A259047 -id:A259047 - cp's OEIS frontend

A371641 The smallest composite number which divides the concatenation of its ascending ordered prime factors, with repetition, when written in base n.

85329, 4, 224675, 4, 1140391, 4, 9, 4, 28749, 4, 841, 4, 9, 4, 239571, 4, 343, 4, 9, 4, 231, 4, 25, 4, 9, 4, 315, 4, 343, 4, 9, 4, 25, 4, 9761637601, 4, 9, 4, 4234329, 4, 715, 4, 9, 4, 609, 4, 49, 4, 9, 4, 195, 4, 25, 4, 9, 4, 1015, 4, 76729, 4, 9, 4, 25, 4, 14332171

Offset: 2

Scott R. Shannon, Mar 30 2024

The number 4 = 2*2 in any base b = 3 + 2*n, n >= 0, will always divide the concatenation of its prime divisors as 4 = "2"_b + "2"_b = "22"_b = 2*(3 + 2*n) + 2 = 8 + 4*n, which is divisible by 4.
The number 9 = 3*3 in any base b = 8 + 6*n, n >= 0, will always divide the concatenation of its prime divisors as 9 = "3"_b + "3"_b = "33"_b = 3*(8 + 6*n) + 3 = 27 + 18*n, which is divisible by 9.
Theorem: if p is prime, then a(p*(m+2)-1) <= p^2 for all m >= 0. Proof: If p is prime and base b = p*(m+2)-1 for some m >= 0, then b > p and p expressed in base b = "p"b and thus "pp"_b = p*(b+1) = p^2*(m+2), i.e., divisible by p^2. - _Chai Wah Wu, Apr 01 2024
a(36) <= 9761637601. a(40) = 4234329. By the theorem above if n+1 is composite, then a(n) <= p^2 where p = A020639(n+1) is the smallest prime factor of n+1. The first few terms where the inequality is strict are: a(406) = 105, a(766) = 105, a(988) = 195, a(1036) = 105, a(1072) = 231, ... - Chai Wah Wu, Apr 11 2024
From Chai Wah Wu, Apr 12 2024: (Start)
Theorem: If n is even, then a(n) >= 9 is odd.
Proof: This is true for n = 2. Let n > 2 be even. Let m be even.
If m=2^k for k>1, then 2 concatenated k times in base n is 2*(n^(k-1)+...+n+1) = 2*(n^k-1)/(n-1).
Since n^k-1 is odd, m does not divide 2*(n^k-1)/(n-1). If m has an odd prime divisor then concatenating the primes in base n will result in an odd number that is not divisible by m.
Finally, a(n) >= 9 since the first odd composite number is 9. (End)

			a(2) = 85329 as 85329 = 3_10 * 3_10 * 19_10 * 499_10 = 11_2 * 11_2 * 10011_2 * 111110011_2 = "111110011111110011"_2 = 255987_10 which is divisible by 85329.
a(10) = 28749 as 28749 =  3_10 * 7_10 * 37_10 * 37_10 = "373737"_10 = 373737_10 which is divisible by 28749. See also A259047.

Michael S. Branicky, Table of n, a(n) for n = 2..129

Cf. A020639, A027746, A259047, A322843, A248915.

has(F,n)=my(f=F[2],t); for(i=1,#f~, my(p=f[i,1],d=#digits(p,n),D=n^d); for(j=1,f[i,2], t=D*t+p)); t%F[1]==0
a(k,lim=10^6,startAt=4)=forfactored(n=startAt,lim, if(vecsum(n[2][,2])>1 && has(n,k), return(n[1]))); a(k,2*lim,lim+1) \\ Charles R Greathouse IV, Apr 11 2024

from itertools import count
from sympy.ntheory import digits
from sympy import factorint, isprime
def fromdigits(d, b):
    n = 0
    for di in d: n *= b; n += di
    return n
def a(n):
    for k in count(4):
        if isprime(k): continue
        sf = []
        for p, e in factorint(k).items():
            sf.extend(e*digits(p, n)[1:])
        if fromdigits(sf, n)%k == 0:
            return k
print([a(n) for n in range(2, 6)]) # Michael S. Branicky, Apr 01 2024

from itertools import count
from sympy import factorint, integer_log
def A371641(n):
    for m in count(4):
        f = factorint(m)
        if sum(f.values()) > 1:
            c = 0
            for p in sorted(f):
                a = pow(n,integer_log(p,n)[0]+1,m)
                for _ in range(f[p]):
                    c = (c*a+p)%m
            if not c:
                return m # Chai Wah Wu, Apr 11 2024

a(36) and beyond from Michael S. Branicky, Apr 27 2024

A248915 Composite numbers which divide the concatenation of their prime factors, with multiplicity, in descending order.

