A336661 -id:A336661 - cp's OEIS frontend

A023087 Numbers k such that k and 3*k are anagrams.

0, 1035, 2475, 10035, 10350, 12375, 14247, 14724, 23751, 24147, 24714, 24750, 24876, 24975, 27585, 28575, 100035, 100350, 102375, 103428, 103500, 107235, 113724, 114237, 123507, 123714, 123750, 123876, 123975, 124137, 128034, 134505, 135045

Offset: 1

David W. Wilson

All terms are divisible by 9. - Eric M. Schmidt, Jul 12 2014
From Petros Hadjicostas, Jul 28 2020: (Start)
This is Schuh's (1968) "treble puzzle" (the treble of k is 3*k). On five pages of his book, he finds the two 4-digit numbers that belong to this sequence (1035 and 2475), the thirteen 5-digit numbers of the sequence and the 104 6-digit numbers of the sequence. Note that if m belongs to the sequence, so does 10*m.
All numbers in this sequence are permutations of numbers that are combinations of numbers from A336661, which is related to another puzzle of Schuh (1968). Before he solved this puzzle, he had to solve the puzzle described in A336661.
For example, 1035 is a permutation of the number 3015 which is a combination of the numbers 301 and 5 that appear in A336661. As another example, note that 12375 and 23751 are both permutations of 31725, which is formed by combining the numbers 31, 72 and 5 from sequence A336661.
If we also admit zeros as initial digits, then we find more solutions to this sequence: 0351, 00351, 01035, 03501, 02475, ... These numbers are also permutations of numbers that can be formed by combining numbers in A336661. (End)

Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968, pp. 25-31.

Zak Seidov and David W. Wilson, Table of n, a(n) for n = 1..10001 (first 3000 terms from Zak Seidov)

Cf. A023086, A023088, A023089, A023090, A023091, A023092, A023093, A336661.

si[n_] := Sort@ IntegerDigits@ n; Flatten@ {0, Table[ Select[ Range[10^d + 8, 4 10^d - 1, 9], si[#] == si[3 #] &], {d, 0, 6}]} (* Giovanni Resta, Mar 20 2017, corrected by Philippe Guglielmetti, Jul 16 2018 *)

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)).

Original entry on oeis.org

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

Petros Hadjicostas, Jul 29 2020

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.

			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).

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

David A. Corneth and Petros Hadjicostas, PARI program.

Cf. A023086, A336661.

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

cp's OEIS Frontend

A023087 Numbers k such that k and 3*k are anagrams.

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