OFFSET
0,12
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
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.
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
EXAMPLE
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)
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 *)
PROG
(PARI) is_ok(k, i, j)=0<i&&j>=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
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Jun 30 2002
STATUS
approved