A018782 Smallest k such that circle x^2 + y^2 = k passes through exactly 4n integer points.

1, 5, 25, 65, 625, 325, 15625, 1105, 4225, 8125, 9765625, 5525, 244140625, 203125, 105625, 27625, 152587890625, 71825, 3814697265625, 138125, 2640625, 126953125, 2384185791015625, 160225, 17850625, 3173828125, 1221025, 3453125

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn

Comments

a(n) is least term of A054994 with exactly n divisors. - Ray Chandler, Jan 05 2012

From Jianing Song, Apr 24 2021: (Start)

a(n) is the smallest k such that A004018(k) = 4n.

Also a(n) is the smallest index of n in A002654.

a(n) is the smallest term in A004613 that has exactly n divisors.

This is a subsequence of A054994. (End)

Examples

			4225 = 5^2 * 13^2 is the smallest number all of whose prime factors are congruent to 1 modulo 4 with exactly 9 divisors, so a(9) = 4225. - _Jianing Song_, Apr 24 2021
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..1432 (a(1433) exceeds 1000 digits).

Crossrefs

Cf. A002144, A002654, A004018, A005179, A054994, A000446, A124980, A016032, A093195.

Programs

Mathematica

(* This program is not convenient to compute huge terms - A054994 is assumed to be computed with maxTerm = 10^16 *) a[n_] := Catch[ For[k = 1, k <= Length[A054994], k++, If[DivisorSigma[0, A054994[[k]]] == n, Throw[A054994[[k]]]]]]; Table[a[n], {n, 1, 28}] (* Jean-François Alcover, Jan 21 2013, after Ray Chandler *)

PARI

primelist(d,r,l) = my(v=vector(l), i=0); if(l>0, forprime(p=2, oo, if(Mod(p,d)==r, i++; v[i]=p; if(i==l, break())))); v
prodR(n, maxf)=my(dfs=divisors(n), a=[], r); for(i=2, #dfs, if( dfs[i]<=maxf, if(dfs[i]==n, a=concat(a, [[n]]), r=prodR(n/dfs[i], min(dfs[i], maxf)); for(j=1, #r, a=concat(a, [concat(dfs[i], r[j])]))))); a
A018782(n)=my(pf=prodR(n, n), a=1, b, v=primelist(4, 1, bigomega(n))); for(i=1, #pf, b=prod(j=1, length(pf[i]), v[j]^(pf[i][j]-1)); if(bJianing Song, Apr 25 2021, following R. J. Mathar's program for A005179.

Formula

A000446(n) = min(a(2n-1), a(2n)) for n > 1.

A124980(n) = min(a(2n-1), a(2n)).

A016032(n) = min(2*a(2n-1), a(2n), a(2n+1)).

A093195(n) = min(a(2n), a(2n+1)).

From Jianing Song, Apr 24 2021: (Start)

If the factorization of n into primes is n = Product_{i=1..r} p_i with p_1 >= p_2 >= ... >= p_r, then a(n) <= (q_1)^((p_1)-1) * (q_2)^((p_2)-1) * ... * (q_r)^((p_r)-1), where q_1 < q_2 < ... < q_r are the first r primes congruent to 1 modulo 4. The smallest n such that the equality does not hold is n = 16.

a(n) <= 5^(n-1) for all n, where the equality holds if and only if n = 1 or n is a prime.

a(p*q) = 5^(p-1) * 13^(q-1) for primes p >= q. (End)