Find the positive integer #n# such that #sum_(k=1)^n floor(log_2 k) = 2018#?

Completing what Oliver started...

Note that:

#8(256) = 2048 > 2018#

So we need less terms than:

#1+2+4+8+16+32+64+128+256 = 511#

and probably more than:

#1+2+4+8+16+32+64+128 = 255#

Start by summing:

#0(1)+1(2)+2(4)+3(8)+4(16)+5(32)+6(64)+7(128)#

#= 0+2+8+24+64+160+384+896 = 1538#

So we need an additional:

#(2018-1538)/8 = 480/8 = 60# terms

So #n = 255+60 = 315#

We have

#log_2 k = floor(log_2 k)# for

#k = 2^j#

We know also that #log_2 k# is a monotonically strict increasing function of #k# so from #j# to #j+1# we have #2^j# equal terms with value #j# so we are looking for

#sum_(k=1)^m k 2^(k-1) + k_2= 2018#

Now considering

#sum_(k=1)^n k x^(k-1) = d/(dx) sum_(k=1)^n x^k -1=d/(dx)((x^(n+1)-1)/(x-1)) -1= ((2-x) x + (m (x-1)-1) x^m)/(x-1)^2#

and for #x = 2#

#sum_(k=1)^m k 2^(k-1)=2^m (m-1)#

Solving for #m#

#2^m (m-1)=2018# we obtain

#m = (Log_e 2 +W(1009 Log_e 2))/Log_e 2 = 8.14232#

so considering #floor(8.14232)= 8# we have

#m_1 = 2^8 = 256# and

#sum_(k=1)^(m_1) floor(log_2 k) = 1546#

Now #2018-1546 = k_2=472 = 8 xx 59# then finally

#n = m_1 + 59 = 256+59=315#