A085645 Smallest number having n divisors ending with 1 or 9.
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
Examples
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
Links
- A. Stenger, An Excess of Divisors
- Wikipedia, Table of divisors
Programs
-
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 *) -
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
Extensions
Corrected and extended by Harvey P. Dale, Jul 18 2003
Terms a(17) and beyond from Andrew Howroyd, Jan 07 2020
Comments