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A117143
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Number of partitions of n in which any two parts differ by at most 3.
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8
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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
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OFFSET
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1,2
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LINKS
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FORMULA
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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
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EXAMPLE
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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).
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MAPLE
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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
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MATHEMATICA
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Table[Count[IntegerPartitions[n], _?(Max[#] - Min[#] <= 3 &)], {n, 30}] (* Birkas Gyorgy, Feb 20 2011 *)
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PROG
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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
(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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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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