This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A130279 #31 Feb 16 2025 08:33:06 %S A130279 1,4,16,36,256,144,4096,576,1296,2304,1048576,3600,16777216,36864, %T A130279 20736,14400,4294967296,32400,68719476736,57600,331776,9437184, %U A130279 17592186044416,129600,1679616,150994944,810000,921600,72057594037927936 %N A130279 Smallest number having exactly n square divisors. %C A130279 A046951(a(n)) = n and A046951(m) <> n for m < a(n); %C A130279 all terms are smooth squares: if prime(k) is a factor of a(n) then also prime(i) are factors, i<k; %C A130279 a(p) = 2^(2*(p-1)) for primes p; %C A130279 if prime(j) is the greatest prime factor of a(n) then a(2*n) = a(n)*prime(j+1)^2; %C A130279 A001221(a(n)) = A122375(n); A001222(a(n)) = 2*A122376(n). %C A130279 a(n+1) is the smallest nonsquarefree number m such that Diophantine equation S(x,y) = (x+y) + (x-y) + (x*y) + (x/y) = m has exactly n solutions, for n >= 0 (A353282); example: a(4) = 36 and 36 is the smallest number m such that equation S(x,y) = m has exactly 3 solutions: (9,1), (8,2), (5,5). - _Bernard Schott_, Apr 13 2022 %C A130279 a(n) is the square of the smallest integer having exactly n divisors (see formula with proof). - _Bernard Schott_, Oct 01 2022 %H A130279 Amiram Eldar, <a href="/A130279/b130279.txt">Table of n, a(n) for n = 1..100</a> %H A130279 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SmoothNumber.html">Smooth Number</a> %F A130279 From _Bernard Schott_, Oct 01 2022: (Start) %F A130279 a(n) = A005179(n)^2. %F A130279 Proof: Suppose a(n) = Product p_i^(2*e_i), where the p_i are primes. Then the n square divisors are all of the form d = Product p_i^(2*k_i) with 0 <= k_i <= e_i. As a(n) = Product (p_i^e_i)^2 = (Product (p_i^e_i))^2, we get that sqrt(a(n)) = Product (p_i^e_i). This is the prime decomposition of sqrt(a(n)). As there is a bijection between prime factors p_i^(2*k_i) and (p_i^k_i), there is also bijection between square divisors of a(n) and divisors of sqrt(a(n)). We conclude that sqrt(a(n)) is the smallest integer that has exactly n divisors. (End) %o A130279 (PARI) a(n) = my(k=1); while(sumdiv(k, d, issquare(d)) != n, k++); k; \\ _Michel Marcus_, Jul 15 2019 %Y A130279 Cf. A001221, A005179, A046951, A046952, A122375, A122376, A353282. %Y A130279 Cf. A357450 (similar, but with odd squares divisors). %K A130279 nonn %O A130279 1,2 %A A130279 _Reinhard Zumkeller_, May 20 2007