025487, A046523). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3, 1).
First differences of A013936. Average value tends towards Pi^2/6 = 1.644934... (A013661, A013679). - Henry Bottomley, Aug 16 2001
We have a(n) = A159631(n) for all n < 125, but a(125) = 2 < 3 = A159631(125). - Steven Finch, Apr 22 2009
Number of 2-generated Abelian groups of order n, if n > 1. - Álvar Ibeas, Dec 22 2014 [In other words, number of order-n abelian groups with rank <= 2. Proof: let b(n) be such number. A finite abelian group is the inner direct product of all Sylow-p subgroups, so {b(n)} is multiplicative. Obviously b(p^e) = floor(e/2)+1 (corresponding to the groups C_(p^r) X C_(p^(e-r)) for 0 <= r <= floor(e/2)), hence b(n) = a(n) for all n. - Jianing Song, Nov 05 2022]
Number of ways of writing n = r*s such that r|s. - Eric M. Schmidt, Jan 08 2015
The number of divisors of the square root of the largest square dividing n. - Amiram Eldar, Jul 07 2020
The number of unordered factorizations of n into cubefree powers of primes (1, primes and squares of primes, A166684). - Amiram Eldar, Jun 12 2025

Examples

			a(16) = 3 because the squares 1, 4, and 16 divide 16.
G.f. = x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + x^10 + ...
		

Crossrefs

One more than A071325.
Differs from A096309 for the first time at n=32, where a(32) = 3, while A096309(32) = 2 (and also A185102(32) = 2).
Sum of the k-th powers of the square divisors of n for k=0..10: this sequence (k=0), A035316 (k=1), A351307 (k=2), A351308 (k=3), A351309 (k=4), A351310 (k=5), A351311 (k=6), A351313 (k=7), A351314 (k=8), A351315 (k=9), A351315 (k=10).
Sequences of the form n^k * Sum_{d^2|n} 1/d^k for k = 0..10: this sequence (k=0), A340774 (k=1), A351600 (k=2), A351601 (k=3), A351602 (k=4), A351603 (k=5), A351604 (k=6), A351605 (k=7), A351606 (k=8), A351607 (k=9), A351608 (k=10).
Cf. A082293 (a(n)==2), A082294 (a(n)==3).

Programs

  • Haskell
    a046951 = sum . map a010052 . a027750_row
    -- Reinhard Zumkeller, Dec 16 2013
    
  • Magma
    [#[d: d in Divisors(n)|IsSquare(d)]:n in [1..120]]; // Marius A. Burtea, Jan 21 2020
    
  • Maple
    A046951 := proc(n)
        local a,s;
        a := 1 ;
        for p in ifactors(n)[2] do
            a := a*(1+floor(op(2,p)/2)) ;
        end do:
        a ;
    end proc: # R. J. Mathar, Sep 17 2012
    # Alternatively:
    isbidivisible := (n, d) -> igcd(n, d) = d and igcd(n/d, d) = d:
    a := n -> nops(select(k -> isbidivisible(n, k), [seq(1..n)])): # Peter Luschny, Jun 13 2025
  • Mathematica
    a[n_] := Length[ Select[ Divisors[n], IntegerQ[Sqrt[#]]& ] ]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Jun 26 2012 *)
    Table[Length[Intersection[Divisors[n], Range[10]^2]], {n, 100}] (* Alonso del Arte, Dec 10 2012 *)
    a[ n_] := If[ n < 1, 0, Sum[ Mod[ DivisorSigma[ 0, d], 2], {d, Divisors @ n}]]; (* Michael Somos, Jun 13 2014 *)
    a[ n_] := If[ n < 2, Boole[ n == 1], Times @@ (Quotient[ #[[2]], 2] + 1 & /@ FactorInteger @ n)]; (* Michael Somos, Jun 13 2014 *)
    a[ n_] := If[ n < 0, 0, SeriesCoefficient[ Sum[ x^k^2 / (1 - x^k^2), {k, Sqrt @ n}], {x, 0, n}]]; (* Michael Somos, Jun 13 2014 *)
    f[p_, e_] := 1 + Floor[e/2]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 15 2020 *)
  • PARI
    a(n)=my(f=factor(n));for(i=1,#f[,1],f[i,2]\=2);numdiv(factorback(f)) \\ Charles R Greathouse IV, Dec 11 2012
    
