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Sequence satisfies: 1 = Sum_{n>=1} n^(n-1) / a(n)^n, with a(1) = 2, by a greedy algorithm.
1

%I #19 Mar 26 2018 23:26:05

%S 2,3,4,5,8,9,15,17,21,28,33,38,50,61,72,78,86,91,110,123,141,161,178,

%T 187,214,230,239,253,278,291,302,314,338,352,398,410,422,448,462,474,

%U 492,548,591,609,661,684,716,757,801,835,856,897,915,969,1005,1061,1085,1108,1153,1206,1249,1286,1315,1338,1386,1412,1552,1585,1615,1645,1669,1731,1771,1798,1830,1873,1902,1972,2023,2051

%N Sequence satisfies: 1 = Sum_{n>=1} n^(n-1) / a(n)^n, with a(1) = 2, by a greedy algorithm.

%C It appears that limit a(n)/n^2 exists and is near exp(-1).

%H Paul D. Hanna, <a href="/A301464/b301464.txt">Table of n, a(n) for n = 1..3000</a>

%e 1 = 1/2 + 2/3^2 + 3^2/4^3 + 4^3/5^4 + 5^4/8^5 + 6^5/9^6 + 7^6/15^7 + 8^7/17^8 + 9^8/21^9 + 10^9/28^10 + 11^10/33^11 + 12^11/38^12 + 13^12/50^13 + 14^13/61^14 + 15^14/72^15 + 16^15/78^16 + 17^16/86^17 + 18^17/91^18 + 19^18/110^19 + 20^19/123^20 + 21^20/141^21 + 22^21/161^22 + 23^22/178^23 + 24^23/187^24 + 25^24/214^25 + 26^25/230^26 + 27^26/239^27 + ... + n^(n-1)/a(n)^n + ...

%o (PARI) /* Must use appropriate precision to obtain N terms: */ N=100;

%o A=[2,3]; for(i=3,N, A=concat(A, ceil((1/(1 - sum(n=1,#A,n^(n-1)/A[n]^n * 1.))*(#A+1)^(#A) )^(1/(#A+1))) ) );A

%K nonn

%O 1,1

%A _Paul D. Hanna_, Mar 26 2018