A122140 Numbers m that divide the sum of cubes of the first m primes A098999(m).
1, 25, 537, 661, 5199, 113253, 240621, 5337048977, 17434578479, 34216676921, 1991831965911, 4495321247369, 22567781434431, 37328883555791, 110447613624133, 188368390470877, 324587968952249, 1983705516917661
Offset: 1
Examples
a(2) = 25 because 25 is the first number n>1 that divides sum of cubes of the first n primes. A098999(25) mod 25 = (2^3 + 3^3 + 5^3 + ... + 89^3 + 97^3) mod 25 = 0.
Links
- OEIS Wiki, Sums of powers of primes divisibility sequences.
Crossrefs
Cf. A098999.
Programs
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Mathematica
s = 0; t = {}; Do[s = s + Prime[n]^3; If[ Mod[s, n] == 0, AppendTo[t, n]], {n, 1000000}]; t
Extensions
a(8) from Donovan Johnson, Oct 15 2012
a(9)-a(10) from Robert Price, Mar 29 2013
a(11) from Paul W. Dyson, Jan 05 2021
a(12) from Bruce Garner, Feb 28 2021
a(13) from Bruce Garner, Apr 06 2021
a(14) from Bruce Garner, May 13 2021
a(15) from Bruce Garner, Jan 08 2022
a(16) from Paul W. Dyson, Jan 17 2022
a(17) from Bruce Garner, Jul 31 2022
a(18) from Paul W. Dyson, Feb 18 2024
Comments