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
LINKS
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
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