A303600 a(1)=2 and a(2)=3, then a(n+1) is the smallest integer larger than a(n) that can be written as the sum of two (not necessarily distinct) earlier terms in exactly one way.

2, 3, 4, 5, 9, 10, 11, 16, 22, 24, 28, 29, 30, 37, 42, 50, 55, 56, 70, 73, 76, 82, 89, 95, 101, 102, 115, 128, 133, 135, 136, 141, 142, 153, 160, 161, 168, 174, 181, 195, 199, 200, 214, 221, 227, 233, 247, 252, 265, 266, 267, 273, 280, 285, 286, 325, 331, 332, 338

Offset: 1

Author

Michel Marcus, Apr 26 2018

Keywords

• nonn

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (Terms 1..100 from David Consiglio, Jr.)
• David A. Corneth, PARI-prog
• Borys Kuca, Structures in Additive Sequences, arXiv:1804.09594 [math.NT], 2018. See V(2,3).

Crossrefs

Cf. A004280 (with first terms 1 and 2).

Programs

Mathematica

Nest[Append[#, Function[{m, s}, First@ SelectFirst[Tally[s], And[First@ # > m, Last@ # < 3] &]] @@ {Max@ #, Sort[Total /@ Tuples[#, {2}]]}] &, {2, 3}, 57] (* Michael De Vlieger, Apr 27 2018 *)

PARI

\\ See PARI link \\ David A. Corneth, Apr 27 2018

Python

terms = [2,3]
while len(terms) < 100:
    options = []
    for x in range(len(terms)):
        for y in range(x,len(terms)):
            options.append(terms[x]+terms[y])
    for y in sorted(options):
        if options.count(y) == 1 and y > max(terms):
            terms.append(y)
            break
print(terms)
# David Consiglio, Jr., Apr 18 2018

Python

from collections import Counter
def A303600_list(num_terms):
    terms = [2, 3]
    sum_counts = Counter([4, 5, 6])
    next_term = 3
    for n in range(2, num_terms):
        next_term += 1
        while sum_counts[next_term] != 1: next_term += 1
        terms.append(next_term)
        for term in terms: sum_counts[term + next_term] += 1
    return terms
print(A303600_list(59))
# David Radcliffe, Jun 19 2025

Extensions

More terms from David Consiglio, Jr., Apr 26 2018