A018293 Divisors of 120.
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Offset: 1
Crossrefs
Programs
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Mathematica
Divisors[120] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
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PARI
divisors(120)
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Divisors[120] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
divisors(120)
Divisors[180] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2012 *)
divisors(180) \\ Charles R Greathouse IV, Jun 21 2017
Divisors[240] (* Harvey P. Dale, Feb 25 2015 *)
divisors(240) \\ Charles R Greathouse IV, Jun 21 2017
The a(0) = 1 through a(11) = 9 partitions (empty columns indicated by dots): () . . . . (3,2) (3,2,1) (4,3) (5,3) (5,4) (6,4) (6,5) (5,2) (4,3,1) (7,2) (7,3) (7,4) (5,2,1) (4,3,2) (5,3,2) (8,3) (5,3,1) (5,4,1) (9,2) (7,2,1) (5,4,2) (4,3,2,1) (6,3,2) (6,4,1) (7,3,1) (5,3,2,1)
Table[Length[Select[IntegerPartitions[n],#=={}||UnsameQ@@#&&!And@@IntegerQ/@(#/Min@@#)||UnsameQ@@#&&!And@@IntegerQ/@(Max@@#/#)&]],{n,0,30}]
From _Gus Wiseman_, Apr 18 2021: (Start) The a(6) = 4 through a(12) = 13 partitions: (6) (7) (8) (9) (10) (11) (12) (3,3) (4,4) (6,3) (5,5) (6,6) (4,2) (6,2) (3,3,3) (8,2) (8,4) (2,2,2) (4,2,2) (4,4,2) (9,3) (2,2,2,2) (6,2,2) (10,2) (4,2,2,2) (4,4,4) (2,2,2,2,2) (6,3,3) (6,4,2) (8,2,2) (3,3,3,3) (4,4,2,2) (6,2,2,2) (4,2,2,2,2) (2,2,2,2,2,2) (End)
with(combinat): with(numtheory): a := proc(n) c := 0: l := sort(convert(divisors(n), list)): for i from 1 to nops(l)-1 do c := c+numbpart(l[i]-1) od: RETURN(c): end: for j from 2 to 100 do printf(`%d,`,a(j)) od: # James Sellers, Jun 21 2003 # second Maple program: a:= n-> max(1, add(combinat[numbpart](d-1), d=numtheory[divisors](n) minus {n})): seq(a(n), n=1..69); # Alois P. Heinz, Feb 15 2023
a[n_] := If[n==1, 1, Sum[PartitionsP[d-1], {d, Most@Divisors[n]}]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Feb 15 2023 *)
The a(5) = 1 through a(12) = 9 partitions: (3,2) (3,2,1) (4,3) (5,3) (5,4) (6,4) (6,5) (7,5) (5,2) (4,3,1) (7,2) (7,3) (7,4) (5,4,3) (5,2,1) (4,3,2) (5,3,2) (8,3) (6,4,2) (5,3,1) (5,4,1) (9,2) (6,5,1) (7,2,1) (5,4,2) (7,3,2) (4,3,2,1) (6,4,1) (7,4,1) (7,3,1) (8,3,1) (5,3,2,1) (9,2,1) (5,4,2,1)
Table[Length[Select[IntegerPartitions[n],#=={}||UnsameQ@@#&&!And@@IntegerQ/@(Max@@#/#)&]],{n,0,30}]
The a(5) = 1 through a(13) = 9 partitions (empty column indicated by dot): (3,2) . (4,3) (5,3) (5,4) (6,4) (6,5) (7,5) (7,6) (5,2) (7,2) (7,3) (7,4) (5,4,3) (8,5) (4,3,2) (5,3,2) (8,3) (7,3,2) (9,4) (9,2) (10,3) (5,4,2) (11,2) (6,4,3) (6,5,2) (7,4,2) (8,3,2)
Table[Length[Select[IntegerPartitions[n],#=={}||UnsameQ@@#&&!And@@IntegerQ/@(#/Min@@#)&&!And@@IntegerQ/@(Max@@#/#)&]],{n,0,30}]
Take[ CoefficientList[ Expand[ Sum[x^k*Product[1 + x^(k*i), {i, 2, 92}], {k, 2, 92}]], x], {2, 81}] (* Robert G. Wilson v, Nov 01 2004 *) Table[If[n==0,0,Length[Select[IntegerPartitions[n],!MemberQ[#,1]&&UnsameQ@@#&&And@@IntegerQ/@(#/Min@@#)&]]],{n,0,30}] (* Gus Wiseman, Apr 18 2021 *)
The a(1) = 1 through a(15) = 6 partitions (A..F = 10..15): 1 2 3 4 5 6 7 8 9 A B C D E F 21 31 41 42 61 62 63 82 A1 84 C1 C2 A5 51 421 71 81 91 632 93 841 D1 C3 621 631 821 A2 931 842 E1 B1 A21 C21 6321 8421
Table[Length[Select[IntegerPartitions[n],#=={}||UnsameQ@@#&&And@@IntegerQ/@(Max@@#/#)&]],{n,0,30}]
seq(n)={Vec(1 + sum(m=1, n, my(u=divisors(m)); x^m*prod(i=1, #u-1, 1 + x^u[i] + O(x^(n-m+1)))))} \\ Andrew Howroyd, Apr 17 2021
Proper divisors of 12 are {1, 2, 3, 4, 6}. Two subsets of this sum to 12: {2, 4, 6} and {1, 2, 3, 6} - more than any smaller number, so 12 is in the sequence.
With[{s = Table[-1 + SeriesCoefficient[Series[Times @@ ((1 + z^#) & /@ Divisors[n]), {z, 0, n}], n], {n, 2520}]}, FirstPosition[s, #][[1]] & /@ Union@ FoldList[Max, s]] (* Michael De Vlieger, Oct 10 2017 *)
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