1,2

For the most part the first differences have the form {2k, 2k, 2k, 2k+1} (A004524). Therefore for the most part the second differences are {0, 0, 1, 1}. See A082915 for the exceptions. - Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 18 2003

a(100)=2574, a(1000)=250731 & a(10000)=25007302 which differs from Sum{i=4..104}A004524(i)=2625, Sum{i=4..1004}A004524(i)=251250 & Sum{i=4..10004}A004524(i)=25012500. - Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 18 2003

A054040(n) <= A011848(n+2), A054040(10000)=25007302 and A011848(n+2)=25007500. - Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 18 2003

Let b(k) = 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k):

.k.......1....2.....3.....4.....5.....6.....7

-------------------------------------------------

b(k)...1.00..1.71..2.28..2.78..3.23..3.64..4.01

For A019529 we have:

n=0: smallest k is a(0) = 1 since 1.00 > 0

n=1: smallest k is a(1) = 2 since 1.71 > 1

n=2: smallest k is a(2) = 3 since 2.28 > 2

n=3: smallest k is a(3) = 5 since 3.23 > 3

n=4: smallest k is a(4) = 7 since 4.01 > 4

For AA054040 we have:

n=1: smallest k is a(1) = 1 since 1.00 >= 1

n=2: smallest k is a(2) = 3 since 2.28 >= 2

n=3: smallest k is a(3) = 5 since 3.23 >= 3

n=4: smallest k is a(4) = 7 since 4.01 >= 4

f[n_] := Block[{k = 0, s = 0}, While[s < n, k++; s = N[s + 1/Sqrt[k], 50]]; k]; Table[f[n], {n, 1, 60}]

Cf. A002387, A004080, A082915, A054044, A011848, A082915.

See A019529 for a different version.

Sequence in context: A077000 A057672 A112509 this_sequence A011848 A131673 A151945

Adjacent sequences: A054037 A054038 A054039 this_sequence A054041 A054042 A054043

nonn

Asher Auel (asher.auel(AT)reed.edu) Apr 13 2000

Definition and offset modified by N. J. A. Sloane, Sep 01 2009