  • PARI
    a(n) = direuler(p=2, n, 1/((1-X^2)*(1-X)))[n]; \\ Michel Marcus, Mar 08 2015
    
  • PARI
    a(n)=factorback(apply(e->e\2+1, factor(n)[,2])) \\ Charles R Greathouse IV, Sep 17 2015
    
  • Python
    from math import prod
    from sympy import factorint
    def A046951(n): return prod((e>>1)+1 for e in factorint(n).values()) # Chai Wah Wu, Aug 04 2024
    
  • Python
    def is_bidivisible(n, d) -> bool: return gcd(n, d) == d and gcd(n//d, d) == d
    def aList(n) -> list[int]: return [k for k in range(1, n+1) if is_bidivisible(n, k)]
    print([len(aList(n)) for n in range(1, 126)])  # Peter Luschny, Jun 13 2025
  • Scheme
    (definec (A046951 n) (if (= 1 n) 1 (* (A008619 (A007814 n)) (A046951 (A064989 n)))))
    (define (A008619 n) (+ 1 (/ (- n (modulo n 2)) 2)))
    ;; Antti Karttunen, Nov 14 2016
    

Formula

a(p^k) = A008619(k) = [k/2] + 1. a(A002110(n)) = 1 for all n. (This is true for any squarefree number, A005117). - Original notes clarified by Antti Karttunen, Nov 14 2016
a(n) = |{(i, j) : i*j = n AND i|j}| = |{(i, j) : i*j^2 = n}|. Also tau(A000188(n)), where tau = A000005.
Multiplicative with p^e --> floor(e/2) + 1, p prime. - Reinhard Zumkeller, May 20 2007
a(A130279(n)) = n and a(m) <> n for m < A130279(n); A008966(n)=0^(a(n) - 1). - Reinhard Zumkeller, May 20 2007
Inverse Moebius transform of characteristic function of squares (A010052). Dirichlet g.f.: zeta(s)*zeta(2s).
G.f.: Sum_{k > 0} x^(k^2)/(1 - x^(k^2)). - Vladeta Jovovic, Dec 13 2002
a(n) = Sum_{k=1..A000005(n)} A010052(A027750(n,k)). - Reinhard Zumkeller, Dec 16 2013
a(n) = Sum_{k = 1..n} ( floor(n/k^2) - floor((n-1)/k^2) ). - Peter Bala, Feb 17 2014
From Antti Karttunen, Nov 14 2016: (Start)
a(1) = 1; for n > 1, a(n) = A008619(A007814(n)) * a(A064989(n)).
a(n) = A278161(A156552(n)). (End)
G.f.: Sum_{k>0}(theta(q^k)-1)/2, where theta(q)=1+2q+2q^4+2q^9+2q^16+... - Mamuka Jibladze, Dec 04 2016
From Antti Karttunen, Nov 12 2017: (Start)
a(n) = A000005(n) - A056595(n).
a(n) = 1 + A071325(n).
a(n) = 1 + A001222(A293515(n)). (End)
L.g.f.: -log(Product_{k>=1} (1 - x^(k^2))^(1/k^2)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Jul 30 2018
a(n) = Sum_{d|n} A000005(d) * A008836(n/d). - Torlach Rush, Jan 21 2020
a(n) = A000005(sqrt(A008833(n))). - Amiram Eldar, Jul 07 2020
a(n) = Sum_{d divides n} mu(core(d)^2), where core(n) = A007913(n). - Peter Bala, Jan 24 2024

Extensions

Data section filled up to 125 terms and wrong claim deleted from Crossrefs section by Antti Karttunen, Nov 14 2016

A007862 Number of triangular numbers that divide n.

