A371641 -id:A371641 - cp's OEIS frontend

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.

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))

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

A372046 Composite numbers that divide the concatenation of the reverse of their ascending order prime factors, with repetition.

Original entry on oeis.org

998, 1636, 9998, 15584, 49447, 99998, 1639964, 2794612, 9999998, 15842836, 1639360636, 1968390098, 27879461212, 65226742928

Offset: 1

Scott R. Shannon, Apr 17 2024

A number 999...9998 will be a term if it has two prime factors 2 and 4999...999. Therefore 999999999999998 and 999...9998 (with 54 9's) are both terms. See A056712.

100000000000 < a(15) <= 999999999999998. Robert P. P. McKone, May 07 2024

			998 is a term as 998 = 2 * 499 = "2" * "994" when each prime factor is reversed. This gives "2994", and 2994 is divisible by 998.
15584 is a term as 15584 = 2 * 2 * 2 * 2 * 2 * 487 = "2" * "2" * "2" * "2" * "2" * "784" when each prime factor is reversed. This gives "22222784", and 22222784 is divisible by 15584.

Cf. A371696, A371695, A371641, A027746, A056712.

a[n_Integer] := Module[{f}, f = Flatten[ConstantArray @@@ FactorInteger[n]]; If[Length[f] < 2, Return[False]]; Mod[FromDigits[StringJoin[StringReverse[IntegerString[#, 10]] & /@ f], 10], n] == 0];
Select[Range[2, 10^5], a] (* Robert P. P. McKone, May 03 2024 *)

from itertools import count, islice
from sympy import factorint
def A372046_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):
                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
A372046_list = list(islice(A372046_gen(),5)) # Chai Wah Wu, Apr 24 2024

a(13)-a(14) from Robert P. P. McKone, May 05 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.

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))

A378893 Numbers that are a proper substring of the concatenation (with repetition) in increasing order of their prime factors.

Original entry on oeis.org

333, 22564, 113113, 210526, 252310, 1143241, 3331233, 3710027, 31373137, 217893044, 433100023, 2263178956

Offset: 1

Scott R. Shannon, Dec 10 2024

			333 is a term as 333 = 3 * 3 * 37 = "3337" when concatenated, which contains "333" as a substring.
113113 is a term as 113113 = 7 * 11 * 13 * 113 = "71113113" when concatenated, which contains "113113" as a substring.
433100023 is a term as 433100023 = 433 * 1000231 = "4331000231" when concatenated, which contains "433100023" as a substring.

Cf. A027746, A378894, A376078, A372309, A371958, A371696, A372046, A371641.

A378894 Numbers, when written in binary, that are a proper substring of the concatenation (with repetition) in increasing order of their prime factors, when written in binary.

Original entry on oeis.org

10, 57, 63, 355, 737, 921, 1526, 13803, 22008, 43364, 44016, 48895, 65151, 88032, 130545, 235929, 255987, 563207, 702460, 1456355, 2799617, 3020897, 3137557, 3774873, 4163463, 5697350, 5995862, 14176747, 42172441, 55933611, 87559273, 93206755, 108530173, 126474397, 180677710, 193337441, 249550095, 259779663, 533713761, 536378647, 715881440, 940339099, 1000732491

Offset: 1

Scott R. Shannon, Dec 10 2024

			10 is a term as 10 = 1010_2 = 2 * 5 = 10_2 * 101_2 = "10101" when concatenated, which contains "1010" as a substring.
63 is a term as 63 = 111111_2 = 3 * 3 * 7 = 11_2 * 11_2 * 111_2 = "1111111" when concatenated, which contains "111111" as a substring.
1526 is a term as 1526 = 10111110110_2 = 2 * 7 * 109 = 10_2 * 111_2 * 1101101_2 = "101111101101" when concatenated, which contains "10111110110" as a substring.

Cf. A027746, A378893, A376078, A372309, A371958, A371696, A372046, A371641.

cp's OEIS Frontend

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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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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A372046 Composite numbers that divide the concatenation of the reverse of their ascending order prime factors, with repetition.

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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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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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A378893 Numbers that are a proper substring of the concatenation (with repetition) in increasing order of their prime factors.

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