A096772 A B3-sequence: a(1) = 1; for n>1, a(n) = smallest number > a(n-1) such that the sums of any three terms are all distinct.

1, 2, 5, 14, 33, 72, 125, 219, 376, 573, 745, 1209, 1557, 2442, 3098, 4048, 5298, 6704, 7839, 10987, 12332, 15465, 19144, 24546, 28974, 34406, 37769, 45864, 50877, 61372, 68303, 77918, 88545, 101917, 122032, 131625, 148575, 171237, 197815, 201454

Offset

1,2

Comments

This is the B3-sequence analog of the Mian-Chowla B2-sequence (A005282): Let a(1)=1; then use the greedy algorithm to choose the smallest a(n) > a(n-1) such that all sums a(i) + a(j) + a(k) are distinct for 1 <= i <= j <= k <= n. The reciprocal sum of the sequence for the first forty terms is 1.837412....

Links

Chai Wah Wu, Table of n, a(n) for n = 1..176

Eric Weisstein's World of Mathematics, B2-Sequence.

Eric Weisstein's World of Mathematics, Mian-Chowla Sequence.

Formula

a(n) = A051912(n-1) + 1. - Peter Kagey, Oct 20 2021

Prog

(Python)
from itertools import count, islice
def A096772_gen(): # generator of terms
    aset1, aset2, aset3, alist = set(), set(), set(), []
    for k in count(1):
        bset2, bset3 = {k<<1}, {3*k}
        if 3*k not in aset3:
            for d in aset1:
                if (m:=d+(k<<1)) in aset3:
                    break
                bset2.add(d+k)
                bset3.add(m)
            else:
                for d in aset2:
                    if (m:=d+k) in aset3:
                        break
                    bset3.add(m)
                else:
                    yield k
                    alist.append(k)
                    aset1.add(k)
                    aset2 |= bset2
                    aset3 |= bset3
A096772_list = list(islice(A096772_gen(), 30)) # Chai Wah Wu, Sep 05 2023

Crossrefs

Row 3 of A347570.

Cf. A005282 (Mian-Chowla B2-sequence). A051912.

Sequence in context: A296502 A036681 A125615 * A090803 A018015 A080039

Adjacent sequences: A096769 A096770 A096771 * A096773 A096774 A096775

Keyword

nonn

Author

Rick L. Shepherd, Aug 15 2004

Status

approved