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A281008 Least positive integer k with exactly n odd divisors greater than sqrt(2*k).

%I A281008 #84 Aug 25 2019 13:40:52
%S A281008 1,3,21,75,105,315,495,945,1575,2835,3465,4095,11025,17955,10395,
%T A281008 23205,17325,24255,31185,36855,51975,61425,45045,108675,143325,121275,
%U A281008 184275,155925,135135,176715,239085,315315,294525,225225,606375,626535,405405,700245,1531530,1351350,2072070,1289925,855855
%N A281008 Least positive integer k with exactly n odd divisors greater than sqrt(2*k).
%C A281008 Conjecture: a(n) is also the smallest number k having n pairs of equidistant subparts in the symmetric representation of sigma(k).
%C A281008 For more information about the "subparts" see A279387.
%C A281008 Observations about the known terms:
%C A281008 Observation 1: terms a(1)-a(51) are divisible by 3.
%C A281008 Observation 2: terms a(3)-a(51) are divisible by 5.
%H A281008 Charles R Greathouse IV, <a href="/A281008/b281008.txt">Table of n, a(n) for n = 0..200</a>
%e A281008 a(3) = 75 because the divisors of 75 are [1, 3, 5, 15, 25, 75], and 75 has three odd divisors greater than the square root of 2*75 = 12.2..., and it is the smallest number with that property.
%e A281008 Other examples (conjectured):
%e A281008 2) The 75th row of A237593 is [38, 13, 7, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 7, 13, 38], and the 74th row of the same triangle is [38, 13, 6, 5, 3, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 3, 5, 6, 13, 38], therefore between both symmetric Dyck paths (described in A237593 and A279387) there are three pairs of equidistant subparts: [38, 38], [21, 21] and [3, 3]. That is the first row with that property, so a(3) = 75. (The diagram of the symmetric representation of sigma(75) is too large to include).
%e A281008 3) The 75th row of A196020 is [149, 73, 47, 0, 25, 19, 0, 0, 0, 5, 0], hence the 75th row of A280850 is [38, 38, 21, 0, 3, 3, 0, 0, 0, 21, 0]. There are three pairs of equidistant subparts [38, 38], [21, 21] and [3, 3]. That is the first row with that property, so a(3) = 75.
%e A281008 4) The 75th row of A237048 is [1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0]. The sum of the even-indexed terms is equal to 3. That is the first row with that property, so a(3) = 75.
%e A281008 5) The 75th row of A261699 is [1, 75, 3, 0, 5, 25, 0, 0, 0, 15, 0]. There are three even-indexed terms that are positive integers: [75, 25, 15]. That is the first row with that property, so a(3) = 75.
%t A281008 cnt[k_] := cnt[k] = DivisorSum[k, Boole[OddQ[#] && #>Sqrt[2k]]&]; a[n_] := a[n] = For[k = 1, True, k++, If[cnt[k]==n, Return[k]]]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 30}] (* _Jean-François Alcover_, Feb 16 2017 *)
%o A281008 (PARI) a(n,{s=0},{q=1},{k=2},{w=1})={if(n<1,return(1));my(z,ii,F,d,L:list,V,p,ans:list);ans=List();if(q<1,q=1);if(k<2,k=2);while(k++,p=sqrt(2*k);F=factor(k);ii=vecsum(F[1,]);F=F[,1]~;L=List([1]);for(i=1,ii,forvec(y=vector(i,t,[1,#F]),d=prod(u=1,#y,F[y[u]]);if((d<=k)&&!(k%d),listput(L,d)),1));V=Set(Vec(L));if(n==sum(u=1,#V,(V[u]>p)&&(V[u]%2==!!w)),if(s,print1(V","));listput(ans,k);if(z++==q,if(#ans==1,return(k),return(Vec(ans))),n++)))} \\ with n>=1, "s" set to 1 also prints the divisors (of "w" version: 1 odd, 0 even) for the first "q" terms from the n-th, resuming their search with k>=2. - _R. J. Cano_, Feb 20 2017
%o A281008 (PARI) a(n)=my(k,s); while(k++, s=sqrtint(2*k); if(sumdiv(k>>valuation(k,2), d, d>s)==n, return(k))) \\ _Charles R Greathouse IV_, Feb 20 2017
%Y A281008 Row 1 of A280849.
%Y A281008 Cf. A000203, A001227, A008585, A067742, A082262, A131576, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A244050, A245092, A261699, A262626, A279387, A280850, A280940, A281005.
%K A281008 nonn
%O A281008 0,2
%A A281008 _Omar E. Pol_, Feb 16 2017
%E A281008 a(10)-a(30) from _Jean-François Alcover_, Feb 16 2017
%E A281008 a(31)-a(43) from _Michael De Vlieger_, Feb 18 2017