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A295572 First differences of A081881.

%I A295572 #39 Jun 06 2020 16:54:32
%S A295572 1,2,6,16,43,117,318,865,2351,6391,17372,47222,128363,348927,948482,
%T A295572 2578241,7008386,19050768,51785356,140767193,382644902,1040136684,
%U A295572 2827384648,7685628310,20891703776,56789538739,154369971201,419621087576,1140648377196,3100603756393
%N A295572 First differences of A081881.
%C A295572 See A081881 and A295571 for discussion.
%C A295572 If the harmonic series is divided into the longest possible consecutive groups so that the sum of each group is <= 1, then a(n) is the number of terms in the n-th group. - _Pablo Hueso Merino_, Feb 16 2020
%H A295572 Jinyuan Wang, <a href="/A295572/b295572.txt">Table of n, a(n) for n = 1..1000</a>
%F A295572 a(1) = 1, a(n) = (max(m) : Sum_{s=r..m} 1/s <= 1)-r+1, r = Sum_{k=1..n-1} a(k). - _Pablo Hueso Merino_, Feb 16 2020
%F A295572 a(n) ~ c * exp(n), where c = (exp(1)-1) * A300897 = 0.290142809280953235916025... - _Vaclav Kotesovec_, Apr 05 2020
%e A295572 From _Pablo Hueso Merino_, Feb 16 2020: (Start)
%e A295572 a(1) = 1 because 1 <= 1, 1 is one term (if you added 1/2 the sum would be greater than 1).
%e A295572 a(2) = 2 because 1/2 + 1/3 = 0.8333... <= 1, 1/2 and 1/3 are two terms (if you added 1/4 the sum would be greater than one).
%e A295572 a(3) = 6 because 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 = 0.9956... <= 1, it is a sum of six terms. (End)
%t A295572 a[1]=1;
%t A295572 a[n_]:= a[n]= Module[{sum = 0}, r = 1 + Sum[a[k], {k, n-1}];
%t A295572    x = r;
%t A295572    While[sum <= 1, sum += 1/x++];
%t A295572    p = x-2;
%t A295572    p -r +1];
%t A295572 Table[a[n], {n, 10}] (* _Pablo Hueso Merino_, Feb 16 2020 *)
%Y A295572 Cf. A081881, A295571, A300897, A331028, A331030.
%K A295572 nonn
%O A295572 1,2
%A A295572 _N. J. A. Sloane_, Nov 30 2017, following a suggestion from Loren Booda
%E A295572 More terms from _Jinyuan Wang_, Feb 20 2020