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%I #14 Aug 29 2024 11:13:06
%S 1,3,5,7,9,11,12,13,15,17,19,20,21,23,24,25,27,28,29,30,31,33,35,36,
%T 37,39,40,41,42,43,44,45,47,48,49,51,52,53,55,56,57,59,60,61,63,65,66,
%U 67,68,69,70,71,72,73,75,76,77,78,79,80,81,83,84,85,87,88,89,90,91,92,93,95,96,97,99,100
%N Complement of A138929: positive integers not of the form 2*p^k, k >= 0, p a prime (also 2).
%C The complement relative to the positive integers of the present sequence is A138929.
%C The sequence includes all positive integers of the forms (i) odd, (ii) 2^k*p, p an odd prime and k>=2, and (iii) 2^e0*p1^e1*p2^e2 *** pk^ek, k >= 2, with odd primes p1, ..., pk, and each exponent from {e0, ..., ek} is >= 1.
%C For a(n) > 1 a regular a(n)-gon, with length ratio (smallest diagonal)/side rho(a(n)) = 2*cos(Pi/a(n)), the inverse of rho(a(n)), which is an element of the algebraic number field Q(rho(a(n))), is in fact a Q(rho(a(n)))-integer. For a(1)=1 rho(1) = -2, and the inverse is not a Q-integer.
%e Even members a(n) of the form (ii) 2^k*p, p an odd prime and k>=2 are: 12, 20, 24, 28, 36, 40, 44, 48, 52, 56, 68, 72, 76, 80, 88, 92, 96, 100,...
%e Even members a(n) of the form (iii), given above, include 30, 42, 60, 66, 70, 78, 84, 90, ...
%e For the regular 5-gon (pentagon) rho(5) = tau = (1 + sqrt(5))/ 2 (the golden section). The number field is Q(rho(5)), and for the inverse one has 1/rho(5) = -1*1 + 1* rho(5) (in the power basis <1, rho(5)>, in which Q(rho(5))-integers have integer coefficients).
%e For the regular 7-gon rho(7) = 2*cos(Pi/7), (approximately 1.801937736) is of degree 3, and 1/rho(7) = 2*1 + 1*rho(7) - 1*rho(7)^2, (approximately 0.5549581320), hence a Q(rho(7))- integer.
%e For Gauss' regular 17-gon rho(17) = 2*cos(Pi/17) (approximately 1.965946199) is of degree 8 and 1/rho(17) = -4*1+ 10*rho(17)^1 + 10*rho(17)^2 - 15*rho(17)^3 -6*rho(17)^4 + 7*rho(17)^5 + 1*rho(17)^6 -1*rho(17)^7, (approximately 0.5086610), hence this is a Q(rho(17))- integer.
%o (Python)
%o from sympy import primepi, integer_nthroot
%o def A230078(n):
%o if n == 1: return 1
%o def f(x): return int(n+1+sum(primepi(integer_nthroot(x>>1,k)[0]) for k in range(1,(x>>1).bit_length())))
%o kmin, kmax = 0,1
%o while f(kmax) > kmax:
%o kmax <<= 1
%o while kmax-kmin > 1:
%o kmid = kmax+kmin>>1
%o if f(kmid) <= kmid:
%o kmax = kmid
%o else:
%o kmin = kmid
%o return kmax # _Chai Wah Wu_, Aug 29 2024
%Y Cf. A138929 (complement), 2*A020513, A230079 (1/rho(a(n))) .
%K nonn,easy,changed
%O 1,2
%A _Wolfdieter Lang_, Nov 02 2013
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