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A085645
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Smallest number having n divisors ending with 1 or 9.
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2
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1, 9, 63, 99, 441, 693, 5103, 1881, 5733, 4851, 35721, 9009, 194481, 56133, 51597, 27027, 2893401, 63063, 2711943423, 81081, 464373, 392931, 670761, 153153, 2528253, 2139291, 693693, 729729, 18983603961, 567567, 1441237924662543, 459459, 4322241, 31827411, 22754277
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OFFSET
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1,2
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COMMENTS
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All a(n) must be in A331029. Also, a(n) cannot be a multiple of either 2 or 5 since removing these factors does not alter the number of divisors ending with 1 or 9. - Andrew Howroyd, Jan 07 2020
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LINKS
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EXAMPLE
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The divisors of 63 are 1, 3, 7, 9, 21 and 63. Three of them end either in 1 or 9. No smaller number satisfies this condition, so a(3) = 63
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MATHEMATICA
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tbl=Table[Length[Select[IntegerDigits/@Divisors[i], Last[ # ]==1||Last[ # ]==9&]], {i, 50000}]; Table[First[Position[tbl, i]], {i, 12}]//Flatten
Module[{nn=3*10^6, lst}, lst=Table[{n, Count[Divisors[n], _?(Mod[#, 10]==1||Mod[#, 10] == 9&)]}, {n, nn}]; Table[SelectFirst[lst, #[[2]]==k&], {k, 18}]][[;; , 1]] (* The program generates the first 18 terms of the sequence. *) (* Harvey P. Dale, Apr 21 2024 *)
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PROG
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(PARI) a(n)={forstep(k=1, oo, 2, if(sumdiv(k, d, abs(d%10-5)==4) == n, return(k)))} \\ Andrew Howroyd, Jan 07 2020
(PARI) \\ faster program: needs lista331029 defined in A331029.
a(n)={my(lim=1); while(1, lim*=10; my(S=lista331029(lim)); for(i=1, #S, my(k=S[i]); if(sumdiv(k, d, abs(d%10-5)==4)==n, return(k))))} \\ Andrew Howroyd, Jan 07 2020
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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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