A237929 Numbers n such that (i) the sum of prime divisors of n (with repetition) is one less than the sum of prime divisors (with repetition) of n+1, and (ii) n and n+1 have the same number of prime divisors (with repetition).
2, 9, 98, 170, 1274, 4233, 4345, 7105, 7625, 14905, 21385, 30457, 34945, 66585, 69874, 77314, 82946, 98841, 175354, 177122, 233090, 236282, 238017, 263145, 265225, 295274, 298082, 322234, 335793, 336106
Offset: 1
Keywords
Examples
For n=98: prime factors = 2,7,7; sum of prime factors = 16; number of prime divisors = 3 For n+1=99: prime factors = 3,3,11; sum of prime factors = 17; number of prime divisors=3.
Links
- Abhiram R Devesh, Table of n, a(n) for n = 1..96049
Crossrefs
Cf. A001414, A006145 Ruth-Aaron numbers (1): sum of prime divisors of n = sum of prime divisors of n+1.
Cf. A228126 Sum of prime divisors of n (with repetition) is one less than the sum of prime divisors (with repetition) of n+1.
Cf. A045920 Numbers n such that factorizations of n and n+1 have same number of primes (including multiplicities).
Programs
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Mathematica
Select[Partition[Table[{n,PrimeOmega[n],Total[Times@@@FactorInteger[n]]},{n,34*10^4}],2,1],#[[1,2]]==#[[2,2]]&[[1,3]]+1==#[[2,3]]&][[;;,1,1]] (* Harvey P. Dale, May 03 2024 *)
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Python
from sympy import primeomega def is_A237929(n): return A001414(n) == A001414(n+1)-1 and primeomega(n) == primeomega(n+1) # David Radcliffe, Aug 08 2025
Comments