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#1
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![]() 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^(2*sigma(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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#2
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![]() Danshaku ![]() ![]() ![]() ![]() Group: Chuunin Posts: 290 Joined: 12-April 06 From: nya ![]() |
Rezultaats : 896 XDD
Bet par teemu runaajot kaads sakars vipaar postot kko taadu? Bet varbuut butu jauka iedeja aizsaakt vienu interesantu topiku, kur kaads uzdod aakiigo uzdevumu un kurs tad pirmais izdomaa ![]() -------------------- skills of a snail
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