A002133 Number of partitions of n with exactly two part sizes.
0, 0, 1, 2, 5, 6, 11, 13, 17, 22, 27, 29, 37, 44, 44, 55, 59, 68, 71, 81, 82, 102, 97, 112, 109, 136, 126, 149, 141, 168, 157, 188, 176, 212, 182, 231, 207, 254, 230, 266, 241, 300, 259, 319, 283, 344, 295, 373, 311, 386, 352, 417, 353, 452, 368, 460, 418, 492, 413
Offset: 1
Examples
a(8) = 13 because we have 71, 62, 611, 53, 5111, 422, 41111, 332, 3311, 311111, 22211, 221111, 2111111.
References
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
- George E. Andrews, Stacked lattice boxes, Ann. Comb. 3 (1999), 115-130. See L_3(n).
- E. T. Bell, The form wx+xy+yz+zu, Bull. Amer. Math. Soc., 42 (1936), 377-380.
- D. Christopher and T. Nadu, Partitions with Fixed Number of Sizes, Journal of Integer Sequences, 15 (2015), #15.11.5.
- W. J. Keith, Partitions into a small number of part sizes, Int. Jour. of Num. Thy., Vol 13 no. 1, 229-241 (2017), doi:10.1142/S1793042117500130
- P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., 19 (1921), 75-113; Coll. Papers II, pp. 303-341.
- N. Benyahia Tani, S. Bouroubi, and O. Kihel, An effective approach for integer partitions using exactly two distinct sizes of parts, Bulletin du Laboratoire 03 (2015), 18-27.
- N. B. Tani and S. Bouroubi, Enumeration of the Partitions of an Integer into Parts of a Specified Number of Different Sizes and Especially Two Sizes, J. Integer Seqs., Vol. 14 (2011), #11.3.6. (This sequence appears as the rightmost column of Table 1 on p. 10.)
Programs
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Maple
g:=sum(sum(x^(i+j)/(1-x^i)/(1-x^j),j=1..i-1),i=1..80): gser:=series(g,x=0,65): seq(coeff(gser,x^n),n=1..60); # Emeric Deutsch, Mar 30 2006 with(numtheory); D00:=n->add(tau(j)*tau(n-j),j=1..n-1); L3:=n->(D00(n)+tau(n)-sigma(n))/2; [seq(L3(n),n=1..60)]; # N. J. A. Sloane, Jun 17 2011 A002133 := proc(n) A055507(n-1)+numtheory[tau](n)-numtheory[sigma](n) ; %/2 ; end proc: # R. J. Mathar, Jun 15 2022 # Using function P from A365676: A002133 := n -> P(n, 2, n): seq(A002133(n), n = 1..59); # Peter Luschny, Sep 15 2023
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Mathematica
nn=50;ss=Sum[Sum[x^(i+j)/(1-x^i)/(1-x^j),{j,1,i-1}],{i,1,nn}];Drop[CoefficientList[Series[ss,{x,0,nn}],x],1] (* Geoffrey Critzer, Sep 13 2012 *) Table[DivisorSigma[0, n] - DivisorSigma[1, n] + Sum[DivisorSigma[0, k]*DivisorSigma[0, n - k], {k, 1, n - 1}], {n, 1, 100}]/2 (* Vaclav Kotesovec, Aug 30 2025 *)
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Python
from sympy import divisor_count, divisor_sigma def A002133(n): return sum(divisor_count(j)*divisor_count(n-j) for j in range(1,(n-1>>1)+1)) + ((divisor_count(n+1>>1)**2 if n-1&1 else 0)+divisor_count(n)-divisor_sigma(n)>>1) # Chai Wah Wu, Sep 15 2023
Formula
G.f.: Sum_{i>=1} Sum_{j=1..i-1} x^(i+j)/((1-x^i)*(1-x^j)). - Emeric Deutsch, Mar 30 2006
Andrews gives a formula which is programmed up in the Maple code below. - N. J. A. Sloane, Jun 17 2011
G.f.: (G(x)^2-H(x))/2 where G(x) = Sum_{k>0} x^k/(1-x^k) and H(x) = Sum_{k>0} x^(2*k)/(1-x^k)^2. More generally, we obtain g.f. for number of partitions of n with m types of parts if we substitute x(i) with -Sum_{k>0}(x^n/(x^n-1))^i in cycle index Z(S(m); x(1),x(2),...,x(m)) of symmetric group S(m) of degree m. - Vladeta Jovovic, Sep 18 2007
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