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Greedy powers of (e/3): Sum_{n>=1} (e/3)^a(n) = 1.
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%I #7 Jun 10 2018 21:12:04

%S 1,24,92,140,171,199,226,251,277,320,363,391,425,449,474,500,524,548,

%T 575,632,673,777,801,836,861,903,932,959,983,1011,1054,1087,1113,1148,

%U 1176,1228,1261,1286,1316,1348,1394,1427,1452,1480,1510,1536,1571,1600

%N Greedy powers of (e/3): Sum_{n>=1} (e/3)^a(n) = 1.

%C The n-th greedy power of x, when 0.5 < x < 1, is the smallest integer exponent a(n) that does not cause the power series Sum_{k=1..n} x^a(k) to exceed unity.

%F a(n) = Sum_{k=1..n} floor(g_k) where g_1=1, g_{n+1}=log_x(x^frac(g_n) - x) (n>0) at x=(e/3) and frac(y) = y - floor(y).

%e a(3)=92 since (e/3) + (e/3)^24 + (e/3)^92 < 1 and (e/3) +(e/3)^24 + (e/3)^91 > 1; since the power 91 makes the sum > 1, then 92 is the 4th greedy power of (e/3).

%p Digits := 1100: summe := 0.0: p := evalf(exp(1)/3.): pexp := p: a := []: for i from 1 to 3000 do: if summe + pexp < 1 then a := [op(a),i]: summe := summe + pexp: fi: pexp := pexp * p: od: a;

%o (PARI) default(realprecision,99); s=1; Le3=1-log(3); for(i=1,50, print1(a=if(i>1,log(s)\Le3,1)","); s-=exp(a*Le3)) \\ _M. F. Hasler_, Sep 28 2009

%Y Cf. A077468 - A077475.

%K easy,nonn

%O 1,2

%A Ulrich Schimke (ulrschimke(AT)aol.com)

%E Some terms corrected (replaced 67,3 with 673 and 153,6 with 1536) by _M. F. Hasler_, Sep 28 2009