Staircase-numbers

I don’t know exactly what a metadrome is – but I know what « numbers with distinct digits in ascending order » are. Those numbers remind me a staircase that goes up and will be named here « up-staircase numbers ».

Let’s build S starting from a(1) = 1 such that [a(n)*a(n+1)] always produces an « up-staircase number ». As usual we want S to be the lexicographically earliest sequence of distinct terms with this property. I hope S begins like this (we decide that the first « up-staircase number » in base 10 is 12):

S = 1, 12, 2, 6, 3, 4, 7, 5, 9, 14, 17, 8, 16, 23, 15, 83, 43, 29, 51,...

Check:
1*12=12
12*2=24
2*6=12
6*3=18
3*4=12
4*7=28
7*5=35
5*9=45
9*14=126
14*17=238
17*8=136
8*16=128
16*23=368
23*15=345
15*83=1245
83*43=3569
43*29=1247
29*51=1479
…

I guess this sequence is finite and will end quite soon.

If we allow any amount of backtracking, what could be the longest such sequence starting with a(1) = 1? 

Same question for the sequence T such that [a(n)*a(n+1)] always forms a « down-staircase number » (that is a « number with distinct digits in descending order », like 320).

________________________

July 14th update:

Allan Wechsler on Math-Fun:

> Next element is 46, I think: 51*46 = 2346.

This sequence is definitely finite, because there are only a finite number of numbers with strictly increasing digits, and each such number has only a finite number of dissections into two factors. I think it becomes infinite if we allow repeated digits in the products, but I am not sure. My instinct is always to investigate logically prior sequences first. In this case, what about A(n) being the smallest k > 1 such that nk is stairstep?

Josh Jordan on Math-Fun:

> If I understand right, the largest possible up-staircase number is 123456789. In that case, the first 21 terms of S are [1, 12, 2, 6, 3, 4, 7, 5, 9, 14, 17, 8, 16, 23, 15, 83, 43, 29, 51, 46, 293], and the sequence ends there, because 293*x* is not an up-staircase number for 2 ≤ *x* < 421355, and 293 * 421355 = 123457015, which is larger than the largest possible up-staircase number. Staircase sequence calculator in Go: https://go.dev/play/p/JcVDRdcE8cg



Commentaires

  1. solal17 juillet 2022 à 07:37

    Variant "double staircase"
    The digits of the product abcdef=[a(n)*a(n+1)] must be :
    - either with the form /\ such that a < b < c < d > e > f, for example 135851
    - or with the form / ‾ \ such that a< b < c <= d > e > f, for example 137761
    There must be at least one digit during the increasing phase and one digit during the decreasing phase of the product's digits..
    So, S is a suitable sequence :
    S= [10, 12, 11, 13, 14, 18, 9, 17, 16, 15, 8, 19, 20, 6, 22, 7, 23, 21, 32, 5, 24, 28, 44, 3, 40, 4, 30, 26, 49, 29, 27, 25, 50, 33, 38, 34, 41, 31, 45, 35, 36, 47, 42, 46, 39, 48, 52, 51, 54, 53, 37, 43, 55, 62, 56, 64, 60, 2, 65, 61, 63, 57, 66, 59, 67, 70, 68, 69, 84, 58, 79, 86, 144, 88, 78, 73, 77, 74, 80, 71, 81, 72, 83, 108, 115, 110, 112, 111, 113, 114, 118, 105, 75, 90, 76, 89, 139, 97, 82, 85, 92, 135, 94, 95, 130, 96, 129, 98, 140, 91, 136, 99, 125, 102, 121, 103, 87, 142, 119, 104, 120, 106, 122, 131, 145, 93, 134, 126, 107, 128, 109, 117, 116, 127, 123, 137, 172, 143, 164, 151, 170, 138, 171, 150, 157, 152, 155, 154, 153, 124, 160, 147, 163, 146, 162, 148, 159, 149, 158, 175, 141, 133, 177, 196, 132, 178, 194, 180, 161, 167, 220, 156, 165, 169, 207, 168, 213, 187, 185, 188, 184, 189, 183, 190, 182, 191, 181, 192, 203, 225, 262, 179, 199, 230, 197, 176, 195, 198, 235, 204, 224]
    Example : a(1) * a(2) = 120 –> 1 < 2 > 0
    a(113) * a(114) = 125 * 102 = 12750 -> 1 < 2 < 7 > 5 >< 0
    a(124) * a(125) = 131 * 145 = 18995 -> 1 < 8 < 9 = 9 >5

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  2. solal17 juillet 2022 à 09:49

