

A281008


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


3



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
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OFFSET

0,2


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.


LINKS

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


EXAMPLE

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 evenindexed 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 evenindexed terms that are positive integers: [75, 25, 15]. That is the first row with that property, so a(3) = 75.


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}] (* JeanFrançois Alcover, Feb 16 2017 *)


PROG

(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 nth, 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


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.
Sequence in context: A342548 A188667 A083564 * A238193 A054646 A228317
Adjacent sequences: A281005 A281006 A281007 * A281009 A281010 A281011


KEYWORD

nonn


AUTHOR

Omar E. Pol, Feb 16 2017


EXTENSIONS

a(10)a(30) from JeanFrançois Alcover, Feb 16 2017
a(31)a(43) from Michael De Vlieger, Feb 18 2017


STATUS

approved



