A350385 - OEIS

%I #25 Jan 08 2022 13:12:25

%S 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,

%T 10,5,2,3,2,2,2,4,3,10,304,4,3,3,1,3,12,6,124

%N 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.

%C 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).

%H Rüdiger Jehn, <a href="https://youtu.be/w2sC1FSpmPA">A new 200 Euro math puzzle</a>, Youtube video, Sep 17 2021.

%H Rüdiger Jehn, <a href="https://arxiv.org/abs/2201.00710">Long Solutions of Sequence A348480 of the On-Line Encyclopedia of Integer Sequences</a>, arXiv:2201.00710 [math.GM], 2022.

%H Rüdiger Jehn, <a href="https://arxiv.org/abs/2104.02482">Porous Numbers</a>, arXiv:2104.02482 [math.GM], 2021.

%e 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.

%o (Python)

%o A348480 = [1, 11, 4399137296449, 767, 4543829, 302306413101798081695809]

%o for m in A348480:

%o     print(bin(m)[2:].count('0'))

%Y Cf. A045572, A333666, A337832, A348480.

%K nonn,base,more

%O 1,3

%A _Ruediger Jehn_, Jan 05 2022