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A067206
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Numbers n such that the digits of n end in phi(n).
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4
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1, 1320, 1640, 1768, 1996, 2640, 3960, 13200, 16400, 19984, 19996, 26400, 39600, 132000, 164000, 199996, 264000, 396000, 1320000, 1640000, 1999936, 2640000, 3960000, 13200000, 16400000, 16666240, 17999488, 18515584, 19999984, 19999996
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OFFSET
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1,2
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COMMENTS
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"(1). If n is in the sequence and 10 divides n then for each natural number k, n*10^k is in the sequence. So since 1320, 1640, 2640, 3960 & 16666240 are in the sequence, for each natural number k, 132*10^k, 164*10^k, 264*10^k, 396*10^k & 1666624*10^k are in the sequence. Hence the sequence is infinite.
"(2). If 5*10^k-1 is prime then 4*(5*10^k-1) is in the sequence. So 4*A093945 is a subsequence of this sequence.
"(3). If p=125*10^k-1 is prime then 16*p is in the sequence. For k = 1, 4, 5, 8, 13, 19, 25, 26, 76, 88, 167, 290, 389, ... p is prime.
"(4). If p=3125*10^k-1 is prime then 64*p is in the sequence. For k = 1, 3, 9, 33, 121, 223, 357, 363, 447, ... p is prime." (End)
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REFERENCES
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Pickover, C. "Wonders of Numbers". Oxford Univ. Press, 2001.
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LINKS
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C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
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EXAMPLE
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The digits of 1768 end in phi(1768) = 768, so 1768 is a term of the sequence.
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MATHEMATICA
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(*returns true if a ends in b, false o.w.*) f[a_, b_] := Module[{c, d, e, g, h, i, r}, r = False; c = ToString[a]; d = ToString[b]; e = StringLength[c]; g = StringPosition[c, d]; h = Length[g]; If[h > 0, i = g[[h]]; If[i[[2]] == e, r = True]]; r]; Select[Range[10^5], f[ #, EulerPhi[ # ]] &]
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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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EXTENSIONS
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STATUS
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approved
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