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).

1, 5, 25, 65, 625, 325, 15625, 1105, 4225, 8125, 9765625, 5525, 244140625, 203125, 105625, 32045, 152587890625, 71825, 3814697265625, 138125, 2640625, 126953125, 2384185791015625, 160225, 17850625, 3173828125, 1221025, 3453125, 37252902984619140625

Offset

1,2

Comments

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.

Links

Jianing Song, Table of n, a(n) for n = 1..1000

Formula

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.

If n = p for prime p or n = pq for primes p >= q, then a(n) = A018782(n).

Example

12 = 3 * 2 * 2, so a(12) = 5^(3-1) * 13^(2-1) * 17^(2-1) = 5525.
15 = 5 * 3, so a(15) = 5^(5-1) * 13^(3-1) = 105625.

Prog

(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

Crossrefs

Cf. A002144, A002654, A018782, A037019, A340624.

Sequence in context: A071383 A088959 A018782 * A146665 A322594 A059302

Adjacent sequences: A340385 A340386 A340387 * A340389 A340390 A340391

Keyword

nonn,easy

Author

Jianing Song, Apr 24 2021

Status

approved