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xD kaads saprot?

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Blackout	Feb 10 2007, 12:25 Post #1
Gimme More Group: weirdo Posts: 294 Joined: 4-November 05 From: Riga	atradu sekojoshu matemaatisku formulu internetaa: "to prove i know , the next line is 1113213211 31131211131221 Btw , interesting facts : If you think that the sequence is "non-mathematical", I derived this mathematical expression that gives the sequence... have fun! (D is a recursive function and t is the term number.) It's a lot easier if you think verbally, isn't it? By the way, % here is a certain non-integer remainder function. 2.1%0.1 would be 0, 2.1%0.2 would be 0.1, 2.1%0.3 would be 0 since 0.3 fits evenly into 2.1, etc.) If you really want conventional operators, you could define % with limits and modular arithmetic... D(t+1) = (sigma(K=1,LOG(D(t)10)-LOG(D(t)10)%1,((D(t)-D(t)%10^(LOG(D(t))- LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)10) -LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t) 10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)* 10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10) %10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t) 10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)-( sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10 ^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)2(D(t) -D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1, LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t) 10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10) -(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))% 10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+1)%((( sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))% 10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t) 10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t) 10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)2(D(t)-D(t)%10^(S -1))%10^S))%10^(K-1))%10^K/10^(K-1)100^(2sigma(N=1,K,(((D(t)-D(t)% 10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1, LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10- (D(t)10-D(t)10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)* 10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)% 10^(R+1)))/10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)% 1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)% 10^(R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t) 10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1)) %10^S+.5)2(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t)) %1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10) %1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^ (R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10% 10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1))%10^ S+1)%(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^ (R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)-(sigma(R=1, LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/ 10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)2(D(t)-D(t)% 10^(S-1))%10^S))%10^(N-1))%10^N+1)%(((D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))% 1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)10)-LOG(D(t) 10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^ R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10- D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^ (S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10- D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)- (sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))% 10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)2 (D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+ sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10 )%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^ R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)* 10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/ 10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1, ABS((D(t)10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)% 10^(R+1)))/10)-(sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t) 10-D(t)10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1))) /10)%10^(S-1))%10^S+.5)2(D(t)-D(t)%10^(S-1))%10^S))%10^(N-1))%10^ N+.5))))/100)+(sigma(K=1,LOG(D(t)10)-LOG(D(t)10)%1,100^(1+sigma(N= 1,K-1,2((((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t) 10%10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)-( sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))% 10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N- 1)+1)%(((sigma(R=1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10% 10^(R+1))%10^(R+2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)-(sigma(R= 1,LOG(D(t)10)-LOG(D(t)10)%1-1,ABS((D(t)10-D(t)10%10^(R+1))%10^(R+ 2)/10-(D(t)10-D(t)10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-1)+ .5)))))/10)" Kaads vispaar saprot par ko iet runa, un kaa viss notiek? >___> --------------------

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Sigfa	Feb 10 2007, 16:42 Post #2
Kami-sama Group: ALK Posts: 2738 Joined: 17-May 04 From: Starp pudelēm un četrām sienām	tev to visu noteikti vajadzeeja rakstiit vienaa rindaa shumin... ar copy+paste tu savu domu nepieraadiisi. Storm, ieliki excelii, lai izreekjina? EDIT: un tu kljuudijies, rezultaats ir 914.458029984138 . This post has been edited by Sigfa: Feb 10 2007, 16:55

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Hydrange xD kaads saprot? Feb 10 2007, 12:25

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