[From the NMBRTHRY list, slightly edited. I reformatted the factorizations. No arithmetic actually checked. Odd abundant numbers are sparse, and finding 4 consecutive abundants means two of them must be odd. -- rcs] Date: Thu, 1 Nov 2007 15:38:16 -0400 From: Bruno Mishutka <bruno.mishutka@googlemail.com> Subject: consecutive abundant numbers To: NMBRTHRY@LISTSERV.NODAK.EDU Snyam Sunder Gupta (http://www.shyamsundergupta.com/canyoufind.htm) has recently asked for the smallest set of four consecutive abundant numbers. I suspect that this is not the smallest m1 such that (m1, m1+1, m1+2, m1+3) are all abundant, but it's the smallest set that I could find. m1 := 141363708067871564084949719820472453374; It is approximately 1.41*10^38. igcd(m1,6);ifactor(m1+1);ifactor(m1+2);ifactor(m1+3); 6 5^3 7^2 11^2 17 23^2 37 43 53 59 73 79 83 281 31696364109343 2^8 337 1638581556795618092601883807266233 3^4 13^2 19 29 31 41 47 61 67 71 151 7160300747074201457 sigma(m1+1)/1.0/(m1+1);sigma(m1+2)/1.0/(m1+2);sigma(m1+3)/1.0/(m1+3); 2.005172014 2.002016877 2.003079741 Note that Paul Erdos ("Note on consecutive abundant numbers", J. London Math. Soc. 10, 128-131 (1935)) has shown that there are two absolute constants c1, c2 such that for all large n there are at least c1 log log log n but not more than c2 log log log n consecutive abundant numbers less than n. Also, the smallest set of two consecutive abundant numbers is (5775, 5776). And the smallest set of three consecutive abundant numbers is (171078830, 171078832, 171078832), due apparently to Laurent Hodges and Michael Reid in 1975 (does anyone have a reference to this?). Yours Sincerely, Bruno