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.

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

Table of n, a(n) for n=1..49.

Rüdiger Jehn, A new 200 Euro math puzzle, Youtube video, Sep 17 2021.

Rüdiger Jehn, Long Solutions of Sequence A348480 of the On-Line Encyclopedia of Integer Sequences, arXiv:2201.00710 [math.GM], 2022.

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

Cf. A045572, A333666, A337832, A348480.

Sequence in context: A203277 A156645 A330084 * A374498 A371259 A037935

Adjacent sequences: A350382 A350383 A350384 * A350386 A350387 A350388

Keyword

nonn,base,more

Author

Ruediger Jehn, Jan 05 2022

Status

approved