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.

32, 18, 3472, 10993850, 2129428800, 546, 5064320, 105263157894736842, 380, 64609423538, 11424, 1673230, 58774271029236501660840264682112, 67650, 122181448512, 1666, 586081355679130611935159482937228562988190880, 210051282051282, 13571630704729343835960800

Offset: 3

Author

Ray Chandler and R. J. Mathar, Apr 23 2009

Keywords

• base
• nonn

Comments

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

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).

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).

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.

Crossrefs

Cf. A159774.