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A147514
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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.
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1
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32, 18, 3472, 10993850, 2129428800, 546, 5064320, 105263157894736842, 380, 64609423538, 11424, 1673230, 58774271029236501660840264682112, 67650, 122181448512, 1666, 586081355679130611935159482937228562988190880, 210051282051282, 13571630704729343835960800
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OFFSET
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3,1
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COMMENTS
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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.
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LINKS
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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