Displaying 1-10 of 11 results found.
Smallest integer with exactly n divisors that are repunits.
+10
10
1, 11, 1111, 111111, 11222211, 111111111111, 1111222222221111, 11223344555544332211, 112244668899998866442211, 112357025813567765307519653211, 112244781144780011109977441077442211, 113491945266228931047738906599340328084311, 113378566812907968345622215431647587096554773311
COMMENTS
Previous name was: Integers whose number of divisors that are repunits sets a new record. From a(1) up to a(18), the terms of these two sequences are exactly the same.
Repunit terms are: R_1, R_2, R_4, R_6, R_12, ... where R_m is A002275(m). It appears that palindromes occur for n = 1 to 9 only. (End)
The indices of the n repunits that divide a(n) are given by the n-th row of A356184. - Bernard Schott, Sep 13 2022
EXAMPLE
111111 has 4 divisors that are repunits: {1, 11, 111, 111111}; also, 111111 is the smallest integer that has at least 4 repunit divisors, hence 111111 is a term.
The 13 repunit divisors of a(13) are R_1, R_2, R_3, R_4, R_5, R_6, R_7, R_8, R_9, R_10, R_12, R_14 and R_18.
MATHEMATICA
repQ[n_] := Union @ IntegerDigits[n] == {1}; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = DivisorSum[n, 1 &, repQ[#] &]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[4, 10^7] (* Amiram Eldar, Sep 05 2022 *)
PROG
(PARI) upto(n) = { l = List(); ulim = n; res = []; reps = vector(logint(n, 10)-1, i, 10^(i+1)\9); for(i = 0, #reps, process(1, i); ); listsort(l, 1); r = 0; for(i = 1, #l, c = f(l[i]); if(c > #res, res = concat(res, vector(c - #res, j, oo)); ); res[c] = min(res[c], l[i]) ); res }
process(n, i) = { if(n <=ulim, listput(l, n); for(j = i + 1, #reps, c = lcm(n, reps[j]); process(c, j) ) ) }
f(n) = my(u = logint(n, 10) + 2); 1 + sum(i = 1, u, n % (10^(i+1)\9) == 0) \\ David A. Corneth, Jan 12 2021, Jan 17 2022, Sep 12 2022
CROSSREFS
Similar, but with divisors that are: A087997 (palindromes), A355699 (repdigits).
a(n) is the smallest number that has exactly n repdigit divisors.
+10
6
1, 2, 4, 6, 12, 24, 72, 66, 666, 132, 1332, 264, 2664, 792, 13320, 3960, 14652, 26664, 48840, 29304, 79992, 341880, 146520, 399960, 1333332, 1025640, 2799720, 8879112, 2666664, 18666648, 7999992, 44395560, 13333320, 93333240, 39999960, 279999720, 269333064
EXAMPLE
72 has 12 divisors: {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72}, only {1, 2, 3, 4, 6, 8, 9} are repdigits; no positive integer smaller than 72 has seven repdigit divisors, hence a(7) = 72.
MATHEMATICA
f[n_] := DivisorSum[n, 1 &, Length[Union[IntegerDigits[#]]] == 1 &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[24, 10^6] (* Amiram Eldar, Jul 15 2022 *)
PROG
(PARI) isrep(n) = 1==#Set(digits(n)); \\ A010785a(n) = my(k=1); while (sumdiv(k, d, isrep(d)) != n, k++); k; \\ Michel Marcus, Jul 15 2022 (Python)
from sympy import divisors
from itertools import count, islice
def c(n): return len(set(str(n))) == 1
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
n, adict = 1, dict()
for k in count(1):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
a(n) is the smallest integer that has exactly n divisors from A333369.
+10
5
1, 3, 9, 15, 45, 105, 195, 315, 945, 900, 1575, 2100, 3900, 6825, 11655, 10500, 6300, 18900, 25200, 35100, 27300, 31500, 44100, 94500, 157500, 107100, 81900, 233100, 220500, 598500, 245700, 333900, 409500, 491400, 900900, 573300, 600600, 1228500, 1669500, 1965600
EXAMPLE
15 has 4 divisors: {1, 3, 5, 15} all of which are in A333369 integers, and no smaller number has this property, hence a(4) = 15.
MATHEMATICA
q[n_] := AllTrue[Tally @ IntegerDigits[n], EvenQ[Plus @@ #] &]; f[n_] := DivisorSum[n, 1 &, q[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[40, 10^7] (* Amiram Eldar, Jul 17 2022 *)
PROG
(PARI) issimber(m) = my(d=digits(m), s=Set(d)); for (i=1, #s, if (#select(x->(x==s[i]), d) % 2 != (s[i] % 2), return (0))); return (1); \\ A333369a(n) = my(k=1); while (sumdiv(k, d, issimber(d)) != n, k++); k; \\ Michel Marcus, Jul 18 2022 (Python)
from sympy import divisors
from itertools import count, islice
def c(n): s = str(n); return all(s.count(d)%2 == int(d)%2 for d in set(s))
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
n, adict = 1, dict()
for k in count(1):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
a(n) is the smallest number that has exactly n odious divisors ( A000069).
