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%I #34 May 11 2020 01:18:00
%S 1,3,8,22,60,163,443,1204,3273,8897,24184,65739,178698,485751,1320408,
%T 3589241,9756569,26521104,72091835,195965925,532690613,1448003214,
%U 3936080824,10699376979,29083922018,79058296722,214902731368,584166189564,1587928337892,4316436745787
%N Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.
%C a(n) is equal to A024581(n) through a(10), and grows very similarly for n > 10.
%C Let b(n) = Sum_{j=1..n} a(n); then for n >= 2 it appears that b(n) = round((b(n-1) + 1/2)*e). Cf. A331030. - _Jon E. Schoenfield_, Jan 14 2020
%F a(n) = min(p): Sum_{b=r+1..p+r} 1/b >= 1, r = Sum_{k=1..n-1} a(k), a(1) = 1.
%e a(1)=1 because 1 >= 1,
%e a(2)=3 because 1/2 + 1/3 + 1/4 = 1.0833... >= 1, etc.
%o (Python)
%o x = 0.0
%o y = 0.0
%o for i in range(1,100000000000000000000000):
%o y += 1
%o x = x + 1/i
%o if x >= 1:
%o print(y)
%o y = 0
%o x = 0
%o (PARI) default(realprecision, 10^5); e=exp(1);
%o lista(nn) = {my(r=1); print1(r); for(n=2, nn, print1(", ", -r+(r=floor(e*r+(e+1)/2+(e-1/e)/(24*(r+1/2)))))); } \\ _Jinyuan Wang_, Mar 31 2020
%Y Cf. A004080, A024581, A136616, A331030.
%Y Some sequences in the same spirit as this: A002387, A004080, A055980, A115515.
%K nonn
%O 1,2
%A _Alejandro Argüelles Trujillo_ and _Pablo Hueso Merino_, Jan 07 2020
%E a(20)-a(21) from _Giovanni Resta_, Jan 14 2020
%E More terms from _Jinyuan Wang_, Mar 31 2020
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