Original entry on oeis.org

378, 12467, 95823, 10715274, 13485829, 111495095, 42002916561, 176685987695

Offset: 1

Paolo P. Lava, Oct 16 2014

Prime numbers are not considered because they trivially satisfy the relation.
For terms in ascending order see A259047 and StackExchange link. [Paolo P. Lava, May 30 2019]
a(9) <= 3953318131772867. - Chai Wah Wu, Apr 12 2024
a(2), the bound for a(9) above, and larger terms may be found using an extension of Andersen's algorithm to arbitrary base and ordering (see links for an implementation and another term). - Michael S. Branicky, Apr 13 2024

			Prime factors of 378 are 2,3,3,3,7; concat(7,3,3,3,2) = 73332 and 73332/378 = 194.

Michael S. Branicky, Python program for Andersen's algorithm extended to arbitrary base/ordering
StackExchange, New term in ascending order

Cf. A037276, A069872, A224930, A240265, A259047.

with(numtheory); P:=proc(q) local a,b,c,d,j,k,n;
for n from 1 to q do if not isprime(n) then a:=ifactors(n)[2]; b:=[]; d:=0;
for k from 1 to nops(a) do b:=[op(b),a[k][1]]; od; b:=sort(b);
for k from nops(a) by -1 to 1 do c:=1; while not b[k]=a[c][1] do c:=c+1; od;
for j from 1 to a[c][2] do d:=10^(ilog10(b[k])+1)*d+b[k]; od; od;
if type(d/n,integer) then print(n); fi;
fi; od; end: P(10^9);

isok(n) = {my(s = ""); my(f = factor(n)); forstep (i=#f~, 1, -1, for (k=1, f[i,2], s = concat(s, Str(f[i,1])))); (eval(s) % n) == 0;} \\ Michel Marcus, Jun 16 2015

a(7)-a(8) from Giovanni Resta, Jun 16 2015

A371696 Composite numbers that divide the reverse of the concatenation of their ascending order prime factors, with repetition.

Original entry on oeis.org

26, 38, 46, 378, 26579, 84941, 178838, 30791466, 39373022, 56405502, 227501395, 904085931, 1657827142

Offset: 1

Scott R. Shannon, Apr 03 2024

a(14) > 10^10, if it exists. - Daniel Suteu, Apr 28 2024

			26 is a term as 26 = 2 * 13 = "213" which is reverse is "312", and 312 is divisible by 26.
227501395 is a term as 227501395 = 5 * 11 * 17 * 23 * 71 * 149 = "511172371149" which in reverse is "941173271115", and 941173271115 is divisible by 227501395.

Cf. A027746, A037276, A259047, A371641, A371666.

from itertools import count, islice
from sympy import isprime, factorint
def ok(k): return not isprime(k) and int("".join(str(p)[::-1]*e for p, e in list(factorint(k).items())[::-1]))%k == 0
def agen(): yield from filter(ok, count(4))
print(list(islice(agen(), 7))) # Michael S. Branicky, Apr 13 2024

from itertools import count, islice
from sympy import factorint
def A371696_gen(startvalue=4): # generator of terms >= startvalue
    for n in count(max(startvalue,4)):
        f = factorint(n)
        if sum(f.values()) > 1:
            c = 0
            for p in sorted(f,reverse=True):
                a = pow(10,len(s:=str(p)),n)
                q = int(s[::-1])
                for _ in range(f[p]):
                    c = (c*a+q)%n
            if not c:
                yield n
A371696_list = list(islice(A371696_gen(),6)) # Chai Wah Wu, Apr 13 2024

a(13) from Daniel Suteu, Apr 28 2024

A371821 Composite numbers which divide the concatenation of their ascending ordered prime factors, with repetition, when written in binary.

