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.

A001401 Number of partitions of n into at most 5 parts.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 84, 101, 119, 141, 164, 192, 221, 255, 291, 333, 377, 427, 480, 540, 603, 674, 748, 831, 918, 1014, 1115, 1226, 1342, 1469, 1602, 1747, 1898, 2062, 2233, 2418, 2611, 2818, 3034, 3266, 3507, 3765, 4033, 4319
Offset: 0

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Comments

a(n) = T_{r}(n) for r large, where T_{r}(n) = number of outcomes in which r indistinguishable dice yield a sum r+n-1.
a(n) = coefficient of q^n in the expansion of (m choose 5)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
For n > 4: also number of partitions of n into parts <= 5: a(n) = A026820(n,5). - Reinhard Zumkeller, Jan 21 2010
Number of different distributions of n+15 identical balls in 5 boxes as x,y,z,p,q where 0 < x < y < z < p < q. - Ece Uslu and Esin Becenen, Jan 11 2016 [i.e., a(n) is the number of partitions of n+15 into 5 distinct parts. - R. J. Mathar, Feb 28 2021]
Tengely and Ulas prove that a(n) is a square only for n=1 and 2027. - Michel Marcus, Feb 11 2021

Examples

			(5 choose 5)_q = 1;
(6 choose 5)_q = q^5 + q^4 + q^3 + q^2 + q + 1;
(7 choose 5)_q = q^10 + q^9 + 2*q^8 + 2*q^7 + 3*q^6 + 3*q^5 + 3*q^4 + 2*q^3 + 2*q^2 + q + 1;
(8 choose 5)_q = q^15 + q^14 + 2*q^13 + 3*q^12 + 4*q^11 + 5*q^10 + 6*q^9 + 6*q^8 + 6*q^7 + 6*q^6 + 5*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.
a(3) = 3, i.e., {1,2,3,4,8}, {1,2,3,5,7}, {1,2,4,5,6}. Number of different distributions of 18 identical balls in 5 boxes as x,y,z,p,q where 0 < x < y < z < p < q. - _Ece Uslu_, Esin Becenen, Jan 11 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 115, row m=5 of Q(m,n) table.
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 2.
  • D. E. Knuth, The Art of Computer Programming, vol. 4, fascicle 3, Generating All Combinations and Partitions, 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).

Crossrefs

a(n) = A008284(n+5, 5), n >= 0.
Cf. A008619, A001400, A001399, A008667 (first differences), A008804.
First differences of A002622.

Programs

  • Maple
    with(combstruct):ZL6:=[S,{S=Set(Cycle(Z,card<6))}, unlabeled]:seq(count(ZL6,size=n),n=0..52); # Zerinvary Lajos, Sep 24 2007
    a:= n-> (Matrix(15, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 1, 0, 0, -1, -1, -1, 1, 1, 1, 0, 0, -1, -1, 1][i] else 0 fi)^n)[1,1]: seq(a(n), n=0..60); # Alois P. Heinz, Jul 31 2008
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=5)},unlabelled]: seq(combstruct[count](B, size=n), n=0..52); # Zerinvary Lajos, Mar 21 2009
  • Mathematica
    CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)*(1 - x^5)), {x, 0, 60} ], x ]
    a[n_] := IntegerPartitions[n, 5] // Length; Table[a[n], {n, 0, 52}] (* Jean-François Alcover, Jul 13 2012 *)
    LinearRecurrence[{1,1,0,0,-1,-1,-1,1,1,1,0,0,-1,-1,1},{1,1,2,3,5,7,10,13,18,23,30,37,47,57,70},60] (* Harvey P. Dale, Jan 05 2019 *)
  • PARI
    a(n)=#partitions(n,,5) \\ Charles R Greathouse IV, Sep 15 2014
    
  • PARI
    a(n) = (n^4 + 30*n^3 + 310*n^2 + 1320*n - 90*n*(n%2) + 2880)\2880 \\ Hoang Xuan Thanh, Aug 12 2025

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)).
a(n) = 1 + (a(n-2) + a(n-3) + a(n-4)) - (a(n-6) + (2*a(n-7)) + a(n-8)) + (a(n-10) + a(n-11) + a(n-12)) - a(n-14). - Norman J. Meluch (norm(AT)iss.gm.com), Mar 09 2000
Let a1(n) = Sum_{i=0..floor(n/3)} (1 + ceiling((n-3*i-1)/2)), a2(n) = Sum_{i=0..floor(n/4)} (1 + ceiling((n-4*i-1)/2) + a1(n-4*i-3)), then a(n) = Sum_{i=0..floor(n/5)} (1 + ceiling((n-5*i-1)/2) + a1(n-5*i-3) + a2(n-5*i-4)). - Jon Perry, Jun 27 2003
(n choose 5)_q=(q^n-1)*(q^(n-1)-1)*(q^(n-2)-1)*(q^(n-3)-1)*(q^(n-4)-1)/((q^5-1)*(q^4-1)*(q^3-1)*(q^2-1)*(q-1)).
a(n) = round(((n+5)^4 + 10*((n+5)^3 + (n+5)^2) - 75*(n+5) - 45*(n+5)*(-1)^(n+5))/2880). - Washington Bomfim, Jul 03 2012
a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) + a(n-10) - a(n-13) - a(n-14) + a(n+15). - David Neil McGrath, Sep 13 2014
a(n+5) = a(n) + A001400(n) = A001400(n)+A026811(n). - Ece Uslu, Esin Becenen, Jan 11 2016
From Vladimír Modrák, Jul 13 2022: (Start)
a(n) = Sum_{k=0..floor(n/5)} Sum_{j=0..floor(n/4)} Sum_{i=0..floor(n/3)} ceiling((max(0, n + 1 - 3*i - 4*j - 5*k))/2).
a(n) = Sum_{j=0..floor(n/5)} Sum_{i=0..floor(n/4)} floor(((max(0, n + 3 - 4*i - 5*j))^2+4)/12). (End)
a(2n) = a(2n-1) + a(n) - a(n-8) = a(n) + Sum_{k=0..n-1} A008804(k). - David García Herrero, Aug 26 2024
a(n) = floor((n^4 + 30*n^3 + 310*n^2 + 1275*n + 45*n*(-1)^n+2880)/2880). - Hoang Xuan Thanh, Aug 12 2025

Extensions

Additional comments from Michael Somos and Branislav Kisacanin (branislav.kisacanin(AT)delphiauto.com)