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

1,2

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

Table of n, a(n) for n=1..35.

A. Stenger, An Excess of Divisors

Wikipedia, Table of divisors

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

Cf. A331029, A331082.

Sequence in context: A292309 A337236 A285456 * A299579 A344526 A269641

Adjacent sequences: A085642 A085643 A085644 * A085646 A085647 A085648

Keyword

base,nonn,changed

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