A001400 Number of partitions of n into at most 4 parts.
1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 54, 64, 72, 84, 94, 108, 120, 136, 150, 169, 185, 206, 225, 249, 270, 297, 321, 351, 378, 411, 441, 478, 511, 551, 588, 632, 672, 720, 764, 816, 864, 920, 972, 1033, 1089, 1154, 1215, 1285, 1350, 1425, 1495
Offset: 0
Examples
(4 choose 4)_q = 1, (5 choose 4)_q = q^4 + q^3 + q^2 + q + 1, (6 choose 4)_q = q^8 + q^7 + 2*q^6 + 2*q^5 + 3*q^4 + 2*q^3 + 2*q^2 + q + 1, (7 choose 4) = q^12 + q^11 + 2*q^10 + 3*q^9 + 4*q^8 + 4*q^7 + 5*q^6 + 4*q^5 + 4*q^4 + 3*q^3 + 2*q^2 + q + 1 so the coefficient of q^0 converges to 1, q^1 to 1, q^2 to 2 and so on. G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 6*x^5 + 9*x^6 + 11*x^7 + ... a(4) = 5, i.e., {1,2,3,8}, {1,2,4,7}, {1,2,5,6}, {2,3,4,5}, {1,3,4,6}. Number of different distributions of 14 identical balls in 4 boxes as x,y,z,p where 0 < x < y < z < p. - _Ece Uslu_, Esin Becenen, Jan 11 2016
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 115, row m=4 of Q(m,n) table; p. 120, P(n,4).
- H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 2.
- G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 275.
- D. E. Knuth, The Art of Computer Programming, vol. 4, Fascicle 3, Generating All Combinations and Partitions, Addison-Wesley, 2005, Section 7.2.1.4., p. 56, exercise 31.
- 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
- T. D. Noe, Table of n, a(n) for n = 0..1000
- Nesrine Benyahia-Tani, Zahra Yahi, and Sadek Bouroubi, Ordered and non-ordered non-congruent convex quadrilaterals inscribed in a regular n-gon. Rostocker Math. Kolloq. 68, 71-79 (2013), g(Z).
- Jonathan Bloom and Nathan McNew, Counting pattern-avoiding integer partitions, arXiv:1908.03953 [math.CO], 2019.
- V. M. Buchstaber and A. V. Ustinov, Coefficient rings of formal group laws, Sbornik: Mathematics, Volume 206, Number 11.
- P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
- Éva Czabarka, Peter Dankelmann, Trevor Olsen, and László A. Székely, Wiener Index and Remoteness in Triangulations and Quadrangulations, arXiv:1905.06753 [math.CO], 2019.
- F. Ellermann, Illustration for A001400, A061924
- C. E. Frasser and G. N. Vostrov, Geodetic Graphs Homeomorphic to a Given Geodetic Graph, arXiv:1611.01873 [cs.DM], 2016. [p. 16, corollary 5]
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 353
- Gerzson Keri and Patric R. J. Östergård, The Number of Inequivalent (2R+3,7)R Optimal Covering Codes, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.7.
- G. Nebe, E. M. Rains and N. J. A. Sloane, Self-Dual Codes and Invariant Theory, Springer, Berlin, 2006.
- Jon Perry, More Partition Functions
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-2,0,0,1,1,-1).
- Index entries for two-way infinite sequences
Crossrefs
Programs
-
Haskell
a001400 n = a001400_list !! n a001400_list = scanl1 (+) a005044_list -- Reinhard Zumkeller, Feb 28 2013
-
Magma
K:=Rationals(); M:=MatrixAlgebra(K,4); q1:=DiagonalMatrix(M,[1,-1,1,-1]); p1:=DiagonalMatrix(M,[1,1,-1,-1]); q2:=DiagonalMatrix(M,[1,1,1,-1]); h:=M![1,1,1,1, 1,1,-1,-1, 1,-1,1,-1, 1,-1,-1,1]/2; G:=MatrixGroup<4,K|q1,q2,h>; MolienSeries(G);
-
Maple
