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A001232
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Numbers k such that 9*k = (k written backwards), k > 0.
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15
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1089, 10989, 109989, 1099989, 10891089, 10999989, 108901089, 109999989, 1089001089, 1098910989, 1099999989, 10890001089, 10989010989, 10999999989, 108900001089, 108910891089, 109890010989, 109989109989, 109999999989, 1089000001089, 1089109891089
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OFFSET
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1,1
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COMMENTS
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This sequence contains the least n-digit non-palindromic number which is a factor of its reversal. Quotient is always 9. - Lekraj Beedassy, Jun 11 2004. (But it contains many other numbers as well. - N. J. A. Sloane, Jul 02 2013)
Nonzero fixed points of the map which sends x to x - reverse(x) if that is nonnegative, otherwise to x + reverse(x). - Sébastien Dumortier, Nov 05 2006. (Clarified comment, see A124074. - Ray Chandler, Oct 11 2017)
Numbers k such that reversal(k)=reversal(k+reversal(k)). Also numbers k such that reversal(k)=reversal(10*k-reversal(k)). - Farideh Firoozbakht, Jun 11 2010
(1) The first digit of any term must be 1, otherwise multiplication by 9 yields one more digit. For the same reason, no "overflow" must occur from the second to the first digit, so the last digit must be 9.
(2) Continuing the reasoning "from right to left" implies that the trailing nonzero digits must be ...9*89, where 9* means any nonnegative number of consecutive digits 9, preceded by a digit 0, which must be preceded by a digit 1. This implies that the initial and also final digits of any term must be 109*89. We might call a term of this form a "primitive" term. So there is exactly one primitive term b(k) = 11*10^(k-2)-11 with k digits, for all k >= 4.
(3) All terms of the sequence are a "symmetric" concatenation of such b(k)'s, "spaced out" with any number of digits 0, also in a symmetrical way: For any n >= 1, let k = (k[1], ..., k[n]) with k[n+1-j] = k[j] >= 4, and m = (m[1], ..., m[n-1]) (possibly of length 0) with m[n-j] = m[j] >= 0, then N = concat(b(k[j])*10^m[j], 1 <= j < n; k[n]) is a term of the sequence, and this yields all terms of the sequence. (For example, with 1089 we also have 1089{0...0}1089 and 1089,001089,001089, etc.) (End)
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REFERENCES
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H. Camous, Jouer Avec Les Maths, "Cardinaux Réversibles", Section I, Problem 6, pp. 27, 37-38; Les Editions d'Organisation, Paris, 1984.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, under #1093.
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LINKS
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N. J. A. Sloane, 2178 And All That, arXiv:1307.0453 [math.NT], 2013; Fib. Quart., 52 (2014), 99-120.
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FORMULA
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Theorem: Terms in this sequence have the form 99*m, where the decimal representation of m contains only 1's and 0's, is palindromic and contains no singleton 1's or 0's. Hence contains Fib(floor(k/2)-1) k-digit terms, k >= 4. - David W. Wilson, Dec 15 1997
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EXAMPLE
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1089*9 = 9801.
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MATHEMATICA
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Rest@Select[FromDigits /@ Tuples[{0, 99}, 11], IntegerDigits[9*#] == Reverse@IntegerDigits[#] &] (* Arkadiusz Wesolowski, Aug 14 2012 *)
okQ[t_]:=t==Reverse[t]&&First[t]!=0&&Min[Length/@Split[t]]>1; 99#&/@Flatten[Table[FromDigits/@Select[Tuples[{0, 1}, n], okQ], {n, 20}]] (* Harvey P. Dale, Jul 03 2013 *)
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PROG
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(PARI) isok(n) = 9*n == eval(concat(Vecrev(Str(n)))); \\ Michel Marcus, Feb 21 2015
(PARI) {A001232_row(n, L(v, s=0)=for(i=1, #v, s*=10^v[i]; i%2 && s+=10^v[i]\900); s)=if(n<4, [], L, Set(apply(L, self()(n, 0)))*99, L=List([[n]]); for(k=4, n\2, listput(L, [k, n-2*k, k]); for(p=0, n\2-k, foreach(self()(n-(k+p)*2, 0), M, listput(L, concat([[k, p], M, [p, k]]))))); L)} \\ List of n-digit terms. - M. F. Hasler, Oct 04 2022
concat(apply(A001232_row, [1..14]))
(Python)
def A001232_row(n, r=11): # list of n-digit terms
L = [] if n<4 else [[n]]
for L1 in range(4, n//2+1):
L.append([L1, n-2*L1, L1])
L.extend([L1, L2]+M+[L2, L1] for L2 in range(n//2-1-L1)
for M in A001232_row(n-(L1+L2)*2, 0))
if not r: return L
def f(L, s=0):
for k, L in enumerate(L):
s *= 10**L
if not k%2: s += 10**(L-2)-1
return r*s
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CROSSREFS
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KEYWORD
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base,nonn,nice
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AUTHOR
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EXTENSIONS
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a(1..10^4) in b-file double-checked with independent code by M. F. Hasler, Oct 04 2022
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STATUS
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approved
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