login
Numbers k such that P(k)^4 | k and P(k+1)^2 | (k+1), where P(k) = A006530(k) is the largest prime dividing k.
3

%I #13 Jun 04 2022 02:47:34

%S 101250,11859210,23049600,32580250,131545575,162364824,969697050,

%T 1176565754,1271688417,1612089680,1862719859,2409451520,2441023914,

%U 3182903731,3697778084,4010283270,4329214629,6666661950,6932744126,7739389944,9188994752,11717364285,17306002674

%N Numbers k such that P(k)^4 | k and P(k+1)^2 | (k+1), where P(k) = A006530(k) is the largest prime dividing k.

%C De Koninck and Moineau (2018) proved that this sequence is infinite assuming the Bunyakovsky conjecture.

%D Jean-Marie De Koninck and Nicolas Doyon, The Life of Primes in 37 Episodes, American Mathematical Society, 2021, p. 232.

%H Daniel Suteu, <a href="/A354566/b354566.txt">Table of n, a(n) for n = 1..1897</a> (terms <= 10^17)

%H Jean-Marie De Koninck and Matthieu Moineau, <a href="http://emis.muni.cz/journals/JIS/VOL21/DeKoninck/dek22.html">Consecutive Integers Divisible by a Power of their Largest Prime Factor</a>, J. Integer Seq., Vol. 21 (2018), Article 18.9.3.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BouniakowskyConjecture.html">Bouniakowsky Conjecture</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bunyakovsky_conjecture">Bunyakovsky conjecture</a>.

%e 101250 = 2 * 3^4 * 5^4 is a term since P(101250) = 5 and 5^4 | 101250, 101251 = 19 * 73^2, P(101251) = 73, and 73^2 | 101251.

%t p[n_] := FactorInteger[n][[-1, 2]]; Select[Range[3*10^7], p[#] > 3 && p[# + 1] > 1 &]

%o (Python)

%o from sympy import factorint

%o def c(n, e): f = factorint(n); return f[max(f)] >= e

%o def ok(n): return n > 1 and c(n, 4) and c(n+1, 2)

%o print([k for k in range(10**6) if ok(k)]) # _Michael S. Branicky_, May 30 2022

%Y Subsequence of A070003, A354558 and A354564.

%Y Cf. A006530, A071178, A354565.

%K nonn

%O 1,1

%A _Amiram Eldar_, May 30 2022