Original entry on oeis.org

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

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Author

Keywords

Comments

Also a(n) is the total number of ways to represent n+1 as a centered polygonal number of the form km(m+1)/2+1 for k>1. - Alexander Adamchuk, Apr 26 2007
Number of oblong numbers that divide 2n. - Ray Chandler, Jun 24 2008
The number of divisors d of 2n such that d+1 is also a divisor of 2n, see first formula. - Michel Marcus, Jun 18 2015
From Gus Wiseman, May 03 2019: (Start)
Also the number of integer partitions of n forming a finite arithmetic progression with offset 0, i.e. the differences are all equal (with the last part taken to be 0). The Heinz numbers of these partitions are given by A325327. For example, the a(1) = 1 through a(12) = 3 partitions are (A = 10, B = 11, C = 12):
1 2 3 4 5 6 7 8 9 A B C
21 42 63 4321 84
321 642
(End)

Crossrefs

Programs

  • Haskell
    a007862 = sum . map a010054 . a027750_row
    -- Reinhard Zumkeller, Jul 05 2014
    
  • Mathematica
    sup=90; TriN=Array[ (#+1)(#+2)/2&, Floor[ N[ Sqrt[ sup*2 ] ] ]-1 ]; Array[ Function[n, 1+Count[ Map[ Mod[ n, # ]&, TriN ], 0 ] ], sup ]
    Table[Count[Divisors[k], ?(IntegerQ[Sqrt[8 # + 1]] &)], {k, 105}] (* _Jayanta Basu, Aug 12 2013 *)
    Table[Length[Select[IntegerPartitions[n],SameQ@@Differences[Append[#,0]]&]],{n,0,30}] (* Gus Wiseman, May 03 2019 *)
  • PARI
    a(n) = sumdiv(n, d, ispolygonal(d, 3)); \\ Michel Marcus, Jun 18 2015
    
  • Python
    from itertools import pairwise
    from sympy import divisors
    def A007862(n): return sum(1 for a, b in pairwise(divisors(n<<1)) if a+1==b)  # Chai Wah Wu, Jun 09 2025

Formula

a(n) = Sum_{d|2*n,d+1|2*n} 1.
G.f.: Sum_{k>=1} x^A000217(k)/(1-x^A000217(k)). - Jon Perry, Jul 03 2004
a(A130317(n)) = n and a(m) <> n for m < A130317(n). - Reinhard Zumkeller, May 23 2007
a(n) = A129308(2n). - Ray Chandler, Jun 24 2008
a(n) = Sum_{k=1..A000005(n)} A010054(A027750(n,k)). - Reinhard Zumkeller, Jul 05 2014
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2. - Amiram Eldar, Dec 31 2023

Extensions

Extended by Ray Chandler, Jun 24 2008

A244964 Number of distinct generalized pentagonal numbers dividing n.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Jul 10 2014

Keywords

Comments

For more information about the generalized pentagonal numbers see A001318.

Examples

			For n = 10 the generalized pentagonal numbers <= 10 are [0, 1, 2, 5, 7]. There are three generalized pentagonal numbers that divide 10; they are [1, 2, 5], so a(10) = 3.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, 1 &, IntegerQ[Sqrt[24*# + 1]] &]; Array[a, 100] (* Amiram Eldar, Dec 31 2023 *)
  • PARI
    a(n) = sumdiv(n, d, issquare(24*d + 1)); \\ Amiram Eldar, Dec 31 2023

Formula

From Amiram Eldar, Dec 31 2023: (Start)
a(n) = Sum_{d|n} A080995(d).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 6 - 2*Pi/sqrt(3) = 2.372401... . (End)
Showing 1-3 of 3 results.