One more cyclic number-Rejoinder 2

The link mentioned in the just preceding blog occurs as follows-

The numbers used for multiplication in the reverse order are 1,....621517,...269324,...441277,....251300,..25130,...2513,...1864802,..1429514,..557296,..1713108.

1000000,100000,10000,1000,100,10..

The number 2513 can be worked from the previous 6 numbers by successively multiplying them with 6,2,1,5,1,7 forming part of 621517 and adding the results.Similarly the number 1864802 can be worked from the previous 6 numbers etcetc-see below-

(1*6)+(621517*2)+(269324*1)+(441277*5)+(251300*1)+(25130*7)=

6+1243034+269324+2206385+251300+175910=4145959.From this result the maximum possible multiple of 2071723 viz 4143446 is deducted producing 2513.

Similarly (621517*6)+(269324*2)+(441277*1)+(251300*5)+(25130*1)+(2513*7)=6008248 which after subtracting 4143446 gives us 1864802.etc etc.

You can similarly get all the subsequent numbers.