Glad Hobo Express
Thursday, December 11, 2025
Tuesday, December 09, 2025
Sunday, December 07, 2025
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 10^3583+10^3034+10^1492+1
3999996 10^3583+10^3034+10^1624+1
3999997 10^3583+10^3035+10^1153+1
3999998 10^3583+10^3036+10^628+1
3999999 10^3583+10^3039+10^217+1
4000000 10^3583+10^3040+10^2776+1
4000001 10^3583+10^3041+10^127+1
4000002 10^3583+10^3041+10^1207+1
4000003 10^3583+10^3041+10^1561+1
4000004 10^3583+10^3044+10^1690+1
4000005 10^3583+10^3044+10^1816+1
...
4000481 10^3583+10^3579+10^1911+1
4000482 10^3583+10^3580+10^1356+1
4000483 10^3583+10^3581+10^457+1
4000484 10^3583+10^3581+10^1299+1
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) = 10^3583+10^3040+10^2776+1 (above)
A157711(5*10^6) ~ 10^4008 [this will take me well into 2026]
Saturday, December 06, 2025
Thursday, December 04, 2025
Tuesday, December 02, 2025
Saturday, November 29, 2025
Thursday, November 27, 2025
Tuesday, November 25, 2025
Sunday, November 23, 2025
Subscribe to:
Comments (Atom)
