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Numbers n which don't have a preimage for A216556, i.e., such that A216587(n)=-1.
+20
3
0, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148
COMMENTS
Concretely, numbers having "20", "30", ..., "90" or "00" as a substring, or starting with "11", "12", ... "19", and the number 0.
Note that A216557 does not necessarily yield 0 for these numbers, e.g., 10 has no preimage for A216556, but 210 does have, and 10 -> 21 -> ... -> 98 -> 109.
FORMULA
A216587(n)=-1 if and only if n is in this sequence.
PROG
(PARI) is_ A216589(n)={n || return(1); n=Vec(Str(n)); n[1]<"2" & 1<#n & n[2]>"0" & return(1); for(i=2, #n, n[i]=="0" || next; n[i-1]=="1" || return(1))}
Concatenate decimal digits of n, each increased by 1.
+10
6
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 22, 23, 24, 25, 26, 27, 28, 29, 210, 31, 32, 33, 34, 35, 36, 37, 38, 39, 310, 41, 42, 43, 44, 45, 46, 47, 48, 49, 410, 51, 52, 53, 54, 55, 56, 57, 58, 59, 510, 61, 62, 63, 64, 65, 66, 67, 68, 69, 610, 71, 72, 73, 74, 75, 76, 77
COMMENTS
In all of the terms, a digit '0' can only occur preceded by a digit '1', and an initial digit '1' can only appear followed by a '0'.
Sequence A216589 lists the complement of the range of this map.
EXAMPLE
a(19) = concat(1+1,9+1) = 210.
MATHEMATICA
Array[FromDigits@ Flatten[IntegerDigits[#] + 1 /. 10 -> {1, 0}] &, 91, 0] (* Michael De Vlieger, Jan 04 2020 *)
PROG
(PARI) A216556(n)={my(t=1); until(n<t*=10, (n+=t)\t%10||n+=(n-t)\(t*=10)*9*t); n}
Number of iterations of A216556 until the initial value n appears as a substring of the iterate; 0 if this will never happen.
+10
5
10, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 19, 28, 37, 46, 55, 64, 73, 82, 90, 0, 9, 19, 28, 37, 46, 55, 64, 73, 82, 0, 89, 9, 19, 28, 37, 46, 55, 64, 73, 0, 79, 89, 9, 19, 28, 37, 46, 55, 64, 0, 69, 79, 89, 9, 19, 28, 37, 46, 55, 0, 59, 69, 79, 89, 9, 19, 28, 37, 46, 0, 49, 59, 69, 79, 89, 9, 19, 28, 37, 0, 39, 49, 59, 69, 79, 89, 9, 19, 28, 0, 29, 39, 49, 59, 69, 79
COMMENTS
Can someone prove (and maybe strengthen) the following conjecture: a(n) = 0 whenever A216587(m) = -1 for all m obtained by concatenating any digit to the left and any digit to the right of n. The nonzero a(n) take only 18 different values: (9, 10, 19, 28, 29, 37, 39, 46, 49, 55, 59, 64, 69, 73, 79, 82, 89, 90). For n < 10^12 the corresponding counts are (108, 75, 829, 388, 306, 326, 302, 289, 291, 277, 303, 265, 315, 254, 327, 245, 339, 2). Specifically a(19) = a(210) = 90.
Nonzero terms are becoming increasingly sparse. For k = 1..12 the number of nonzero a(n) for n < 10^k is (10, 92, 247, 489, 797, 1194, 1678, 2236, 2860, 3565, 4359, 5421). (End)
FORMULA
a(n)=0 for all numbers having "20", "30", ..., "90" or "00" or "111", "222", ... "999" as a substring.
EXAMPLE
a(211) = 9 since under the action of A216556, 211 -> 322 -> 433 -> 544 -> 655 -> 766 -> 877 -> 988 -> 1099 -> 211010, which contains the substring 211. a(111) = 0 since if some number has "111" as its substring, then its preimage for A216556 (cf. A216587) contains at least the substring "00" (e.g., A216587(21110) = 1009), and has in turn no preimage under A216556. Therefore 111 cannot occur as a substring in the orbit of any number under A216556.
PROG
(PARI) A216557(n, L=9e9, f)={my(s=Mod(n, 10^#Str(n)), t=n); n && until(20>t\=10, t%1000%111||return; t%10 || t%100==10 || return); for(i=1, L, t=n= A216556(n); until(!t\=10, s==t && return(i))); f} \\ 3rd (optional) argument f allows to specify a return value (e.g., f=[] or -1) in case no result is found within the limit of L iterations. If the zero result is deduced from the initial value (cf. FORMULA) the function returns an empty result (which also evaluates to 0). [PARI syntax updated Jan 02 2020]
Indices n for which A216557(n)=0, i.e., n does not reappear as substring in its orbit under A216556.
+10
1
20, 30, 40, 50, 60, 70, 80, 90, 100, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 155, 156, 157, 158, 159
COMMENTS
E. Angelini asked on the SeqFan list whether n=127 is in this sequence. The proof of the affirmative answer was given in the second reply to the post. (But "Lemma 2" should read: There is no n in the range of A216556 such that ...". Indeed, 2016,...,9016 also yield 127 as substring, but are not in A216556.) The same reasoning can be applied to terms 122,...,129.
EXAMPLE
The number 0 is not in this sequence, because repeated application of A216556 yields 0 > 1 > ... > 9 > 10 which contains 0 as substring, after A216557(0)=10 iterations. a(1)=20 is the least number which does not occur as a substring in its orbit under A216556. This is the case since no term in (the range of) A216556 may have "20" as a substring. As explained in A216556 (and obvious from the definition), the digit "0" can only occur preceded by the digit "1". This also explains a(2)=30 through a(9)=100. In A216557 it is explained why a(10)=111 and all numbers having this or 222,...,999 as substring are in this list. To see why a(11)=112 and the following terms are in this list, observe that 112 itself has no preimage under A216556 (cf. A216587), and consider all possibilities for "patterns" in which it might occur: Obviously not followed by a digit 0, but possibly by a digit among 1,...,9; and maybe preceded by a digit among 2,...,9. However, all these cases lead to a preimage which has "00" as substring, and is therefore not in the range of A216556.
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