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

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

Author

Reinhard Zumkeller, Jun 30 2002

Keywords

• nonn

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)
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Joseph W. Andrushkiw, Roman I. Andrushkiw and Clifton E. Corzatt, Representations of Positive Integers as Sums of Arithmetic Progressions, Mathematics Magazine, Vol. 49, No. 5 (Nov., 1976), pp. 245-248.
• M. A. Nyblom and C. Evans, On the enumeration of partitions with summands in arithmetic progression, Australian Journal of Combinatorics, Vol. 28 (2003), pp. 149-159.
• Vladimir Shevelev, Representation of positive integers by the form x1...xk+x1+...+xk, arXiv:1508.03970 [math.NT], 2015.

Crossrefs

Cf. A001620, A067432, A066938, A072668, A006093, A072671, A161840, A260803, A260804.

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