A117143 Number of partitions of n in which any two parts differ by at most 3.
1, 2, 3, 5, 7, 10, 13, 17, 22, 27, 33, 41, 48, 57, 68, 78, 90, 105, 118, 134, 153, 170, 190, 214, 235, 260, 289, 315, 345, 380, 411, 447, 488, 525, 567, 615, 658, 707, 762, 812, 868, 931, 988, 1052, 1123, 1188, 1260, 1340, 1413, 1494, 1583, 1665, 1755, 1854
Offset: 1
Examples
a(6) = 10 because we have [6], [4,2], [4,1,1], [3,3], [3,2,1], [3,1,1,1], [2,2,2], [2,2,1,1], [2,1,1,1,1] and [1,1,1,1,1,1] ([5,1] does not qualify).
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
- Jonathan Bloom, Nathan McNew, Counting pattern-avoiding integer partitions, arXiv:1908.03953 [math.CO], 2019.
- Index entries for linear recurrences with constant coefficients, signature (1,1,1,-2,-2,1,1,1,-1).
Programs
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Magma
[(2*Floor((n+2)/3)*(14*Floor((n+2)/3)^2-(10*n+21)*Floor((n+2)/3)+2*(n^2+5*n+7))-(1-(-1)^Floor((n+2)/3))*(-1)^(n+2-Floor((n+2)/3)))/16: n in [1..60]]; // Vincenzo Librandi, May 12 2015
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Maple
g:=sum(x^k/(1-x^k)/(1-x^(k+1))/(1-x^(k+2))/(1-x^(k+3)),k=1..85): gser:=series(g,x=0,65): seq(coeff(gser,x^n),n=1..59); with(combinat): for n from 1 to 7 do P:=partition(n): A:={}: for j from 1 to nops(P) do if P[j][nops(P[j])]-P[j][1]<4 then A:=A union {P[j]} else A:=A fi od: print(A); od: # this program yields the partitions -
Mathematica
Table[Count[IntegerPartitions[n], ?(Max[#] - Min[#] <= 3 &)], {n, 30}] (* _Birkas Gyorgy, Feb 20 2011 *)
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PARI
Vec(x*(x^5-x^4-x^3+x+1)/((x-1)^4*(x+1)*(x^2+x+1)^2) + O(x^100)) \\ Colin Barker, Mar 05 2015
Formula
G.f.: sum(x^k/[(1-x^k)(1-x^(k+1))(1-x^(k+2))(1-x^(k+3))], k=1..infinity). More generally, the g.f. of the number of partitions in which any two parts differ by at most b is sum(x^k/product(1-x^j, j=k..k+b), k=1..infinity).
G.f.: x*(x^5-x^4-x^3+x+1) / ((x-1)^4*(x+1)*(x^2+x+1)^2). - Colin Barker, Mar 05 2015
a(n)=(2*floor((n+2)/3)*(14*floor((n+2)/3)^2-(10*n+21)*floor((n+2)/3)+2*(n^2+5*n+7))-(1-(-1)^floor((n+2)/3))*(-1)^(n+2-floor((n+2)/3)))/16. - Luce ETIENNE, May 12 2015