cp's OEIS Frontend

A340388 Let n = p_1*p_2*...*p_k be the prime factorization of n, with the primes sorted in descending order. Then a(n) = 5^(p_1 - 1)*13^(p_2 - 1)*17^(p_3 - 1)*...*A002144(k)^(p_k - 1).

%I A340388 #17 Apr 26 2021 06:28:22
%S A340388 1,5,25,65,625,325,15625,1105,4225,8125,9765625,5525,244140625,203125,
%T A340388 105625,32045,152587890625,71825,3814697265625,138125,2640625,
%U A340388 126953125,2384185791015625,160225,17850625,3173828125,1221025,3453125,37252902984619140625
%N A340388 Let n = p_1*p_2*...*p_k be the prime factorization of n, with the primes sorted in descending order. Then a(n) = 5^(p_1 - 1)*13^(p_2 - 1)*17^(p_3 - 1)*...*A002144(k)^(p_k - 1).
%C A340388 Analog of A037019: this is an easy way to produce a number k such that A002654(k) = n, or equivalently, a number k whose prime factors are all congruent to 1 modulo 4 and with exactly n divisors.
%H A340388 Jianing Song, <a href="/A340388/b340388.txt">Table of n, a(n) for n = 1..1000</a>
%F A340388 By definition a(n) >= A018782(n) for all n. Note that a(16) = 32045 is strictly larger than A018782(16) = 27625. The "exceptional" numbers k such that a(k) > A018782(k) are listed in A340624.
%F A340388 If n = p for prime p or n = pq for primes p >= q, then a(n) = A018782(n).
%e A340388 12 = 3 * 2 * 2, so a(12) = 5^(3-1) * 13^(2-1) * 17^(2-1) = 5525.
%e A340388 15 = 5 * 3, so a(15) = 5^(5-1) * 13^(3-1) = 105625.
%o A340388 (PARI) a(n) = my(f=factor(n), w=omega(n), p=1, product=1); forstep(i=w, 1, -1, for(j=1, f[i,2], p=nextprime(p+1); while(!(p%4==1), p=nextprime(p+1)); product *= p^(f[i,1]-1))); product
%Y A340388 Cf. A002144, A002654, A018782, A037019, A340624.
%K A340388 nonn,easy
%O A340388 1,2
%A A340388 _Jianing Song_, Apr 24 2021