0,1

a(n) exists for every n, since the sum of the inverses of the primes is infinite.

Heuristic: square the first several primes and then add successive primes until the number is abundant. - Fred Schneider (frederick.william.schneider(AT)gmail.com), Sep 20 2006

For terms 3 4 and 5, squaring only the first two will be part of the minimal solution: 49061132957714428902152118459264865645885092682687973 = 11^2 * 13^2 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 127 * 131 * 137 - Fred Schneider (frederick.william.schneider(AT)gmail.com), Sep 20 2006

7970466327524571538225709545434506255970026969710012787303278390616918473506860039424701 = 13^2 * 17^2 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 127 * 131 * 137 * 139 * 149 * 151 * 157 * 163 * 167 * 173 * 179 * 181 * 191 * 193 * 197 * 199 * 211 * 223 * 227 - Fred Schneider (frederick.william.schneider(AT)gmail.com), Sep 20 2006

a(0) = 12 = first abundant number; a(1)=945=first odd abundant number; a(5) is the first abundant number not divisible by 2,3,5,7 or 11

Cf. A005101, A005231.

Sequence in context: A159870 A114809 A114371 this_sequence A079916 A123283 A004812

Adjacent sequences: A047799 A047800 A047801 this_sequence A047803 A047804 A047805

nonn

Ulrich Schimke (ulrschimke(AT)aol.com)

2 more terms from Fred Schneider (frederick.william.schneider(AT)gmail.com), Sep 20 2006