Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #22 Dec 23 2018 03:43:25
%S 102564,128205,153846,179487,205128,230769,102564102564,128205128205,
%T 153846153846,179487179487,205128205128,230769230769,
%U 102564102564102564,128205128205128205,153846153846153846,179487179487179487
%N Numbers m with the property that shifting the rightmost digit of m to the left end multiplies the number by 4.
%C a(13) <= 102564102564102564. - _Donovan Johnson_, Jun 06 2009
%C 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 Wikipedia, <a href="https://en.wikipedia.org/wiki/Parasitic_number">Parasitic number</a>.
%F 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 (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 Cf. A146088 (k=2), A146561 (k=3), this sequence (k=4), A146754 (k=5).
%K nonn,base
%O 1,1
%A _N. J. A. Sloane_, based on correspondence from William A. Hoffman III (whoff(AT)robill.com), Apr 10 2009
%E a(7)-a(12) from _Donovan Johnson_, Jun 06 2009
%E More terms from _Hagen von Eitzen_, Jun 26 2009