A385584 a(n) is the number of pairs (p, t) such that p is a pyramidal number, t is a triangular number, p + t <= n and t <= p.

1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 7, 8, 8, 9, 9, 9, 10, 10, 10, 10, 12, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 19, 19, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 25, 26, 26, 27, 27, 27, 28, 29, 29, 29, 30, 30, 30, 30, 30, 31, 31, 31

Offset: 0

Author

Robert G. Wilson v, Jul 03 2025

Keywords

• nonn

Comments

This sequence is nondecreasing.

Examples

			[n] #  solutions
----------------------------------------------------
[0] 1 [(0, 0)]
[1] 2 [(0, 0), (1, 0)]
[2] 3 [(0, 0), (1, 0), (1, 1)]
[3] 3 [(0, 0), (1, 0), (1, 1)]
[4] 4 [(0, 0), (1, 0), (1, 1), (4, 0)]
[5] 5 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1)]
[6] 5 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1)]
[7] 6 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1), (4, 3)]
[8] 6 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1), (4, 3)]
[9] 6 [(0, 0), (1, 0), (1, 1), (4, 0), (4, 1), (4, 3)]
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Crossrefs

Cf. A000217, A000292, A001399, A001477, A027568.

Programs

Python

def a(n: int) -> int:
    count = 0
    for p in range(n + 1):
        pv = p * (p + 1) * (p + 2) // 6
        if pv > n: break
        for t in range(n - p + 1):
            tv = t * (t + 1) // 2
            if pv + tv <= n and tv <= pv:
                count += 1
    return count
print([a(n) for n in range(74)])  # Peter Luschny, Jul 10 2025

Formula

a(n) = card({(t in A000217, p in A000292) : t <= p, t + p <= n}). - Peter Luschny, Jul 10 2025

Extensions

New name and two terms (n=4 and n=20) corrected by Peter Luschny, Jul 10 2025