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 1-2 of 2 results.

A072670 Number of ways to write n as i*j + i + j, 0 < i <= j.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 2, 0, 2, 0, 2, 1, 1, 0, 3, 1, 1, 1, 2, 0, 3, 0, 2, 1, 1, 1, 4, 0, 1, 1, 3, 0, 3, 0, 2, 2, 1, 0, 4, 1, 2, 1, 2, 0, 3, 1, 3, 1, 1, 0, 5, 0, 1, 2, 3, 1, 3, 0, 2, 1, 3, 0, 5, 0, 1, 2, 2, 1, 3, 0, 4, 2, 1, 0, 5, 1, 1, 1, 3, 0, 5, 1, 2, 1, 1, 1, 5, 0, 2, 2, 4, 0, 3, 0, 3, 3
Offset: 0

Views

Author

Reinhard Zumkeller, Jun 30 2002

Keywords

Comments

a(n) is the number of partitions of n+1 with summands in arithmetic progression having common difference 2. For example a(29)=3 because there are 3 partitions of 30 that are in arithmetic progressions: 2+4+6+8+10, 8+10+12 and 14+16. - N-E. Fahssi, Feb 01 2008
From Daniel Forgues, Sep 20 2011: (Start)
a(n) is the number of nontrivial factorizations of n+1, in two factors.
a(n) is the number of ways to write n+1 as i*j + i + j + 1 = (i+1)(j+1), 0 < i <= j. (End)
a(n) is the number of ways to write n+1 as i*j, 1 < i <= j. - Arkadiusz Wesolowski, Nov 18 2012
For a generalization, see comment in A260804. - Vladimir Shevelev, Aug 04 2015
Number of partitions of n into 3 parts whose largest part is equal to the product of the other two. - Wesley Ivan Hurt, Jan 04 2022

Examples

			a(11)=2: 11 = 1*5 + 1 + 5 = 2*3 + 2 + 3.
From _Daniel Forgues_, Sep 20 2011 (Start)
Number of nontrivial factorizations of n+1 in two factors:
  0 for the unit 1 and prime numbers
  1 for a square: n^2 = n*n
  1 for 6 (2*3), 10 (2*5), 14 (2*7), 15 (3*5)
  1 for a cube: n^3 = n*n^2
  2 for 12 (2*6, 3*4), for 18 (2*9, 3*6) (End)
		

Crossrefs

Programs

  • Maple
    0, seq(ceil(numtheory:-tau(n+1)/2)-1, n=1..100); # Robert Israel, Aug 04 2015
  • Mathematica
    p2[n_] := 1/2 (Length[Divisors[n]] - 2 + ((-1)^(Length[Divisors[n]] + 1) + 1)/2); Table[p2[n + 1], {n, 0, 104}] (* N-E. Fahssi, Feb 01 2008 *)
    Table[Ceiling[DivisorSigma[0, n + 1]/2] - 1, {n, 0, 104}] (* Arkadiusz Wesolowski, Nov 18 2012 *)
  • PARI
    is_ok(k,i,j)=0=i&&k===i*j+i+j;
    first(m)=my(v=vector(m,z,0));for(l=1,m,for(j=1,l,for(i=1,j,if(is_ok(l,i,j),v[l]++))));concat([0],v); /* Anders Hellström, Aug 04 2015 */
    
  • PARI
    a(n)=(numdiv(n+1)+issquare(n+1))/2-1 \\ Charles R Greathouse IV, Jul 14 2017

Formula

a(n) = A038548(n+1) - 1.
From N-E. Fahssi, Feb 01 2008: (Start)
a(n) = p2(n+1), where p2(n) = (1/2)*(d(n) - 2 + ((-1)^(d(n)+1)+1)/2); d(n) is the number of divisors of n: A000005.
G.f.: Sum_{n>=1} a(n) x^n = 1/x Sum_{k>=2} x^(k^2)/(1-x^k). (End)
lim_{n->infinity} a(A002110(n)-1) = infinity. - Vladimir Shevelev, Aug 04 2015
a(n) = A161840(n+1)/2. - Omar E. Pol, Feb 27 2019
Sum_{k=1..n} a(k) ~ n * (log(n) + 2*gamma - 3) / 2, where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 14 2024

A038549 Least number with exactly n divisors that are at most its square root.

Original entry on oeis.org

1, 4, 12, 24, 36, 60, 192, 120, 180, 240, 576, 360, 1296, 900, 720, 840, 9216, 1260, 786432, 1680, 2880, 15360, 3600, 2520, 6480, 61440, 6300, 6720, 2359296, 5040, 3221225472, 7560, 46080, 983040, 25920, 10080, 206158430208, 32400, 184320
Offset: 1

Views

Author

Keywords

Comments

Least number of identical objects that can be arranged in exactly n ways in a rectangle, modulo rotation.
Smallest number which has n distinct unordered factorizations of the form x*y. - Lekraj Beedassy, Jan 09 2008
Note that an upper bound on a(n) is 3*2^(n-1), which is attained at n = 4 and the odd primes in A005382 (primes p such that 2p-1 is also prime). - T. D. Noe, Jul 13 2013

Crossrefs

Cf. A038548 (records), A072671, A004778, A086921.
Cf. A227068 (similar, but with limit < sqrt).

Programs

  • Haskell
    import Data.List (elemIndex)
    import Data.Maybe (fromJust)
    a038549 = (+ 1) . fromJust . (`elemIndex` a038548_list)
    -- Reinhard Zumkeller, Dec 26 2012
  • Mathematica
    nn = 18; 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 (* T. D. Noe, Jul 10 2013 *)

Formula

a(n) = min(A005179(2n-1), A005179(2n)).

Extensions

More terms from David W. Wilson.
Showing 1-2 of 2 results.