A351841 Lexicographically earliest infinite sequence of distinct positive integers such that a(1) = 1 and a(n) + a(n+1) is divisible by the last digit of a(n+1). No term ending in zero is allowed in the sequence.

1, 11, 13, 21, 3, 31, 23, 41, 43, 27, 9, 33, 37, 47, 51, 39, 17, 19, 53, 61, 29, 71, 73, 67, 59, 49, 7, 77, 81, 63, 91, 83, 57, 69, 93, 101, 79, 111, 121, 113, 97, 119, 103, 107, 109, 87, 123, 129, 131, 133, 141, 151, 143, 137, 157, 149, 117, 99, 153, 127, 89, 161, 147, 159, 171, 181, 173, 163

Offset

1,2

Comments

Only terms ending in 1, 3, 7 or 9 are in the sequence.

Proof. If a(n-1) is odd, then a(n) cannot be even; otherwise, their sum would be odd and hence not divisible by a(n)'s last, even digit. For sake of contradiction, let a(n) be the first term in the sequence that ends in 5; then a(n-1) + a(n) ends in 0 and a(n-1) ends in 5 (a contradiction to a(n)'s being the first such term) or a(n-1) + a(n) ends in 5, which means a(n-1) ends in 0 (not allowed). - Michael S. Branicky, Feb 21 2022

Links

Carole Dubois, Table of n, a(n) for n = 1..4016

Example

a(1) + a(2) = 1 + 11 = 12 which is divisible by 1 (the last digit of 11);
a(2) + a(3) = 11 + 13 = 24 which is divisible by 3 (the last digit of 13);
a(3) + a(4) = 13 + 21 = 34 which is divisible by 1 (the last digit of 21);
a(4) + a(5) = 21 + 3 = 24 which is divisible by 3 (the last digit of 3);
a(5) + a(6) = 3 + 31 = 34 which is divisible by 1 (the last digit of 31);
a(6) + a(7) = 31 + 23 = 54 which is divisible by 3 (the last digit of 23); etc.

Mathematica

nn = 67; f[n_] := (5 n + Mod[3 n + 2, 4] - 4)/2; c[_] = 0; a[1] = c[1] = u = 1; Do[While[c[Set[v, f[u]]] != 0, u++]; j = v; While[Nand[c[#2] == 0, Mod[#1 + #2, Mod[#2, 10]] == 0] & @@ {a[n - 1], Set[k, f[j]]}, j++]; Set[{a[n], c[k]}, {k, n}], {n, 2, nn}]; Array[a[#] &, nn] (* Michael De Vlieger, Feb 21 2022 *)

Prog

(Python)
def aupton(terms):
    alst, aset, minunused = [1], {1}, 2
    while len(alst) < terms:
        an = minunused
        while an in aset or (an%10 == 0 or (alst[-1]+an)%(an%10)): an += 1
        alst.append(an); aset.add(an)
        while minunused in aset: minunused += 1
    return alst
print(aupton(78)) # Michael S. Branicky, Feb 21 2022

Crossrefs

Cf. A351840.

Sequence in context: A089774 A346287 A330975 * A257864 A248599 A366379

Adjacent sequences: A351838 A351839 A351840 * A351842 A351843 A351844

Keyword

base,nonn

Author

Eric Angelini and Carole Dubois, Feb 21 2022

Status

approved