A230078 Complement of A138929: positive integers not of the form 2*p^k, k >= 0, p a prime (also 2).

1, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 33, 35, 36, 37, 39, 40, 41, 42, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 56, 57, 59, 60, 61, 63, 65, 66, 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

Offset

1,2

Comments

The complement relative to the positive integers of the present sequence is A138929.

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.

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.

Links

Table of n, a(n) for n=1..76.

Example

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,...
Even members a(n) of the form (iii), given above, include 30, 42, 60, 66, 70, 78, 84, 90, ...
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).
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.
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.

Prog

(Python)
from sympy import primepi, integer_nthroot
def A230078(n):
    if n == 1: return 1
    def f(x): return int(n+1+sum(primepi(integer_nthroot(x>>1, k)[0]) for k in range(1, (x>>1).bit_length())))
    kmin, kmax = 0, 1
    while f(kmax) > kmax:
        kmax <<= 1
    while kmax-kmin > 1:
        kmid = kmax+kmin>>1
        if f(kmid) <= kmid:
            kmax = kmid
        else:
            kmin = kmid
    return kmax # Chai Wah Wu, Aug 29 2024

Crossrefs

Cf. A138929 (complement), 2*A020513, A230079 (1/rho(a(n))) .

Sequence in context: A162495 A107315 A340855 * A275669 A191275 A260392

Adjacent sequences: A230075 A230076 A230077 * A230079 A230080 A230081

Keyword

nonn,easy,changed

Author

Wolfdieter Lang, Nov 02 2013

Status

approved