A339061 Least integer j such that H(k+j)>=n+1, where k is the least integer to satisfy H(k)>=n, and H(k) is the sum of the first k terms of the harmonic series.

1, 3, 7, 20, 52, 144, 389, 1058, 2876, 7817, 21250, 57763, 157017, 426817, 1160207, 3153770, 8572836, 23303385, 63345169, 172190019, 468061001, 1272321714, 3458528995, 9401256521, 25555264765, 69466411833, 188829284972, 513291214021

Offset: 0

Author

Matthew J. Bloomfield, Dec 21 2020

Keywords

• nonn

Examples

			Define H(0)=0, H(k) = Sum_{i=1..k} 1/i for k=1,2,3,...
a(0)=1: To reach n+1 from n=0 requires 1 additional term of the harmonic partial sum: H(0+1) = H(0) + 1/1 = H(1) = 1.
a(1)=3: To reach n+1 from n=1 requires 3 additional terms of the harmonic partial sum: H(1+3) = H(1) + 1/(1+1) + 1/(1+2) + 1/(1+3) = H(4) = 2.08333....
a(2)=7: To reach n+1 from n=2 requires 7 additional terms of the harmonic partial sum: H(4+7) = H(4) + 1/(4+1) + 1/(4+2) + ... + 1/(4+6) + 1/(4+7) = H(11) = 3.01987....
a(3)=20: To reach n+1 from n=3 requires 20 additional terms of the harmonic partial sum: H(11+20) = H(11) + 1/(11+1) + 1/(11+2) + ... + 1/(11+19) + 1/(11+20) = H(31) = 4.02724....
		

Crossrefs

First differences of A004080.

Cf. A001113 (e), A001620 (gamma).

Cf. A001008/A002805 (harmonic numbers).

Some sequences in the same spirit as this: A331028, A002387, A004080.

Programs

R

#set size of search space
Max=10000000
#initialize sequence to empty
seq=vector(length=0)
#initialize partial sum to 0
partialsum=0
k=1
n=1
for(i in 1:Max){
   partialsum=partialsum+1/i
   if(partialsum>=n){
      seq=c(seq, k)
      k=0
      n=n+1
   }
   k=k+1
}
#print sequence numbers below Max
seq

Formula

a(n) ~ (e-1)*e^(n-gamma), where e is Euler's number and gamma is the Euler-Mascheroni constant.

Conjecture: a(n) = floor(1/2 + e^(n-gamma+1)) - floor(1/2 + e^(n-gamma)) for n > 1 where e is Euler's number and gamma is the Euler-Mascheroni constant. - Adam Hugill, Nov 06 2022