project-euler/Yet to solve/792 Too Many Twos.sf

#!/usr/bin/ruby

# Too Many Twos
# https://projecteuler.net/problem=792

# See also:
#   https://oeis.org/A059304

func g(n) {
    return 0 if (n < 0)
    return 1 if (n == 0)
    __FUNC__(n-1)*(8 - 4/n)
}

func Sg(n) {
    return 1 if (n == 0)
    return g(n)+g(0) if (n == 1)
    return g(n)+g(1)+g(0) if (n == 2)
    return g(n)+g(2)+g(1)+g(0) if (n == 3)
    #__FUNC__(n-1) + g(n)

    #assert_eq((8 - 4/n), g(n)/g(n-1))
    #assert_eq((8 - 4/(n-1)), g(n-1)/g(n-2))

    (g(n)/g(n-1) * __FUNC__(n-1)) - ((n-2) * __FUNC__(n-2))
}

#~ say 10.of(g).acc
#~ say 10.of(Sg)

#say 10.of {|n| g(n) / g(n-1) }
#say 10.of {|n| g(n+1) / g(n) }

func S2(n) {
    sum(1..n, {|k|
        3 * (-2)**k * binomial(2*k, k)
    })
}

func f(n) {
    return 4/3 if (n == 0)
    (-1)**n * (2**n * (2*n)!) / ((n!)**2)
}

func S(N) {
    sum(1..N, {|n|
        #(-1)**n * (2**n * (2*n)!) / ((n!)**2)
        f(n)
    })
}

func f2(n) {
    return 1 if (n == 0)
    __FUNC__(n-1) * (8 - 4/n)
}

func S4(n) {
    return  4  if (n == 0)
    #return f(1) if (n == 1)
    #return f(1) if (n == 2)
    #n*__FUNC__(n-1) - (n - 1)*__FUNC__(n-2)
    __FUNC__(n-1) + 3*f(n)
}

func u5(n) {

    return 2 if (n == 0)

    __FUNC__(n-1).valuation(2) - ((-1)**n + (n + valuation((2*n)!, 2)) - valuation(((n!)**2), 2))
}

#~ func S(n) {
    #~ 2**n * n * binomial(2*n, n)
#~ }

func u2(n) {
    3*S(n) + 4 -> valuation(2)
}

func u(n) {
    S4(n)  -> valuation(2)
}

func U(N) {
    sum(1..N, {|n|
        u(n**3)
    })
}

say S(4)
say u(4)
say u(20)
say U(5)
#say U(1e4)

assert_eq(u(4), 7)
assert_eq(u(20), 24)
assert_eq(U(5), 241)