OFFSET
1,1
COMMENTS
Equivalently, subsequence of terms of A339744 excluding terms whose prime factor set has already been encountered.
a(n) = A005117(n + 1)^2 when A005117(n + 1) is prime. Proof: if A005117(n + 1) is a prime p then rad(A005117(n + 1))^2 = rad(p)^2 = p^2 and so integers whose prime factors set is the same as the prime factors set of A005117(n + 1) = p are p^m where m >= 1. p^2 > sigma(p^1) = p + 1 but p^2 < sigma(p^2) = p^2 + p + 1. Q.E.D. - David A. Corneth, Dec 19 2020
From Bernard Schott, Jan 19 2021: (Start)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Michel Marcus)
FORMULA
a(n) <= A005117(n+1)^2. - David A. Corneth, Dec 19 2020
EXAMPLE
n a(n) prime factor set
1 4 [2] A000079
2 9 [3] A000244
3 25 [5] A000351
4 18 [2, 3] A033845
5 49 [7] A000420
6 80 [2, 5] A033846
7 121 [11] A001020
8 169 [13] A001022
9 112 [2, 7] A033847
10 135 [3, 5] A033849
11 289 [17] A001026
12 361 [19] A001029
13 441 [3, 7] A033850
14 352 [2, 11] A033848
15 529 [23] A009967
16 416 [2, 13] A288162
17 841 [29] A009973
18 360 [2, 3, 5] A143207
PROG
(PARI) u(n) = {my(fn=factor(n)[, 1]); for (k = n, n^2, my(fk = factor(k)); if (fk[, 1] == fn, if (factorback(fk[, 1])^2 < sigma(fk), return (k)); ); ); }
lista(nn) = {for (n=2, nn, if (issquarefree(n), print1(u(n), ", "); ); ); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Dec 17 2020
STATUS
approved