+10
5
1, 2, 4, 8, 16, 28, 64, 56, 84, 112, 1024, 168, 4096, 448, 336, 728, 36309, 672, 57057, 1456, 1344, 7168, 105105, 2184, 6384, 24150, 5376, 5208, 405405, 4368, 389025, 11648, 20020, 72618, 10416, 8736, 927675, 114114, 48300, 24024, 855855, 17472, 1426425, 40040
COMMENTS
a(n) <= 2^(n-1) with equality for n = 1, 2, 3, 4, 5, 7, 11, 13 up to a(44).
EXAMPLE
a(6) = 28 since 28 has 6 divisors {1, 2, 4, 7, 14, 28} that have all an odd number of 1's in their binary expansion: 1, 10, 100, 111, 1110 and 11100; also, no positive integer smaller than 28 has six divisors that are odious.
MATHEMATICA
f[n_] := DivisorSum[n, 1 &, OddQ[DigitCount[#, 2, 1]] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[20, 10^6] (* Amiram Eldar, Jul 21 2022 *)
PROG
(PARI) isod(n) = hammingweight(n) % 2; \\ A000069a(n) = my(k=1); while (sumdiv(k, d, isod(d)) != n, k++); k; \\ Michel Marcus, Jul 22 2022 (Python)
from sympy import divisors
from itertools import count, islice
def c(n): return bin(n).count("1")&1
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
n, adict = 1, dict()
for k in count(1):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
a(n) is the smallest number that has exactly n evil divisors ( A001969).
+10
4
1, 3, 6, 12, 18, 45, 30, 135, 72, 60, 90, 765, 120, 1575, 270, 180, 600, 3465, 480, 13545, 360, 540, 1530, 10395, 1260, 720, 3150, 1980, 1080, 49725, 1440, 45045, 2520, 3060, 6930, 2160, 3780, 58905, 27090, 6300, 5040, 184275, 4320, 135135, 6120, 7920, 20790, 329175, 7560, 8640
EXAMPLE
a(4) = 18 since 18 has six divisors: {1, 2, 3, 6, 9, 18} of which four {3, 6, 9, 18} have an even number of 1's in their binary expansion: 11, 110, 1001 and 10010 respectively; also, no positive integer smaller than 18 has exactly four divisors that are evil.
MAPLE
# output in unsorted b-file style
A356019_list := [seq(0, i=1..1000)] ;
for n from 1 do
if op(evd+1, A356019_list) <= 0 then printf("%d %d\n", evd, n) ;
end if;
end if;
MATHEMATICA
f[n_] := DivisorSum[n, 1 &, EvenQ[DigitCount[#, 2, 1]] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[50, 10^6] (* Amiram Eldar, Jul 23 2022 *)
PROG
(Python)
from sympy import divisors
from itertools import count, islice
def c(n): return bin(n).count("1")&1 == 0
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
n, adict = 0, dict()
for k in count(1):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
a(n) is the smallest number with exactly n divisors that are Brazilian.
+10
3
1, 7, 14, 24, 40, 48, 60, 84, 140, 144, 120, 168, 252, 700, 240, 336, 560, 360, 420, 672, 1120, 2304, 960, 720, 1008, 1080, 840, 2184, 1800, 1260, 2016, 5376, 8960, 2160, 1680, 2880, 4032, 3600, 7056, 19600, 3960, 2520, 3360, 6480, 9072, 9900, 6300, 11520, 16128
COMMENTS
Primes can be partitioned into Brazilian primes and non-Brazilian primes. If two distinct primes each larger than 11 are in the same category then the larger one has a multiplicity that is smaller than or equal to that of the smaller prime. - David A. Corneth, Jan 24 2021
EXAMPLE
Of the eight divisors of 24, three are Brazilian numbers: 8, 12 and 24, and there is no smaller number with three Brazilian divisors, hence a(3) = 24.
MATHEMATICA
brazQ[n_] := Module[{b = 2, found = False}, While[b < n - 1 && Length[Union[IntegerDigits[n, b]]] > 1, b++]; b < n - 1]; d[n_] := DivisorSum[n, 1 &, brazQ[#] &]; m = 30; s = Table[0, {m}]; c = 0; n = 1; While[c < m, i = d[n]; If[i < m && s[[i + 1]] == 0, c++; s[[i + 1]] = n]; n++]; s (* Amiram Eldar, Jan 21 2021 *)
PROG
(PARI) isokb(n) = for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d), return(1))); \\ A125134isok(k, n) = sumdiv(k, d, isokb(d)) == n;
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Jan 23 2021
a(n) is the smallest number that has exactly n nonpalindromic divisors ( A029742).
+10
3
1, 10, 20, 30, 48, 72, 60, 140, 144, 120, 210, 180, 300, 240, 560, 504, 360, 420, 780, 1764, 900, 960, 720, 1200, 840, 1560, 2640, 1260, 1440, 2400, 3900, 3024, 1680, 3120, 2880, 4800, 7056, 3600, 2520, 3780, 3360, 5460, 6480, 16848, 6300, 8820, 7200, 9240, 6720, 12480, 5040
EXAMPLE
48 has 10 divisors: {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}, only 12, 16, 24 and 48 are nonpalindromic; no positive integer smaller than 48 has four nonpalindromic divisors, hence a(4) = 48.
