Convergents of e
C_K_Yang czp001 liuguangxi Burloq NaryaTheGreat madbat2 guadruss Min_25 smutley llhuii Philippe_57721 Mushin Apprentice56 faust gerrob zilet hallvabo yehuju stuvtro Mart rijman harvey frack mathpseudo Quandray mihtanat leojay nireal kwisatz petr0v sinan Phicar hervas ThunderLord benito255 faridprog tehron kjangwa dmitrijs Zeta2 LKM Chaosdreamer telipinu lesnik7 R2D2 djcomidi lordoric HvT tujv Madvillain Ch0W UnTaran Caesum bai.li walek20 clytorock s_ha_dum st0le mtoader Buri totoiste Hertz teebee Ryan dloser jablonskim Spaulding Jinx Juampi yachoor Mr_KaLiMaN Erkman84 CommComm nielkh phoenix1204 TheHiveMind elasolova
Public  12/02/09  10xp  Programming  77.8% 
Square root of 2 can be represented as [1;(2)] where (2) means 2 repeats ad infinitum. In a similar way, sqrt(23)=[4;(1,3,1,8)] (Note:Continued Fraction)
Surprisingly,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
The first ten terms in the sequence of convergents for e are:
2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
Find the sum of digits in the numerator of the 1000th convergent of the continued fraction for e.
Surprisingly,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
The first ten terms in the sequence of convergents for e are:
2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
Find the sum of digits in the numerator of the 1000th convergent of the continued fraction for e.
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