A371113 The sequence is a succession of triples of nonnegative integers. In each triple, the digits form a palindromic pattern. The sequence starts with a(0)=0 and is always extended with the smallest available integer not yet present in the sequence.

0, 1, 10, 2, 3, 32, 4, 5, 54, 6, 7, 76, 8, 9, 98, 11, 12, 111, 13, 14, 131, 15, 16, 151, 17, 18, 171, 19, 20, 291, 21, 22, 212, 23, 24, 232, 25, 26, 252, 27, 28, 272, 29, 30, 392, 31, 33, 313, 34, 35, 343, 36, 37, 363, 38, 39, 383, 40, 41, 404, 42, 43, 424, 44, 45, 444, 46, 47, 464, 48, 49, 484, 50, 51, 505

Offset

0,3

Comments

The sequence is a permutation of the nonnegative integers: there is no constraint on terms a(3k) and a(3k+1), which are always the smallest nonnegative integers that did not occur earlier. Therefore each nonnegative integer will occur exactly once. - M. F. Hasler, Apr 18 2024

From M. F. Hasler, Apr 20 2024: (Start)

See A369856 for the inverse permutation.

Conjecture 1: The only fixed points of this permutation are n = 0 and 1.

Conjecture 2: For all k >= 0, a(3k+2) <= R(concatenate(a(3k), a(3k+1), 1)), where R = A004086 = reverse.

Conjecture 3: For all k >= 0, a(3k+1) - a(3k) = 1 or 2. (Verified up to k = 3333. a(3k+1) - a(3k) = 2 for k = 15, 36, 62, 81, 110, 129, 138, 163, 182, ...)

(End)

Links

Table of n, a(n) for n=0..74.

Formula

a(3k) = 2k + o(log k) ; a(3k) + 1 <= a(3k+1) <= a(3k) + 2 (conjectured). - M. F. Hasler, Apr 20 2024

Example

Triples begin (0,1,10), (2,3,32), (4,5,54), (6,7,76), (8,9,98), ... where the palindromic pattern is clearly visible in each triple.

Mathematica

a[0]=0; a[n_]:=a[n]=(k=1; While[MemberQ[Array[a, n-1], k]|| If[Mod[n, 3]==2, !PalindromeQ[Flatten[IntegerDigits/@Join[{a[n-2]}, {a[n-1]}, {k}]]]], k++]; k); Array[a, 100, 0]

Prog

(PARI) A371113_upto(N)={my(L, U=[-1], F=fromdigits); vector(N, n, [L, N]=[N, if(n%3, U[1]+1, L=eval(Vec(Str(L, N))); N=0; for(k=1, #L, L[k] && Vecrev(n=concat(L, Vecrev(L, k)))==n && !setsearch(U, n=F(Vecrev(L, k))) && n>U[1] && [N=n; break]); if(!N, L=Vecrev(L, -1-#L); until(!setsearch(U, N=F(L))&& N>U[1], L[1]++)); N)]; U=setunion(U, [N]); while(#U>1 && U[1]+1==U[2], U=U[^1]); N)} \\ M. F. Hasler, Apr 19 2024
(Python)
def A371113(n):
    try: return A371113.terms[n]
    except AttributeError: A371113.terms = []; A371113.least_unused = 0
    except IndexError: pass # just need to compute more terms
    while len(T := A371113.terms) <= n:
        if len(T)%3 < 2:
            T.append(x := A371113.least_unused)
            while x+1 in T: x += 1
            A371113.least_unused = x+1
        else:
            for i in range(1, len(suffix := (str(T[-2])+str(T[-1]))[::-1])+1):
                if suffix[-i] != '0' and (s := suffix[:-i-1:-1] + suffix
                    )==s[::-1] and (x := int(suffix[-i:])) not in T: break
            else: x = int('1'+suffix); assert x not in T
            T.append(x); assert x > A371113.least_unused
    return T[n] # M. F. Hasler, Apr 19 2024

Crossrefs

Cf. A238880 (analog for pairs instead of triples), A369856 (inverse permutation), A004086 (read n backwards).

Sequence in context: A336954 A322467 A342078 * A049296 A220468 A169851

Adjacent sequences: A371110 A371111 A371112 * A371114 A371115 A371116

Keyword

nonn,base

Author

Eric Angelini and Giorgos Kalogeropoulos, Apr 09 2024

Status

approved