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

1, 3, 21, 75, 105, 315, 495, 945, 1575, 2835, 3465, 4095, 11025, 17955, 10395, 23205, 17325, 24255, 31185, 36855, 51975, 61425, 45045, 108675, 143325, 121275, 184275, 155925, 135135, 176715, 239085, 315315, 294525, 225225, 606375, 626535, 405405, 700245, 1531530, 1351350, 2072070, 1289925, 855855

Offset: 0

Author

Omar E. Pol, Feb 16 2017

Keywords

• nonn

Comments

Conjecture: a(n) is also the smallest number k having n pairs of equidistant subparts in the symmetric representation of sigma(k).

For more information about the "subparts" see A279387.

Observations about the known terms:

Observation 1: terms a(1)-a(51) are divisible by 3.

Observation 2: terms a(3)-a(51) are divisible by 5.

Examples

			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.
Other examples (conjectured):
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).
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.
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.
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.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..200

Crossrefs

Row 1 of A280849.

Cf. A000203, A001227, A008585, A067742, A082262, A131576, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A244050, A245092, A261699, A262626, A279387, A280850, A280940, A281005.

Programs

Mathematica

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 *)

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

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

Extensions

a(10)-a(30) from Jean-François Alcover, Feb 16 2017

a(31)-a(43) from Michael De Vlieger, Feb 18 2017