A239480 Palindromes such that additive and multiplicative persistences coincide.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 99, 101, 111, 121, 131, 141, 151, 161, 171, 202, 212, 222, 262, 282, 303, 313, 393, 404, 424, 454, 474, 525, 545, 565, 585, 595, 636, 656, 676, 757, 838, 858, 959, 1001, 1111, 1221, 1331, 1441, 1991, 2002, 2112, 2552

Offset: 1

Author

Arkadiusz Wesolowski, Mar 20 2014

Keywords

• nonn
• base

Comments

Palindromes k for which A031286(k) = A031346(k).

Examples

			99 -> 18 -> 9 has additive persistence 2. 99 -> 81 -> 8 has multiplicative persistence 2. The palindromic number 99 is therefore in the sequence.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Additive Persistence
• Eric Weisstein's World of Mathematics, Multiplicative Persistence
• Index entries for sequences related to palindromes

Crossrefs

Cf. A002113, A031286, A031346, A239427.

Programs

PARI

for(n=0, 2552, s=Vec(Str(n)); if(s==vecextract(s, "-1..1"), v=n; a=0; while(n>9, a++; n=sumdigits(n)); n=v; m=0; while(n>9, m++; d=digits(n); n=prod(k=1, #d, d[k])); n=v; if(a==m, print1(n, ", "))));

Python

from math import prod
from itertools import count, islice, product
def A031286(n):
    ap = 0
    while n > 9: n, ap = sum(map(int, str(n))), ap+1
    return ap
def A031346(n):
    mp = 0
    while n > 9: n, mp = prod(map(int, str(n))), mp+1
    return mp
def is_pal(n): return (s:=str(n)) == s[::-1]
def pals(base=10): # all d-digit palindromes
    digits = "".join(str(i) for i in range(base))
    for d in count(1):
        for p in product(digits, repeat=d//2):
            if d > 1 and p[0] == "0": continue
            left = "".join(p); right = left[::-1]
            for mid in [[""], digits][d%2]: yield int(left + mid + right)
def ok(n): return is_pal(n) and A031286(n) == A031346(n)
def agen(): yield from filter(ok, pals())
print(list(islice(agen(), 20))) # Michael S. Branicky, Jun 22 2023

Formula

A002113 INTERSECT A239427.