A336670 Numbers that have decimal expansion c(1)c(2)...c(n) with distinct digits that satisfy c(1) <> 0, c(1) is the largest digit, and for each i in 1..n there is j in {0, 1} such that c(i) == 2*c(i-1) + j (mod 10) (with c(0): = c(n)).

0, 9, 63, 512, 874, 5012, 7513, 8624, 9874, 62513, 75013, 86374, 98624, 625013, 875124, 986374, 8750124, 9875124, 86251374, 86375124, 87513624, 98750124, 862501374, 863750124, 875013624, 986251374, 986375124, 987513624, 9862501374, 9863750124, 9875013624

Offset: 1

Author

Petros Hadjicostas, Jul 29 2020

Keywords

• nonn
• base
• fini
• full

Comments

This is one of Schuh's examples of a puzzle tree.

Putting the number on a circle and going clockwise, we observe that a 0 is followed by a 1; a 1 is followed by a 2 or 3; a 2 is followed by a 4 or 5; a 3 is followed by a 6 or 7; a 4 is followed by an 8 or 9; a 5 is followed by a 0 or 1; a 6 is followed by a 2 or 3; a 7 is followed by a 4 or 5; an 8 is followed by a 6 or 7; and a 9 is followed by an 8. (These observations assume the number has at least two digits.)

Schuh (pp. 31-35) uses the solution to this problem to solve the "doubles puzzle": find all numbers (with no initial 0) that are written with the same digits as their double (the double of k is 2*k). These numbers are listed in A023086.

The number 0 has been included here for two reasons: (i) we may assume that it satisfies the conditions of the problem vacuously, and (ii) its inclusion allows Schuh to solve the "doubles puzzle". The numbers in A023086 are all permutations of combinations of numbers in this sequence.

Examples

			In all the cases below, the first digit must be the largest and all the digits must be distinct.
9 belongs to this list because c(1) = 9 = c(0) and 9 == 2*9 + 1 (mod 10).
63 belongs to this list because c(1) = 6, c(2) = 3 = c(0), 6 == 2*3 (mod 10), and 3 == 2*6 + 1 (mod 10).
512 belongs to this list because 5 == 2*2 + 1 (mod 10), 1 == 2*5 + 1 (mod 10), and 2 == 2*1 (mod 10).
5012 belongs to this list because 5 == 2*2 + 1 (mod 10), 0 == 2*5 (mod 10), 1 == 2*0 + 1 (mod 10), and 2 == 2*1 (mod 10).
62513 belongs to this list because 6 == 2*3 (mod 10), 2 == 2*6 (mod 10), 5 == 2*2 + 1 (mod 10), 1 = 2*5 + 1 (mod 10), and 3 = 2*1 + 1 (mod 10).
		

References

• Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968, pp. 31-35.

Links

• David A. Corneth and Petros Hadjicostas, PARI program.

Crossrefs

Cf. A023086, A336661.

Programs

PARI

\\ See the Corneth and Hadjicostas link. David A. Corneth, Jul 30 2020