Original entry on oeis.org

85329, 177904587, 333577497

Offset: 1

Scott R. Shannon, Apr 07 2024

The base 2 version of A259047. Assuming a(4) exists it is greater than 10^10.
a(4) <= 55133857902732922904331439521901. - Chai Wah Wu, Apr 12 2024
a(1), a(3), the bound on a(4) above, and larger terms can be generated using an adaptation of the method of J. K. Andersen referenced in A259047; see linked Python program for an implementation and two more terms. - Michael S. Branicky, Apr 12 2024

			177904587 is a term as 177904587 = 3_10 * 7_10 * 103_10 * 233_10 * 353_10 = 11_2 * 111_2 * 1100111_2 * 11101001_2 * 101100001_2 = "11111110011111101001101100001"_2 = 533713761_10, which is divisible by 177904587.

Michael S. Branicky, Python program generating terms in A371821

Cf. A027746, A004676, A259047, A371641.

from itertools import count, islice
from sympy import factorint
def A371821_gen(startvalue=1): # generator of terms >= startvalue
    for n in count(max(startvalue,1)):
        f = sorted(factorint(n,multiple=True))
        if len(f) > 1:
            c = 0
            for p in f:
                c = ((c<A371821_list = list(islice(A371821_gen(),3)) # Chai Wah Wu, Apr 11 2024

from sympy import factorint, isprime
def ok(n): return not isprime(n) and int("".join(bin(p)[2:]*e for p, e in factorint(n).items()), 2)%n == 0 # Michael S. Branicky, Apr 12 2024

A371695 The smallest composite number that divides the reverse of the concatenation of its ascending ordered prime factors, with repetition, when written in base n.

Original entry on oeis.org

623, 4, 114, 4, 57, 4, 9, 4, 26, 4, 185, 4, 9, 4, 1718, 4, 343, 4, 9, 4, 70, 4, 25, 4, 9, 4, 195, 4, 226, 4, 9, 4, 25, 4, 123, 4, 9, 4, 654, 4, 862, 4, 9, 4, 42, 4, 49, 4, 9, 4, 3385, 4, 25, 4, 9, 4, 238, 4, 202, 4, 9, 4, 25, 4, 453, 4, 9, 4, 2435, 4, 721, 4, 9, 4, 49, 4, 70, 4, 9, 4, 186

Offset: 2

Scott R. Shannon, Apr 03 2024

See A371641 for an explanation of multiple terms being 4 and 9. The largest number in the first 10000 terms is a(5980) = 1030778.

			a(2) = 623 as 623 = 7_10 * 89_10 = 111_2 * 1011001_2 = "1111011001"_2 which when reversed is "1001101111"_2 = 623_10 which is divisible by 623.
a(4) = 114 as 114 = 2_10 * 3_10 * 19_10 = 2_4 * 3_4 * 103_4 = "23103"_4 which when reversed is "30132"_4 = 798_10 which is divisible by 114.

Scott R. Shannon, Table of n, a(n) for n = 2..10000

Cf. A020639, A371641, A027746, A259047, A322843, A248915, A371948.

from itertools import count
from sympy.ntheory import digits
from sympy import factorint, isprime
def fromdigits(d, b):
    n = 0
    for di in d: n *= b; n += di
    return n
def a(n):
    for k in count(4):
        if isprime(k): continue
        sf = []
        for p, e in list(factorint(k).items())[::-1]:
            sf.extend(e*digits(p, n)[1:][::-1])
        if fromdigits(sf, n)%k == 0:
            return k
print([a(n) for n in range(2, 83)]) # Michael S. Branicky, Apr 16 2024

If n+1 is composite, then a(n) <= A020639(n+1)^2. The numbers n where n+1 is composite and a(n) < A020639(n+1)^2 are 288, 298, 340, 360, 376, 516, 526, 550, 582, 736, ... and appear to be identical to A371948. - Chai Wah Wu, Apr 16 2024

A371699 The smallest composite number which divides the concatenation of its descending ordered prime factors, with repetition, when written in base n.