MATHEMATICA
f[n_] := DivisorSum[n, 1 &, ! PalindromeQ[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[50, 10^5] (* Amiram Eldar, Jul 14 2022 *)
PROG
(PARI) isnp(n) = my(d=digits(n)); d!=Vecrev(d); \\ A029742a(n) = my(k=1); while (sumdiv(k, d, isnp(d)) != n, k++); k; \\ Michel Marcus, Jul 14 2022 (Python)
from sympy import divisors
from itertools import count, islice
def c(n): s = str(n); return s != s[::-1]
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
n, adict = 0, dict()
for k in count(1):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
a(n) is the smallest integer that has exactly n Lucas divisors ( A000032).
+10
3
1, 2, 4, 12, 36, 252, 2772, 52668, 1211364, 35129556, 1089016236, 44649665676, 2098534286772, 417608323067628, 88115356167269508, 24760415083002731748, 7948093241643876891108, 4140956578896459860267268
COMMENTS
The new Lucas numbers that appear at each step are in A356063.
EXAMPLE
36 is divisible by 1, 2, 3, 4, 18, which are all Lucas numbers, and no integer < 36 has 5 divisors that are Lucas numbers, hence a(5) = 36.
a(n) is the smallest integer that has exactly n divisors whose decimal digits are in strictly increasing order.
+10
3
1, 2, 4, 6, 16, 12, 54, 24, 36, 48, 72, 180, 144, 360, 336, 468, 504, 936, 1008, 1512, 2520, 3024, 5040, 6552, 7560, 22680, 13104, 19656, 49140, 105840, 39312, 78624, 98280, 248976, 334152, 196560, 393120, 668304, 1244880, 1670760, 1867320, 4520880, 3341520, 3734640
COMMENTS
This sequence is finite since A009993 is finite with 511 nonzero terms, hence the last term is a(511) = lcm of the 511 nonzero terms of A009993.
EXAMPLE
For n=7, the divisors of 54 are {1, 2, 3, 6, 9, 18, 27, 54} of which 7 have their digits in strictly increasing order (all except 54). No integer < 54 has 7 such divisors, so a(7) = 54.
MATHEMATICA
s[n_] := DivisorSum[n, 1 &, Less @@ IntegerDigits[#] &]; seq[len_, nmax_] := Module[{v = Table[0, {len}], n = 1, c = 0, i}, While[c < len && n < nmax, i = s[n]; If[i <= len && v[[i]] == 0, v[[i]] = n; c++]; n++]; v]; seq[25, 10^4] (* Amiram Eldar, Sep 16 2022 *)
PROG
(PARI) isok(d) = Set(d=digits(d)) == d; \\ A009993f(n) = sumdiv(n, d, isok(d)); \\ A357171a(n) = my(k=1); while (f(k) !=n, k++); k; \\ Michel Marcus, Sep 16 2022 (Python)
from sympy import divisors
from itertools import count, islice
def c(n): s = str(n); return s == "".join(sorted(set(s)))
def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))
def agen():
n, adict = 1, dict()
for k in count(1):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
Integers whose number of divisors that are Zuckerman numbers sets a new record.
+10
2
1, 2, 4, 6, 12, 24, 72, 144, 360, 432, 1080, 2016, 2160, 6048, 8064, 15120, 24192, 48384, 88704, 120960, 241920, 266112, 532224, 1064448, 1862784, 2661120, 3725568, 5322240, 7451136, 10450944, 19160064, 20901888, 28740096, 38320128, 57480192, 99283968, 114960384
COMMENTS
A Zuckerman number is a number that is divisible by the product of its digits ( A007602). The terms in this sequence are not necessarily Zuckerman numbers. For example a(7) = 72 has product of digits = 14 and 72/14 = 36/7 = 5.142...
The first seven terms are the first seven terms of A087997, then A087997(8) = 66 while a(8) = 144.
EXAMPLE
The 8 divisors of 24 are all Zuckerman numbers, and also, 24 is the smallest integer that has at least 8 divisors that are Zuckerman numbers, hence 24 is a term.
MATHEMATICA
zuckQ[n_] := (prod = Times @@ IntegerDigits[n]) > 0 && Divisible[n, prod]; s[n_] := DivisorSum[n, 1 &, zuckQ[#] &]; smax = 0; seq = {}; Do[s1 = s[n]; If[s1 > smax, smax = s1; AppendTo[seq, n]], {n, 1, 10^5}]; seq (* Amiram Eldar, Jan 14 2021 *)
PROG
(PARI) isokz(n) = iferr(!(n % vecprod(digits(n))), E, 0); \\ A007602lista(nn) = {my(m=0); for (n=1, nn, my(x = sumdiv(n, d, isokz(d)); ); if (x > m, m = x; print1(n, ", ")); ); } \\ Michel Marcus, Jan 15 2021
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