A110875 Minimum positive integer such that length of the gap described at A109322 is exactly n (in contrast to A109322 where the gap length is >= n).

2, 16, 9, 64, 49, 872, 481, 1768, 423, 2980, 1333, 49180, 5335, 46666, 4425, 86815, 8763, 1109259, 14089, 658513, 29883, 137539, 22825, 10927365, 259843, 1667974, 46773, 7698572, 40291, 16048081, 178705, 16039804, 1135023, 132082042, 661285, 525395164

Offset: 1

Author

Bojan Basic (bbasic(AT)ptt.yu), Sep 18 2005

Keywords

• nonn

Comments

Conjectures and open problems: 1) It is not known whether the sequence is infinite; 2) It is conjectured that for every n there is corresponding a(n). If Conjecture 2) were proved, Conjecture 1) would follow as a direct consequence.

a(50) > 10^10. - Donovan Johnson, Jan 25 2012

Note that the sequence appears to undulate with terms that satisfy a(2n-1) < a(2n) < a(2n+1). Is there an explanation? - Michel Marcus, Nov 21 2013

Examples

			a(2)=16 because 16,17 are not contained in values of sigma(k) and 15,18 are; namely: sigma(8)=15 and sigma(10)=18, where sigma(k)=sum of all positive divisors of k.
		

Links

• Donovan Johnson, Table of n, a(n) for n = 1..49
• Art of Problem Solving, Gaps in {sigma(n)}...

Crossrefs

Cf. A109322, A109323, A109324.

Cf. A231965 (analog for sigma(n) - n).

Programs

PARI

oksuccs(v, vi, n) = {for (i = 1, n-1, if (! vecsearch(v, vi+i, ) , return (0));); return(! vecsearch(v, vi-1) && !vecsearch(v, vi+n));}
a(n) = {v = readvec("suntouch2.log"); for (i=1, #v, vi = v[i]; if (oksuccs(v, vi, n), return(vi)););} \\ where file read by readvec is the second column of b-file.  Michel Marcus, Nov 21 2013