Original entry on oeis.org

42, 4, 42, 4, 374, 4, 9, 4, 378, 4, 609, 4, 9, 4, 3525, 4, 343, 4, 9, 4, 70, 4, 25, 4, 9, 4, 195, 4, 343, 4, 9, 4, 25, 4, 130, 4, 9, 4, 366, 4, 3562, 4, 9, 4, 42, 4, 49, 4, 9, 4, 474, 4, 25, 4, 9, 4, 238, 4, 1131, 4, 9, 4, 25, 4, 555, 4, 9, 4, 14405, 4, 12207

Offset: 2

Chai Wah Wu, Apr 12 2024

Conjecture: a(n) <= A371641(n) for n >= 2.

			a(2) = 42 since 42 = 7*3*2 = 111_2 * 11_2 * 10 _2 and 42 divides 126 = 1111110_2.
a(10) = 378 since 278 = 7*3*3*3*2 and 278 divides 73332.

Chai Wah Wu, Table of n, a(n) for n = 2..10000

Cf. A020639, A027746, A259047, A322843, A248915, A371641, A371821.

from itertools import count
from sympy import factorint, integer_log
def A371699(n):
    for m in count(4):
        f = factorint(m)
        if sum(f.values()) > 1:
            c = 0
            for p in sorted(f,reverse=True):
                a = pow(n,integer_log(p,n)[0]+1,m)
                for _ in range(f[p]):
                    c = (c*a+p)%m
            if not c:
                return m

If p is prime, then a(p*(m+2)-1) <= p^2 for all m >= 0.
If n+1 is composite, then a(n) <= q^2, where q = A020639(n+1) is the smallest prime factor of n+1. This implies that if n > 2 is odd, then a(n) = A371641(n) = 4.
The first few terms n where n+1 is composite and a(n) < A020639(n+1)^2 are a(288) = 70, a(298) = 42, a(340) = 42, a(360) = 182, ...
If n is even, then a(n) >= 9. This is true as it is easy to verify that a(n) cannot be equal to 4, 6 or 8 in this case.
Suppose n>2 is even. 2 concatenated twice in base n is 2(n+1) which is not divisible by 4.
Next, 3 concatenated by 2 is 3*n+2 which is not divisible by 6. Finally 2 concatenated 3 times is 2(n^3-1)/(n-1) which is not divisible by 8 since n^3-1 is odd.
This implies that if n = 6*k+2 for some k > 0, then a(n) = A371641(n) = 9.

A371900 Numbers k such that k+1 is composite and A371641(k) != p^2 where p = A020639(k+1) is the smallest prime factor of k+1.

Original entry on oeis.org

406, 766, 988, 1036, 1072, 1138, 1246, 1396, 1402, 1456, 1500, 1642, 1738, 1762, 1768, 1816, 1918, 1926, 1942, 2076, 2116, 2158, 2182, 2278, 2506, 2716, 2746, 2812, 2866, 2920, 2992, 3076, 3148, 3172, 3286, 3316, 3382, 3496, 3568, 3682, 3706, 3712, 3742, 3762

Offset: 1

Chai Wah Wu, Apr 11 2024

If k+1 is composite, then A371641(k) <= A020639(k+1)^2. This sequence lists numbers k where the inequality is strict.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A020639, A027746, A259047, A322843, A248915, A371641.

from itertools import count, islice
from sympy import isprime, primefactors, factorint, integer_log
def A371900_gen(startvalue=2): # generator of terms >= startvalue
    for n in count(max(startvalue,2)):
        if not isprime(n+1):
            q = min(primefactors(n+1))
            for m in range(4,q**2):
                f = factorint(m)
                if sum(f.values()) > 1:
                    c = 0
                    for p in sorted(f):
                        a = pow(n,integer_log(p,n)[0]+1,m)
                        for _ in range(f[p]):
                            c = (c*a+p)%m
                    if not c:
                        yield n
                        break
A371900_list = list(islice(A371900_gen(),30))

A371948 Numbers k such that k+1 is composite and A371699(k) != p^2 where p = A020639(k+1) is the smallest prime factor of k+1.

