A395817 Strictly increasing minimal sequence where consecutive terms can be added digit-by-digit without carrying in base 10.

1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25, 30, 31, 32, 33, 34, 35, 40, 41, 42, 43, 44, 45, 50, 100, 101, 102, 103, 104, 105, 110, 111, 112, 113, 114, 115, 120, 121, 122, 123, 124, 125, 130, 131, 132, 133, 134, 135, 140, 141, 142, 143, 144

Offset

1,2

Comments

Many integers (e.g., 6, 7, 8, 9, 15, 16...) are never present because the greedy behavior and the strictly increasing condition bypass them to avoid carries.

A subset of nonnegative integers with no decimal digits > 5 (A007092).

Links

Vincenzo Manto, Table of n, a(n) for n = 1..10000

Formula

a(1)=1; a(n) = min { x > a(n-1) | d_i(x) + d_i(a(n-1)) <= 9 }.

Example

a(5) = 5 because a(4) = 4 and 4 is the smallest k such that 4 + k doesn't carry: 4 + 5 = 9 (no carry).
a(15) = 24 because a(14) = 23 and 24 is the smallest k such that 23 + k doesn't carry: 23 + 24 = 47 (no carry).
a(56) = 150 because a(55) = 144 and 150 is the smallest k such that 144 + k doesn't carry: 144 + 150 = 294 (no carry).

Mathematica

s={1}; k=2; Do[While[Max[IntegerDigits[k]]>5||Max[IntegerDigits[s[[-1]]]+Take[IntegerDigits[k], -IntegerLength[s[[-1]]]]]> 9, k++]; AppendTo[s, k]; k++, {i, 58}]; s

Prog

(Python)
def has_carry(a, b):
    while a or b:
        if (a % 10) + (b % 10) > 9:
            return True
        a //= 10; b //= 10
    return False
def sequence(n):
    terms = [1]
    while len(terms) < n:
        k = terms[-1] + 1
        while has_carry(terms[-1], k):
            k += 1
        terms.append(k)
    return terms

Crossrefs

Cf. A052413, A007092.

Sequence in context: A391164 A037473 A007092 * A241213 A362931 A375514

Adjacent sequences: A395811 A395814 A395816 * A395818 A395819 A395820

Keyword

nonn,base

Author

Vincenzo Manto and James C. McMahon, May 07 2026

Status

approved