The four-millionth term of A157711

The sum of the first 3583 prime counts of A383675 is 3998725. The sum of the first 3584 prime counts is 4000485. Hence, the four-millionth term of A157711 is a 3584-digit integer:

3998726 10^3583+10^118+10^44+1

3998727 10^3583+10^147+10^131+1

3998728 10^3583+10^208+10^96+1

3998729 10^3583+10^210+10^88+1

3998730 10^3583+10^211+10^61+1

...

3999995 

3999996 

3999997 

3999998 

3999999 

4000000   to be determined (Dec 5 or 6)

4000001 

4000002 

4000003 

4000004 

4000005 

...

4000481 

4000482 

4000483 

4000484 

4000485 10^3583+10^3581+10^3547+1

A157711(1*10^6) = 10^1793+10^673+10^615+1 [June 19]

A157711(2*10^6) = 10^2535+10^1160+10^398+1 [July 21]

A157711(3*10^6) = 10^3103+10^2747+10^859+1 [September 10]

A157711(4*10^6) =   to be determined (above)

A157711(5*10^6) ~ 10^4008 [this will take me well into 2026]

I'll continue to update (here, until the next millionth is reached) my current plot of A383675:

A383675 to n=3595; max=(3593,2895) [updated December 3]  click to enlarge