OFFSET
1,2
COMMENTS
Since every multiple of 6 (other than 6 itself) is an abundant number, the maximum length of consecutive runs of deficient numbers is 5.
All terms are congruent to 1 modulo 6.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
EXAMPLE
115 is a term since all of 115, 116, 117, 118 and 119 are deficient.
2989 is not a term since 2989 + 3 = 2992 is an abundant number.
MATHEMATICA
Position[Partition[DivisorSigma[-1, Range[600]], 5, 1], _?(Max[#] < 2 &), 1] // Flatten (* Amiram Eldar, Mar 21 2024 *)
PROG
(PARI) isA343302(k) = if(k%6!=1, 0, for(i=0, 4, if( sigma(k+i) >= 2*(k+i), return(0) )); 1)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Apr 11 2021
STATUS
approved