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A146569 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 4.

%I A146569 #32 May 06 2025 10:59:34
%S A146569 0,102564,128205,153846,179487,205128,230769,102564102564,
%T A146569 128205128205,153846153846,179487179487,205128205128,230769230769,
%U A146569 102564102564102564,128205128205128205,153846153846153846,179487179487179487
%N A146569 Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 4.
%C A146569 a(13) <= 102564102564102564. - _Donovan Johnson_, Jun 06 2009
%C A146569 The condition is equivalent to constraining the numbers to be of the form 10*m+d with a k-digit number m and a nonzero digit d such that 4*(10*m+d) = 10^k * d + m, i.e., 39*m = (10^k - 4)*d. Checking modulo 13, this implies k = 5 (mod 6). Also, m >= 10^(k-1) implies d >= 4. Each such k and d leads to a solution. - _Hagen von Eitzen_, Jun 26 2009
%H A146569 Wikipedia, <a href="https://en.wikipedia.org/wiki/Parasitic_number">Parasitic number</a>.
%F A146569 If n = 6*k + r with 1 <= r <= 6, then a(n) = (10^(6*k) - 1)/(10^6 - 1)*a(r) as well as a(n) = floor((r + 3)/39*10^(6*(k+1))). - _Hagen von Eitzen_, Jun 26 2009
%o A146569 (PARI) a(n) = local(r=(n-1)%6+1,k=(n-r)/6);floor((r+3)/39*10^(6*(k+1))) \\ _Hagen von Eitzen_, Jun 26 2009
%Y A146569 Cf. A146088 (k=2), A146561 (k=3), this sequence (k=4), A146754 (k=5), A291354 (k=6), A291215 (k=7), A291321 (k=8), A291353 (k=9).
%Y A146569 All these are subsequences of A034089 (except for an initial 0 in some of these).
%Y A146569 Cf. A092697, A097717.
%K A146569 nonn,base
%O A146569 0,2
%A A146569 _N. J. A. Sloane_, based on correspondence from William A. Hoffman III (whoff(AT)robill.com), Apr 10 2009
%E A146569 a(7)-a(12) from _Donovan Johnson_, Jun 06 2009
%E A146569 More terms from _Hagen von Eitzen_, Jun 26 2009
%E A146569 a(0) = 0 prefixed for consistency with A146088 by _M. F. Hasler_, May 03 2025