A375380 Lexicographically earliest sequence S of distinct nonnegative integers such that 9 out of the last 10 digits of S always sum to a square.

1000000000, 0, 1, 2, 3, 4, 6, 5, 7, 8, 9, 10, 11, 13, 12, 15, 14, 17, 16, 21, 18, 19, 20, 27, 22, 25, 23, 30, 24, 26, 31, 28, 29, 32, 34, 33, 36, 38, 35, 37, 40, 39, 41, 42, 43, 44, 47, 45, 50, 46, 51, 48, 52, 54, 55, 49, 57, 53, 56, 58, 59, 60, 64, 61, 62, 63

Offset

1,1

Comments

The smallest available 10-digit integer to start S with is a(1) = 1000000000.

The sequence is infinite as there are infinitely many integers whose 9 of the last 10 digits sum to a square.

Links

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Example

We start with a(1) = 1000000000:
S = 1000000000,
The last 10 digits of S are [1000000000]; we discard a 0 and the sum of the 9 remaining digits = 1 (a square). We try to extend S with 0:
S = 1000000000,0,
The last 10 digits of S are [0000000000]; we discard a 0 and the sum of the 9 remaining digits = 0 (a square). We try to extend S with 1:
S = 1000000000,0,1,
The last 10 digits of S are [0000000001]; we discard a 0 and the sum of the 9 remaining digits = 1 (a square). We try to extend S with 2:
S = 1000000000,0,1,2,
The last 10 digits of S are [0000000012]; we discard 2 and the sum of the 9 remaining digits = 1 (a square). We try to extend S with 3:
S = 1000000000,0,1,2,3,
The last 10 digits of S are [0000000123]; we discard 2 and the sum of the 9 remaining digits = 4 (a square). We try to extend S with 4:
S = 1000000000,0,1,2,3,4,
The last 10 digits of S are [0000001234]; we discard 1 and the sum of the 9 remaining digits = 9 (a square). We try to extend S with 5:
S = 1000000000,0,1,2,3,4,5,
The last 10 digits of S are [0000012345]; as no square can be reached, whatever we discard, we don’t extend S with 5. We try to extend S with 6:
S = 1000000000,0,1,2,3,4,6,
The last 10 digits of S are [0000012346]; we discard a 0 and the sum of the 9 remaining digits = 16 (a square).
Etc.

Prog

(Python)
from itertools import count, islice, combinations
def f(k, last10):
    d = list(map(int, str(k)))
    if len(d) >= 10: return d[-10:]
    else: return last10[len(d):] + d
def c(k, last10):
    for pick9 in combinations(f(k, last10), 9):
        if sum(pick9) in {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}:
            return True
    return False
def agen(): # generator of terms
    an, seen, last10, mink = 1000000000, set(), [], 0
    while True:
        yield an
        seen.add(an)
        last10 = f(an, last10)
        an = next(k for k in count(mink) if k not in seen and c(k, last10))
        while mink in seen: mink += 1
print(list(islice(agen(), 66))) # Michael S. Branicky, Aug 14 2024

Crossrefs

Cf. A352000.

Sequence in context: A136953 A136961 A307478 * A100909 A260519 A260520

Adjacent sequences: A375377 A375378 A375379 * A375381 A375382 A375383

Keyword

base,nonn

Author

Eric Angelini, Aug 14 2024

Extensions

a(14) and beyond from Michael S. Branicky, Aug 14 2024

Status

approved