    Variant "descending staircase"
    The digits of the product abcdef=[a(n)*a(n+1)] must be :
    such that a > b > c > d > e > f, for example 754210
    The sequence S starting with 9 is longer than those starting with 1, 2, 3, 4, 5, 6, 7 or 8.
    As the increasing staircase sequence, the decreasing staircase sequence seems to be finite.
    S=[9, 6, 5, 2, 10, 3, 7, 12, 8, 4, 13, 40, 16, 20, 21, 30, 14, 15, 28, 19, 39, 25, 26, 29, 34, 215, 294, 328, 2669, 37, 23, 27, 32, 135, 46, 142, 60, 72, 106, 70, 76, 115, 54, 18, 24, 35, 152, 635, 138, 45, 17, 43, 147, 430, 178, 49244, 1780]

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  3. MFH18 juillet 2022 à 01:06

    Yes, any sequence of distinct (strictly) "up" or "down" or "up-then-down" staircase numbers is obviously finite because they are all smaller than or equal to 12345678987654321. If we allow repeated digits (e.g. 1112333) there are infinitely many such numbers, but the sequence might still be finite if it reaches a point where no successor of this form can be found for the last a(n). Then one might impose the additional restriction that a(n) must be chosen so that the sequence can be extended infinitely. But then it becomes more complicated to compute the terms - backtracking becomes involved and one might never know whether a term (after a(1)=1) is correct -- unless we can find an "upper bound" which does allow an infinite extension (even if that is not necessarily the "smallest" one).

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  4. solal18 juillet 2022 à 09:53

    Variant "double staircase b"
    The digits of the product abcdef=[a(n)*a(n+1)] must be :
    - either with the form \/ such that a > b > c > d < e < f, for example 873125
    - or with the form \_/ such that a > b > c >= d < e < f, for example 975548
    There must be at least one digit during the increasing phase and one digit during the decreasing phase of the product's digits..
    So, S is a suitable sequence :
    S= [11, 19, 16, 13, 8, 26, 4, 27, 12, 9, 23, 18, 6, 17, 24, 21, 5, 41, 15, 7, 29, 14, 22, 28, 31, 33, 25, 37, 55, 39, 52, 2, 51, 63, 32, 64, 49, 42, 48, 66, 46, 44, 47, 67, 3, 34, 62, 68, 59, 36, 56, 38, 53, 76, 43, 73, 58, 87, 35, 61, 69, 45, 141, 54, 57, 71, 75, 95, 77, 81, 78, 103, 72, 74, 82, 79, 65, 93, 99, 83, 86, 88, 97, 84, 102, 89, 104, 92, 94, 96, 91, 106, 191, 158, 133, 153, 139, 231, 179, 182, 229, 184, 174, 117, 172, 119, 171, 118, 178, 236, 138, 146, 144, 149, 203, 202, 159, 195, 273, 193, 109, 294, 107, 197, 211, 98, 311, 164, 257, 125, 161, 188, 112, 269, 162, 319, 131, 157, 134, 151, 207, 194, 222, 239, 173, 123, 245, 221, 145, 283, 143, 213, 142, 148, 136, 228, 137, 147, 205, 249, 121, 339, 122, 247, 127, 252, 124, 244, 169, 181, 177, 114, 272, 199, 152, 204, 201, 156, 129, 163, 253, 206, 258, 196, 266, 241, 167, 301, 168, 186, 113, 85, 111, 289, 105, 287, 223, 189, 217, 185, 233]

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  5. solal18 juillet 2022 à 22:24

    I tried to search the longest sequence by changing its start point. (from 1 to 1000)

    So, for the up staircase sequence, I found a 38 terms sequence :
    38 : [327, 14, 2, 6, 3, 4, 7, 5, 9, 15, 23, 16, 8, 17, 21, 18, 13, 12, 29, 43, 83, 149, 232, 58, 41, 87, 27, 47, 31, 19, 24, 52, 28, 56, 26, 48, 72, 173]

    And for the down staircase sequence, I found a 66 terms sequence :
    66 : [478, 195, 5, 2, 10, 3, 7, 6, 9, 8, 4, 13, 40, 16, 20, 21, 30, 14, 15, 28, 19, 39, 25, 26, 29, 34, 215, 294, 328, 2669, 37, 23, 27, 32, 135, 46, 142, 60, 12, 35, 18, 24, 36, 120, 61, 140, 38, 230, 41, 181, 410, 211, 31, 210, 47, 160, 52, 145, 58, 17, 43, 147, 430, 178, 49244, 1780]

    "up-then-down" staircase and "down-then-up" staircase sequences, although finite as MFH argues, are much longer.

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