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