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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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2
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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FORMULA
| G:=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).
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EXAMPLE
| 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
| 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
| Table[Count[IntegerPartitions[n], _?(Max[#] - Min[#] <= 3 &)], {n,
30}] (Gyorgy Birkas)
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CROSSREFS
| Cf. A117142.
Sequence in context: A194205 A136413 A177337 * A177332 A115001 A008766
Adjacent sequences: A117140 A117141 A117142 * A117144 A117145 A117146
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KEYWORD
| nonn
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 27 2006
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