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Search: id:A067734
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| A067734 |
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Number of ways writing n as a product of decimal digits of some other number which has no digits equal to 1. |
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+0
10
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| 0, 1, 1, 2, 1, 3, 1, 4, 2, 2, 0, 7, 0, 2, 2, 7, 0, 7, 0, 5, 2, 0, 0, 17, 1, 0, 3, 5, 0, 8, 0, 13, 0, 0, 2, 21, 0, 0, 0, 12, 0, 8, 0, 0, 5, 0, 0, 38, 1, 3, 0, 0, 0, 15, 0, 12, 0, 0, 0, 24, 0, 0, 5, 24, 0, 0, 0, 0, 0, 6, 0, 58, 0, 0, 3, 0, 0, 0, 0, 26, 5, 0, 0, 24, 0, 0, 0, 0, 0, 24, 0, 0, 0, 0, 0, 82, 0
(list; graph; listen)
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OFFSET
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1,4
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COMMENT
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For n=36, this was given as an exercise for children of age 14 years.
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EXAMPLE
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To n=36, 21 other numbers belong without digit 1 of which digit-product equals 36: 49, 66, 94, 229, 236, 263, 292, 326, 334, 343, 362, 433, 623, 632, 922, 2233, 2323, 2332, 3223, 3232, 3322. If digit=1 was permitted then infinite number of solutions would exist, like 1111141111131111 etc. If n has prime divisor larger than 7, i.e. it has a two or more decimal-digit p-divisor, like 11, then no solutions exist at all. Largest solution is a [decimal] number glued from not necessarily distinct prime-factors, like 36=3.2.2.2
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MATHEMATICA
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id1[x_] := IntegerDigits[x]; id2[x_] := DeleteCases[id1[x], 1] f[x_] := Apply[Times, IntegerDigits[x]]; k=0; Do[s=f[n]; If[Equal[s, 36]&&!Greater[Length[id1[n]], Length[id2[n]]], k=k+1; Print[{k, n}]], {n, 1, 3400}]
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CROSSREFS
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Cf. A000073, A001222, A002473, A067734, A068183-A068187, A068189-A068191.
Sequence in context: A056538 A120385 A132460 this_sequence A067004 A117920 A079617
Adjacent sequences: A067731 A067732 A067733 this_sequence A067735 A067736 A067737
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KEYWORD
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base,nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Jan 28 2002
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