A345112 a(n) is the number of steps to reach a palindrome > n under repeated applications of the map x -> A345111(x) starting with n, or 0 if no palindrome is ever reached.

1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 19, 1, 1, 1, 1, 1, 1, 3, 1, 19

Offset

1,5

Comments

Is a(49) = 0? If a(49) != 0, it is > 100000 (10^5) (see A345115).

Links

Table of n, a(n) for n=1..48.

Example

For n = 39: The trajectory of 39 under the given map starts 39, 132, 453, 987, 1866, 10527, 15798, 73779, 111576, 227337, 500709, 507804, 585849, 1444344, 5887785, 14765640, 62422041, 86642457, 153067035, 683737386, reaching the palindrome 683737386 after 19 iterations, so a(39) = 19.

Prog

(PARI) eva(n) = subst(Pol(n), x, 10)
rot(vec) = if(#vec < 2, return(vec)); my(s=concat(Str(2), ".."), v=[]); s=concat(s, Str(#vec)); v=vecextract(vec, s); v=concat(v, vec[1]); v
a(n) = my(x=n, i=0); while(1, x=x+eva(rot(digits(x))); i++; if(digits(x)==Vecrev(digits(x)), break)); i
(Python)
def pal(s): return s == s[::-1]
def rotl(s): return s[1:] + s[0]
def A345111(n): return n + int(rotl(str(n)))
def a(n):
    i, iter, seen = 0, n, set()
    while not (iter > n and pal(str(iter))) and iter not in seen:
        seen.add(iter)
        i, iter = i+1, A345111(iter)
    return i if iter > n and pal(str(iter)) else 0
print([a(n) for n in range(1, 49)]) # Michael S. Branicky, Jun 09 2021

Crossrefs

Cf. A016016, A345110, A345111, A345113, A345114, A345115.

Sequence in context: A038082 A107740 A016016 * A063059 A214564 A102675

Adjacent sequences: A345109 A345110 A345111 * A345113 A345114 A345115

Keyword

nonn,base,more

Author

Felix Fröhlich, Jun 09 2021

Status

approved