prog.sf

#!/usr/bin/ruby

# b(2^n - 1) = 0, for all n >= 1.

# Number of 0's minus number of 1's among the edge truncated binary representations of the first n natural numbers.

# By "edge truncated" we mean removing the first and last binary digits.

# See also:
#   https://oeis.org/A308430    (same process on primes)

func b(n) is cached {
    return 0 if (n == 0)
    return 0 if (n == 1)
    var t = n.as_bin.chars.slice(1, -1)
    b(n-1) + (t.count('0') - t.count('1'))
}

say 128.of(b)

__END__

[0, 0, 0, 0, 1, 2, 1, 0, 2, 4, 4, 4, 4, 4, 2, 0, 3, 6, 7, 8, 9, 10, 9, 8, 9, 10, 9, 8, 7, 6, 3, 0, 4, 8, 10, 12, 14, 16, 16, 16, 18, 20, 20, 20, 20, 20, 18, 16, 18, 20, 20, 20, 20, 20, 18, 16, 16, 16, 14, 12, 10, 8, 4, 0, 5, 10, 13, 16, 19, 22, 23, 24, 27, 30, 31, 32, 33, 34, 33, 32, 35, 38, 39, 40, 41, 42, 41, 40, 41, 42, 41, 40, 39, 38, 35, 32, 35, 38, 39, 40, 41, 42, 41, 40, 41, 42, 41, 40, 39, 38, 35, 32, 33, 34, 33, 32, 31, 30, 27, 24, 23, 22, 19, 16, 13, 10, 5, 0]

for k in (1 .. (2**(ilog2(1e4)))) {
    say b(k)
}

for k in (1..1e7) {
    if (b(k) <= 0) {
        say [k, factor_exp(k+1)]
    }
}

__END__
#say 101.of{|n| b(n+1) }
for n in (1..10000) {
    say b(n)
}

__END__

#~ say b(2**1 - 1)
#~ say b(2**2 - 1)
#~ say b(2**3 - 1)
#~ say b(2**4 - 1)
#~ say b(2**5 - 1)
#~ say b(2**6 - 1)

#~ __END__

#~ __END__

for k in (1..1e5) {
    if (b(k) == 0) {
        say [k, factor_exp(k+1)]
    }
}


__END__
func A070939(n) {
    n.as_bin.len
}

func A000120(n) {
    n.popcount
}

func b(n) is cached {

    return 0 if (n == 1)

    b(n-1) + (A070939(n) - 2*A000120(n) + 2)
}

#say 101.of(b)

for k in (1..1000) {
    say b(k)
}


#b(n) = Sum_{k=2..n} (A070939(k) - 2*A000120(k) + 2).



#a(n) = Sum_{k=2..n} (A035100(k) - 2*A014499(k) + 2) = Sum_{k=2..n} (A070939(A000040(k)) - 2*A000120(A000040(k)) + 2). - ~~~~



__END__

func g(n) {
    var t = n.as_bin.chars.slice(1, -1)
    t.count('0') - t.count('1')
}

#say 20.of(g)

for k in (1..10) {
    say [g(k.prime), k]
}

__END__

func f(n) is cached {
    n == 0 && return 0
    #n ==  && return 0
    #var t = n.as_bin.chars.slice(1, -1)
    #f(n-1) + t.count('0') - t.count('1')
    f(n-1) + g(n.prime)

    #f(n-1) + (1+ilog2(prime(n))) - 2*popcount(prime(n)) + 2
}

say 200.of(f)

__END__
#say f(16381.primepi)
#say (1..20 -> grep{2**_ -1 -> is_prime }.each{say f(primepi(2**_ - 1)) })

#__END__
#say 40.of(f)
for k in (1..1e4) {
    #say f(k)
    if (f(k) == 0) {
        say [k, f(k)]
    }

     #   say k.prime
   # }
}