experimental-projects/oeis-research/Joseph Biberstine/Sum of remainders of the n-th prime mod k, for k = 1..n/prog.sf

#!/usr/bin/ruby

# Sum of remainders of the n-th prime mod k, for k = 1,2,3,...,n.
# https://oeis.org/A099726

# Formula:
#   a(n) = n*p - A024916(p) + Sum_{k=n+1..p} k*floor(p/k), where p = prime(n).

# Some large terms:
#   a(10^1)  = 30
#   a(10^2)  = 2443
#   a(10^3)  = 248372
#   a(10^4)  = 25372801
#   a(10^5)  = 2437160078
#   a(10^6)  = 252670261459
#   a(10^7)  = 24690625139657
#   a(10^8)  = 2516604108737704
#   a(10^9)  = 249876964098609078
#   a(10^10) = 24994462548503343285
#   a(10^11) = 2513170619960257982344

func T(n) {     # Sum_{k=1..n} k = n-th triangular number
    n.faulhaber(1)
}

func S(n) {     # Sum_{k=1..n} sigma(k) = Sum_{k=1..n} k*floor(n/k)
    #var s = n.isqrt
    #sum(1..s, {|k| T(idiv(n,k)) + k*idiv(n,k) }) - T(s)*s
    n.dirichlet_sum({1}, {_}, {_}, {.faulhaber(1)})
}

func g(a,b) {
    var total = 0
    while (a <= b) {
        var t = idiv(b, a)
        var u = idiv(b, t)
        total += t*(T(u) - T(a-1))
        a = u+1
    }
    return total
}

func a(n) { # sub-linear formula
    var p = n.prime
    n*p - S(p) + g(n+1, p)
}

func a_linear(n) { # just for testing
    var p = n.prime
    sum(1..n, {|k| p % k })
}

assert_eq(a.map(1..100), a_linear.map(1..100))

say a.map(1..58)   #=> [0, 1, 3, 5, 7, 7, 14, 18, 28, 30, 31, 26, 38, 45, 63, 71, 93, 75, 96, 115, 101, 142, 161, 167, 152, 159, 203, 224, 219, 222, 216, 250, 263, 296, 341, 320, 319, 349, 433, 427, 496, 419, 487, 481, 538, 537, 495, 631, 635, 676, 697, 777, 665, 820, 784, 874, 929, 856]

for k in (1..9) {
    say "a(10^#{k}) = #{a(10**k)}"
}

__END__

# PARI/GP program (linear time)

a(n) = my(p=prime(n)); sum(k=1, n, p%k);

# PARI/GP program (sublinear time)

T(n) = n*(n+1)/2;
S(n) = my(s=sqrtint(n)); sum(k=1, s, T(n\k) + k*(n\k)) - s*T(s); \\ A024916
g(a,b) = my(s=0); while(a <= b, my(t=b\a); my(u=b\t); s += t*(T(u) - T(a-1)); a = u+1); s;
a(n) = my(p=prime(n)); n*p - S(p) + g(n+1, p);

# PARI/GP program (sublinear time) -- shorter, but slower

T(n) = n*(n+1)/2;
g(a,b) = my(s=0); while(a <= b, my(t=b\a); my(u=b\t); s += t*(T(u) - T(a-1)); a = u+1); s;
a(n) = my(p=prime(n)); n*p - g(1, p) + g(n+1, p);