A001400 := n->if n mod 2 = 0 then round(n^2*(n+3)/144); else round((n-1)^2*(n+5)/144); fi; with(combstruct):ZL5:=[S,{S=Set(Cycle(Z,card<5))}, unlabeled]:seq(count(ZL5,size=n),n=0..55); # Zerinvary Lajos, Sep 24 2007 A001400:=-(-z**8+z**9+2*z**4-z**7-1-z)/(z**2+1)/(z**2+z+1)/(z+1)**2/(z-1)**4; # [conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for an initial 1] B:=[S,{S = Set(Sequence(Z,1 <= card),card <=4)},unlabelled]: seq(combstruct[count](B, size=n), n=0..55); # Zerinvary Lajos, Mar 21 2009
-
Mathematica
CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)), {x, 0, 65} ], x ] LinearRecurrence[{1, 1, 0, 0, -2, 0, 0, 1, 1, -1}, {1, 1, 2, 3, 5, 6, 9, 11, 15, 18}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2012 *) a[n_] := Sum[Floor[(n - j - 3*k + 2)/2], {j, 0, Floor[n/4]}, {k, j, Floor[(n - j)/3]}]; Table[a[n], {n, 0, 55}] (* L. Edson Jeffery, Jul 31 2014 *) a[ n_] := With[{m = n + 5}, Round[ (2 m^3 - 3 m (5 + 3 (-1)^m)) / 288]]; (* Michael Somos, Dec 29 2014 *) a[ n_] := With[{m = Abs[n + 5] - 5}, Sign[n + 5] Length[ IntegerPartitions[ m, 4]]]; (* Michael Somos, Dec 29 2014 *) a[ n_] := With[{m = Abs[n + 5] - 5}, Sign[n + 5] SeriesCoefficient[ 1 / ((1 - x) (1 - x^2) (1 - x^3) (1 - x^4)), {x, 0, m}]]; (* Michael Somos, Dec 29 2014 *) Table[Length@IntegerPartitions[n, 4], {n, 0, 55}] (* Robert Price, Aug 18 2020 *)
-
PARI
a(n) = round(((n+4)^3 + 3*(n+4)^2 -9*(n+4)*((n+4)% 2))/144) \\ Washington Bomfim, Jul 03 2012
-
PARI
{a(n) = n+=5; round( (2*n^3 - 3*n*(5 + 3*(-1)^n)) / 288)}; \\ Michael Somos, Dec 29 2014
-
PARI
a(n) = #partitions(n,,4); \\ Ruud H.G. van Tol, Jun 02 2024
Formula
G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)).
a(n) = 1 + (a(n-2) + a(n-3) + a(n-4)) - (a(n-5) + a(n-6) + a(n-7)) + a(n-9). - Norman J. Meluch (norm(AT)iss.gm.com), Mar 09 2000
P(n, 4) = (1/288)*(2*n^3 + 6*n^2 - 9*n - 13 + (9*n+9)*pcr{1, -1}(2, n) - 32*pcr{1, -1, 0}(3, n) - 36*pcr{1, 0, -1, 0}(4, n)) (see Comtet).
Let c(n) = Sum_{i=0..floor(n/3)} (1 + ceiling((n-3*i-1)/2)), then a(n) = Sum_{i=0..floor(n/4)} (1 + ceiling((n-4*i-1)/2) + c(n-4*i-3)). - Jon Perry, Jun 27 2003
Euler transform of finite sequence [1, 1, 1, 1].
(n choose 4)_q = (q^n-1)*(q^(n-1)-1)*(q^(n-2)-1)*(q^(n-3)-1)/((q^4-1)*(q^3-1)*(q^2-1)*(q-1)).
a(n) = round(((n+4)^3 + 3*(n+4)^2 - 9*(n+4)*((n+4) mod 2))/144). - Washington Bomfim, Jul 03 2012
a(n) = a(n-1) + a(n-2) - 2*a(n-5) + a(n-8) + a(n-9) - a(n-10). - David Neil McGrath, Sep 12 2014
a(n) = -a(-10-n) for all n in Z. - Michael Somos, Dec 29 2014
a(n) - a(n+1) - a(n+3) + a(n+4) = 0 if n is odd, else floor(n/4) + 2 for all n in Z. - Michael Somos, Dec 29 2014
a(n) = n^3/144 + n^2/24 - 7*n/144 + 1 + floor(n/4)/4 + floor(n/3)/3 + (n+5)*floor(n/2)/8 + floor((n+1)/4)/4. - Vaclav Kotesovec, Aug 18 2015
a(6*n) - a(6*n+1) - a(6*n+4) + a(6*n+5) = n+1. - Richard Turk, Apr 19 2016
From Vladimír Modrák and Zuzana Soltysova, Dec 09 2020: (Start)
a(n) = round((n + 3)^2/12) + Sum_{i=0..floor(n/4)} round((n - 4*i - 1)^2/12).
a(n) = floor(((n + 3)^2 + 4)/12) + Sum_{i=0..floor(n/4)} floor(((n - 4*i - 1)^2 + 4)/12). (End)
a(n) - a(n-3) = A008642(n). - R. J. Mathar, Jun 23 2021
a(n) - a(n-2) = A025767(n). - R. J. Mathar, Jun 23 2021
a(n) = round((2*n^3 + 30*n^2 + 135*n + 175)/288 + (-1)^n*(n+5)/32). - Dave Neary, Oct 28 2021
From Vladimír Modrák, Jul 13 2022: (Start)
a(n) = Sum_{j=0..floor(n/4)} Sum_{i=0..floor(n/3)} ceiling((max(0,n + 1 - 3*i - 4*j))/2).
a(n) = Sum_{i=0..floor(n/4)} floor(((n + 3 - 4*i)^2 + 4)/12). (End)
a(n) = floor(((n+4)^2*(n+7) - 9*(n+4)*(n mod 2) + 32)/144). - Vladimír Modrák, Mar 23 2025
Comments