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A146088
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Numbers m with the property that shifting the rightmost digit of m to the left end doubles the number.
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4
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105263157894736842, 157894736842105263, 210526315789473684, 263157894736842105, 315789473684210526, 368421052631578947, 421052631578947368, 473684210526315789, 105263157894736842105263157894736842, 157894736842105263157894736842105263
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| The sequence is infinite, since repeating 105263157894736842 any number of times (e.g. 105263157894736842105263157894736842) gives another number with the same property.
A number N = 10n+m is in the sequence iff 2N = m*10^d+n, where d is the number of digits of n = [N/10]. This is equivalent to 19n = m(10^d-2), i.e. 10^d=2 (mod 19) and n = m(10^d-2)/19, m=2..9 (to ensure that n has d digits). Thus for each d = 18j-1, j=1,2,3... we have exactly 8 solutions which are the j-fold repetition of one among {a(1),...,a(8)}. [From M. F. Hasler (www.univ-ag.fr/~mhasler), May 04 2009]
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FORMULA
| a(n) = ((10^d-2)/19*10+1)m, where m=(n-1)%8+2 is the trailing digit and d=(n+7)\8*18-1 is the number of other digits. [From M. F. Hasler (www.univ-ag.fr/~mhasler), May 04 2009]
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PROG
| Contribution from M. F. Hasler (www.univ-ag.fr/~mhasler), May 04 2009: (Start)
(PARI) A146088(n) = ((10^((n+7)\8*18-1)-2)/19*10+1)*((n-1)%8+2)
/* or a more experimental approach: */ for(d=1, 99, Mod(10, 19)^k-2 & next; for(m=2, 9, print1(", ", m*(10^k-2)/19, m))) \\\\ (End)
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CROSSREFS
| Cf. A092697, A146561, A146569, A146754.
Sequence in context: A115499 A104837 A008923 * A092697 A097717 A128857
Adjacent sequences: A146085 A146086 A146087 * A146089 A146090 A146091
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KEYWORD
| nonn,base
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AUTHOR
| N. J. A. Sloane, based on correspondence from William A. Hoffman III (whoff(AT)robill.com), Apr 10 2009
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EXTENSIONS
| More terms from M. F. Hasler (www.univ-ag.fr/~mhasler), May 04 2009
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