A147514 - OEIS

A147514

Least number m, written in base 10, such that m/2 is obtained merely by shifting the leftmost digit of m to the right end, and 2m by shifting the rightmost digit of m to the left end, digits defined in base n.

%I #4 Mar 31 2012 08:15:15

%S 32,18,3472,10993850,2129428800,546,5064320,105263157894736842,380,

%T 64609423538,11424,1673230,58774271029236501660840264682112,67650,

%U 122181448512,1666,586081355679130611935159482937228562988190880,210051282051282,13571630704729343835960800

%N Least number m, written in base 10, such that m/2 is obtained merely by shifting the leftmost digit of m to the right end, and 2m by shifting the rightmost digit of m to the left end, digits defined in base n.

%C Serves as an extension to A159774, which misses proper representation for solutions beyond base 12.

%C Algorithm: write m in base b with LSB d_0, k middle digits d_m, and MSB digit d_e as m=d_0+d_m*b+d_e*b^(k+1).

%C Demand m/2 = d_e+d_0*b_d_m*b^2 and 2*m=d_m+d_e*b^k+d_0*b^(k+1). Mix these to obtain m*(2b-1)=2*d_e*(b^(k+2)-1).

%C Loop over (outer loop) k=0,1,2... and (inner loop d_e=0.. b-1 to obtain integer m to be checked against the condition.

%Y Cf. A159774.

%K base,nonn

%O 3,1

%A _Ray Chandler_ and _R. J. Mathar_, Apr 23 2009