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A350385
Minimum number of zeros that need to be added to x_n ones such that a combination of these zeros and ones can make a number b with the property gcd(b, rev(b)) = digitsum(b) = x_n where x_n is coprime to 10.
1
0, 1, 36, 1, 12, 66, 3, 3, 6, 4, 2, 3, 4, 10, 75, 16, 7, 3, 3, 7, 2, 5, 4, 3, 3, 6, 2, 2, 2, 10, 10, 5, 2, 3, 2, 2, 2, 4, 3, 10, 304, 4, 3, 3, 1, 3, 12, 6, 124
OFFSET
1,3
COMMENTS
Only for numbers x_n coprime to 10 (A045572, i.e., numbers ending with 1,3,7 or 9) do there exist numbers b such that gcd(b, rev(b)) = x_n and digitsum(b) = x_n (rev(b) is the digit reversal of b, e.g., rev(123) = 321). If b must consist only of zeros and ones, the smallest values of b that satisfy these two constraints are converted to decimal and form sequence A348480. The question arose: How many zeros are needed for each x_n to find a matching number b? In most cases just a few zeros are enough, but some numbers, such as 7, 11, 13 and 37, require more zeros than ones and the corresponding b is called a "long solution". x_n = 101 requires 304 zeros because 101 is a porous number (see A337832).
LINKS
Rüdiger Jehn, A new 200 Euro math puzzle, Youtube video, Sep 17 2021.
Rüdiger Jehn, Porous Numbers, arXiv:2104.02482 [math.GM], 2021.
EXAMPLE
a(2) = 1 because x_2 = 3 and if you add 1 zero to 3 ones you can form b = 1011 for which gcd(b,rev(b)) = digitsum(b) = 3.
PROG
(Python)
A348480 = [1, 11, 4399137296449, 767, 4543829, 302306413101798081695809]
for m in A348480:
print(bin(m)[2:].count('0'))
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Ruediger Jehn, Jan 05 2022
STATUS
approved