A354566 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.

101250, 11859210, 23049600, 32580250, 131545575, 162364824, 969697050, 1176565754, 1271688417, 1612089680, 1862719859, 2409451520, 2441023914, 3182903731, 3697778084, 4010283270, 4329214629, 6666661950, 6932744126, 7739389944, 9188994752, 11717364285, 17306002674

Offset: 1

Author

Amiram Eldar, May 30 2022

Keywords

• nonn

Comments

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

Examples

			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.
		

References

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

Links

• Daniel Suteu, Table of n, a(n) for n = 1..1897 (terms <= 10^17)
• Jean-Marie De Koninck and Matthieu Moineau, Consecutive Integers Divisible by a Power of their Largest Prime Factor, J. Integer Seq., Vol. 21 (2018), Article 18.9.3.
• Eric Weisstein's World of Mathematics, Bouniakowsky Conjecture.
• Wikipedia, Bunyakovsky conjecture.

Crossrefs

Subsequence of A070003, A354558 and A354564.

Cf. A006530, A071178, A354565.

Programs

Mathematica

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

Python

from sympy import factorint
def c(n, e): f = factorint(n); return f[max(f)] >= e
def ok(n): return n > 1 and c(n, 4) and c(n+1, 2)
print([k for k in range(10**6) if ok(k)]) # Michael S. Branicky, May 30 2022