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Search: id:A099500
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| A099500 |
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Number of distinct products of subsets of integers in the interval [n^2+1, (n+1)^2-1] which are twice a square. |
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6
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| 1, 1, 1, 1, 2, 1, 2, 4, 2, 4, 4, 4, 8, 4, 32, 8, 16, 16, 32, 16, 32, 128, 32, 64, 59, 124, 499, 32, 478, 122, 248, 1886, 239, 1980, 240, 944, 500, 6181, 3848, 970, 3776, 3844, 6167, 13774, 3806, 27985, 24562, 4016, 102193, 13224, 105671, 12565, 399445, 29436
(list; graph; listen)
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OFFSET
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1,5
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COMMENT
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Granville and Selfridge prove Kaplansky's conjecture that a(n) > 0 for all n. Interestingly, the first 24 terms are powers of 2, but the 25th term is not.
a(n) <= A120253(n) = 2^A120254(n). The sequences are equal iff each subset has a distinct product. - Martin Fuller (martin_n_fuller(AT)btinternet.com), Jun 13 2006
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LINKS
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Andrew Granville and John Selfridge, Product of integers in an interval, modulo squares (pdf), Electronic Journal of Combinatorics, Volume 8(1), 2001.
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EXAMPLE
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a(5) = 2 because the interval [26,35] contains two sets of such integers: {32} and {27,28,30,35}.
a(25) < A120253(25) because there are distinct subsets with the same product. The smallest is 627*630*646*648*650*660*663*672 = 627*637*640*646*648*660*663*675 = 2*125707982400^2.
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MATHEMATICA
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Needs["DiscreteMath`Combinatorica`"]; Table[lst=Range[n^2+1, (n+1)^2-1]; x=Times@@lst; {p, e}=Transpose[FactorInteger[x]]; Do[If[e[[i]]==1 && p[[i]]!=2, lst=DeleteCases[lst, _?(Mod[ #, p[[i]]]==0&)]], {i, Length[p]}]; i=1; cnt=0; xLst={}; While[i<2^Length[lst], ss=NthSubset[i, lst]; x=Times@@ss; If[Mod[x, 2]==0 && IntegerQ[Sqrt[x/2]] && !MemberQ[xLst, x], cnt++; AppendTo[xLst, x] (*; Print[{n, ss}]*)]; i++ ]; cnt, {n, 31}]
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CROSSREFS
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Cf. A099501 (size of the subset having the least number of integers).
Cf. A120253 number of subsets whose product is twice a square.
Sequence in context: A099470 A099254 A121339 this_sequence A120253 A060547 A079878
Adjacent sequences: A099497 A099498 A099499 this_sequence A099501 A099502 A099503
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Oct 20 2004
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EXTENSIONS
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More terms from Martin Fuller (martin_n_fuller(AT)btinternet.com), Jun 13 2006
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