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A085645
Smallest number having n divisors ending with 1 or 9.
2
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
OFFSET
1,2
COMMENTS
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
EXAMPLE
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
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Sequence in context: A292309 A337236 A285456 * A299579 A344526 A269641
KEYWORD
base,nonn
AUTHOR
Lekraj Beedassy, Jul 11 2003
EXTENSIONS
Corrected and extended by Harvey P. Dale, Jul 18 2003
Terms a(17) and beyond from Andrew Howroyd, Jan 07 2020
STATUS
approved