OFFSET
1,5
COMMENTS
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, Jun 13 2006
LINKS
Andrew Granville and John Selfridge, Product of integers in an interval, modulo squares, Electronic Journal of Combinatorics, Volume 8(1), 2001.
EXAMPLE
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.
MATHEMATICA
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Oct 20 2004
EXTENSIONS
More terms from Martin Fuller, Jun 13 2006
STATUS
approved