A248915 -id:A248915 - cp's OEIS frontend

A259047 Composite numbers which divide the concatenation of their prime factors, with multiplicity, in ascending order.

28749, 21757820799, 4373079629403

Offset: 1

Giovanni Resta, Jun 17 2015

a(2) found by Jens Kruse Andersen, who also cleverly derived 119 large terms of the sequence from the factorization of numbers of the form 10^k+1 (see Links).

10^13 < a(4) <= 7810053011863508278028459 (the smallest of J. K. Andersen's large terms).

			4373079629403 is equal to 3*367*2713*1464031 and it is a divisor of 336727131464031, hence it is in the sequence.

Michael S. Branicky, Python program for Andersen's algorithm extended to arbitrary base/ordering.
Carlos Rivera, Puzzle 472. What is the second solution?, The Prime Puzzles & Problems Connection.
StackExchange, New term in ascending order.

Cf. A248915.

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

Original entry on oeis.org

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

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))

cp's OEIS Frontend

A259047 Composite numbers which divide the concatenation of their prime factors, with multiplicity, in ascending order.

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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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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.

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