OFFSET
1,1
COMMENTS
Equivalently, this is a list of integers m for which there is exactly one integer k satisfying 1+1/2+...+1/m < log(k) < 1+1/2+...+1/(m+1). - Clark Kimberling May 30 2013.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..400
FORMULA
Conjecturally, a(n) = floor(n/(2 - exp(g)) - 1/2 + exp(g)/(24n)), where g is the Euler-Mascheroni constant. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Aug 11 2007
MATHEMATICA
Flatten[Position[Differences[Table[Floor[E^Sum[1/k, {k, 1, m}]], {m, 1, 500}]], 1]] (* Clark Kimberling, May 30 2013 *)
PROG
(PARI) lista(n) = {old = 1; expo = exp(old); for (i=2, n, new = old + 1/i; expn = exp(new); if (floor(expn)==ceil(expo), print1(i-1, ", ")); old = new; expo = expn; ); } \\ Michel Marcus, Mar 21 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved