Jonatan Djurachkovitch - cp's OEIS frontend

User: Jonatan Djurachkovitch

Jonatan Djurachkovitch has authored 1 sequences.

A382630 a(n) is the number of ways that n can be written as b+c*d, where b, c and d are positive integers and b < c < d.

0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 2, 1, 1, 1, 2, 0, 2, 1, 2, 2, 3, 0, 3, 2, 2, 1, 3, 1, 4, 2, 3, 3, 3, 1, 4, 3, 3, 1, 4, 2, 5, 3, 3, 4, 5, 1, 5, 3, 4, 3, 5, 2, 4, 3, 5, 5, 6, 1, 6, 5, 4, 4, 5, 3, 6, 4, 5, 3, 6, 2, 7, 6, 5, 5, 6, 4, 7, 3, 6, 6, 7, 2, 6, 6, 6, 4, 7, 3, 7

Offset: 0

Jonatan Djurachkovitch, Apr 01 2025

nonn

24 is the largest number for which a(n) = 0.
From David A. Corneth, Apr 04 2025: (Start)
As b + c*d = n where b < c < d we have b is at its largest when c = b+1 and d = b+2 and so b <= sqrt(n + 2) - 2.
Proof: c = b + 1 and d = b + 2 gives b + c*d = b + (b + 1)*(b + 2) = b^2 + 4*b + 2 <= n.
Adding two to both sides gives b^2 + 4*b + 4 = (b + 2)^2 <= n + 2 and so b <= sqrt(n + 2) - 2.
Once b is chosen, c*d = n - b and so we can look at divisor pairs (c, d) of n - b having b < c < d. (End)

			13 can be written as 1 + 3 * 4 and as 1 + 2 * 6, so a(13) = 2.

Robert Israel, Table of n, a(n) for n = 0..10000
Skolornas Matematiktävling, Finaltävling i Linköping den 20 november 2021 (Swedish)

N:= 100: # for a(0) .. a(N)
V:= Array(0..N):
for b from 1 while b + (b+1)*(b+2) <= N do
  for c from b+1 while b + c*(c+1) <= N do
    for d from c+1 do
      x:= b+c*d;
      if x > N then break fi;
      V[x]:= V[x]+1
od od od:
convert(V,list); # Robert Israel, May 04 2025

a(n) = {
	my(res = 0);
	for(b = 1, sqrtint(n + 2) - 2,
		d = divisors(n - b);
		ind = setsearch(d, b+1, 1 - ((n-b) % (b+1) == 0));
		res += max(0, #d\2 - ind + 1);	
	);
	res
} \\ David A. Corneth, Apr 04 2025

first(n) = {
	my(res = vector(n));
	for(b = 1, sqrtint(n + 2),
		for(c = b + 1, sqrtint(n - b - 1),
			for(d = c + 1, (n - b) \ c,
				res[b + c*d]++
			);
		);
	);
	concat(0, res)
} \\ David A. Corneth, Apr 04 2025

# returns the first 100 terms as a list
import numpy as np
terms = 100
a = np.zeros(terms)
for d in range(3,terms // 2):
    for c in range(2,d):
        for b in range(1,c):
            val = b + c*d
            if val < terms:
                a[val] += 1
print(a)

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User: Jonatan Djurachkovitch

A382630 a(n) is the number of ways that n can be written as b+c*d, where b, c and d are positive integers and b < c < d.

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