A331029 -id:A331029 - cp's OEIS frontend

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

Lekraj Beedassy, Jul 11 2003

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

			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

A. Stenger, An Excess of Divisors
Wikipedia, Table of divisors

Cf. A331029, A331082.

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 *)

a(n)={forstep(k=1, oo, 2, if(sumdiv(k, d, abs(d%10-5)==4) == n, return(k)))} \\ Andrew Howroyd, Jan 07 2020

\\ 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

Corrected and extended by Harvey P. Dale, Jul 18 2003

Terms a(17) and beyond from Andrew Howroyd, Jan 07 2020

A331082 Smallest number having exactly n divisors ending with 3 or 7.

Original entry on oeis.org

3, 21, 63, 189, 567, 693, 3969, 2079, 5733, 6237, 413343, 9009, 74529, 43659, 51597, 27027, 1750329, 63063, 26040609, 81081, 464373, 670761, 219667417263, 153153, 2528253, 819819, 693693, 729729, 160137547184727, 567567, 6036849, 459459, 37614213, 19253619

Offset: 1

Andrew Howroyd, Jan 08 2020

A term cannot be a multiple of either 2 or 5 since removing these factors does not alter the number of divisors ending with 3 or 7.

All terms must be in A331029.

			a(1) = 3: divisors of 3 are 1, 3*. (* marks divisors ending in 3 or 7.)
a(2) = 21: divisors of 21 are 1, 3*, 7*, 21.
a(3) = 63: divisors of 63 are 1, 3*, 7*, 9, 21, 63*.
a(4) = 189: divisors of 189 are 1, 3*, 7*, 9, 21, 27*, 63*, 189.

Cf. A085645, A331029.

a(n)={forstep(k=1, oo, 2, if(sumdiv(k, d, abs(d%10-5)==2) == n, return(k)))}

\\ 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)==2)==n, return(k))))}

from sympy import divisors
def count37(iterable): return sum(i%10 in [3, 7] for i in iterable)
def a(n):
  m = 2
  while count37(divisors(m)) != n: m += 1
  return m
print([a(n) for n in range(1, 17)]) # Michael S. Branicky, Mar 21 2021

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A085645 Smallest number having n divisors ending with 1 or 9.

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A331082 Smallest number having exactly n divisors ending with 3 or 7.

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