Original entry on oeis.org

288, 298, 340, 360, 376, 516, 526, 550, 582, 736, 778, 792, 802, 816, 892, 988, 1002, 1006, 1072, 1138, 1146, 1198, 1206, 1246, 1270, 1338, 1342, 1348, 1356, 1390, 1402, 1456, 1500, 1516, 1536, 1576, 1632, 1642, 1702, 1726, 1738, 1750, 1768, 1816, 1828, 1842

Offset: 1

Chai Wah Wu, Apr 13 2024

nonn

If k+1 is composite, then A371699(k) <= A020639(k+1)^2. This sequence lists numbers k where the inequality is strict.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A020639, A027746, A259047, A322843, A248915, A371641, A371699, A371900.

from itertools import count, islice
from sympy import isprime, primefactors, factorint, integer_log
def A371948_gen(startvalue=2): # generator of terms >= startvalue
    for n in count(max(startvalue,2)):
        if not isprime(n+1):
            q = min(primefactors(n+1))
            for m in range(4,q**2):
                f = factorint(m)
                if sum(f.values()) > 1:
                    c = 0
                    for p in sorted(f,reverse=True):
                        a = pow(n,integer_log(p,n)[0]+1,m)
                        for _ in range(f[p]):
                            c = (c*a+p)%m
                    if not c:
                        yield n
                        break
A371948_list = list(islice(A371948_gen(), 30))

A349705 Numbers k such that the concatenation in increasing order of their prime factors, with multiplicity, is congruent to 1 (mod k).

Original entry on oeis.org

36, 39, 66, 1435, 5714, 6410, 13861, 22564, 27346, 33137, 45542, 79260, 171860, 268218, 442068, 486127, 675423, 2287527, 3710027, 9610766, 14318290, 26293568, 29361702, 49703324, 227358366, 433100023, 442960845, 479174118, 1221238938, 1243718114, 4053362596, 8620689655

Offset: 1

J. M. Bergot and Robert Israel, Nov 25 2021

			a(3) = 66 is a term because the concatenation of its prime factors is 2311 and 2311 == 1 (mod 66).

Martin Ehrenstein, Table of n, a(n) for n = 1..39

Cf. A037276, A259047.

filter:= proc(n) local L,t;
  lcat(map(t -> t[1]$t[2], sort( ifactors(n)[2], (a,b) -> a[1] < b[1]))) mod n = 1;
end proc:
select(filter, [$1..10^7]);

upto=10^5;a={};Do[If[Mod[FromDigits[Flatten[Map[IntegerDigits[ConstantArray[First[#],Last[#]]]&,FactorInteger[k]]]],k]==1,AppendTo[a,k]],{k,upto}];a (* Paolo Xausa, Nov 26 2021 *)

from sympy import factorint
def ok(k): return int("".join(map(str, factorint(k, multiple=True))))%k == 1
print([k for k in range(2, 10**5) if ok(k)]) # Michael S. Branicky, Nov 26 2021

from itertools import count, islice
from sympy import factorint
def A349705_gen(startvalue=1): # generator of terms >= startvalue
    for k in count(max(startvalue,1)):
        c = 0
        for d in sorted(factorint(k,multiple=True)):
            c = (c*10**len(str(d)) + d) % k
        if c == 1:
            yield k
A349705_list = list(islice(A349705_gen(),10)) # Chai Wah Wu, Feb 28 2022

a(28)-a(32) from Martin Ehrenstein, Nov 27 2021

cp's OEIS Frontend

A371641 The smallest composite number which divides the concatenation of its ascending ordered prime factors, with repetition, when written in base n.

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A248915 Composite numbers which divide the concatenation of their prime factors, with multiplicity, in descending order.

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A371696 Composite numbers that divide the reverse of the concatenation of their ascending order prime factors, with repetition.

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A371821 Composite numbers which divide the concatenation of their ascending ordered prime factors, with repetition, when written in binary.

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A371695 The smallest composite number that divides the reverse of the concatenation of its ascending ordered prime factors, with repetition, when written in base n.

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A371699 The smallest composite number which divides the concatenation of its descending ordered prime factors, with repetition, when written in base n.

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A371900 Numbers k such that k+1 is composite and A371641(k) != p^2 where p = A020639(k+1) is the smallest prime factor of k+1.

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A371948 Numbers k such that k+1 is composite and A371699(k) != p^2 where p = A020639(k+1) is the smallest prime factor of k+1.