Search: id:A143578 Results 1-1 of 1 results found. %I A143578 %S A143578 1,2,3,5,7,11,13,15,17,19,23,29,31,35,37,41,43,47,53,59,61,67,71,73,79, %T A143578 83,89,95,97,101,103,107,109,113,119,127,131,137,139,143,149,151,157, %U A143578 163,167,173,179,181,191,193,197,199,209,211,223,227,229,233,239,241,251, 257,263,269,271,277,281,283,287,293 %N A143578 A positive integer n is included if j+n/j divides k+n/k for every divisor k of n, where j is the largest divisor of n that is <= sqrt(n). %C A143578 This sequence trivially contains all the primes. %C A143578 There is no term <= 5*10^7 with bigomega(n)>2, i.e. with more than 2 prime factors. - M. F. Hasler, Aug 25 2008. Compare A142591. %C A143578 If it is always true that the terms have <= 2 prime divisors, then this sequence is equal to {1} U primes U {pq: p, q prime, p+q | p^2-1}. - David Wilson, Aug 25 2008 %H A143578 Leroy Quet, Home Page (listed in lieu of email address) %e A143578 The divisors of 35 are 1,5,7,35. The sum of the two middle divisors is 5+7 = 12. 12 divides 7 + 35/7 = 5+35/5 = 12, of course. And 12 divides 1 + 35/1 = 35 +35/35 = 36. So 35 is in the sequence. %t A143578 a = {}; For[n = 1, n < 200, n++, b = Max[Select[Divisors[n], # <= Sqrt[n] &]]; If[ Length[Union[Mod[Divisors[n] + n/Divisors[n], b + n/b]]] == 1, AppendTo[a, n]]]; a [From Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Aug 29 2008] %o A143578 (PARI code from M. F. Hasler, Aug 25 2008) isA143578(n)={ local( d=divisors(n), j=(1+#d)\2, r=d[ j ]+d[ 1+#d-j ]); for( k=1, j, ( d[k]+d[ #d+1-k] ) % r & return ); 1 } %o A143578 for(n=1,300,isA143578(n) && print1(n",")) %Y A143578 Cf. A063655, A142591. %Y A143578 Sequence in context: A117287 A121615 A097605 this_sequence A086070 A117093 A062063 %Y A143578 Adjacent sequences: A143575 A143576 A143577 this_sequence A143579 A143580 A143581 %K A143578 nonn %O A143578 1,2 %A A143578 Leroy Quet Aug 24 2008 %E A143578 More terms from M. F. Hasler, Aug 25 2008 and Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Aug 29 2008 Search completed in 0.001 seconds