Jelly, 1011 bytes
V+€J%ÞfÞ8ḢV+€J%ÞfÞẒƇḢ
A monadic Link that accepts a list of pairwise coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
V+€J%ÞfÞ8ḢV+€J%ÞfÞẒƇḢ - Link: Coprimes
V - Digit-smash {Coprimes}
(e.g. [2, 11, 17] -> 21117 -- N.B. V >= Product+Max so is sufficient*)
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in:
Coprimes Ƈ - {Comprimes} for which:
Ẓ - is prime?
Ḣ - head
* Thanks to Leo for this proof that the digit concatenation performed by V is guaranteed to be greater than or equal to the product plus the maximum that we may need to search through:
The concatenation \$a|b\$ of two positive numbers \$a\$ and \$b\$ is equal to \$a\times c + b\$, where \$c\$ is the first power of \10ドル\$ larger than \$b\$ (i.e. \$c = 10^{\lfloor 1 + \log_{10}{b} \rfloor}\$).
Since \$c \ge b + 1\$ we get \$a|b \ge a \times (b + 1) + b = a \times b + a + b\$.
As such the concatenation of two numbers is always at least as large as their product plus their sum. This then extends to more terms as each of the operations, concatenation and taking a product, may be applied in succession.
Jelly, 10 bytes
V+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of pairwise coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
V+€J%ÞfÞ8Ḣ - Link: Coprimes
V - Digit-smash {Coprimes}
(e.g. [2, 11, 17] -> 21117 -- N.B. V >= Product+Max so is sufficient*)
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
* Thanks to Leo for this proof that the digit concatenation performed by V is guaranteed to be greater than or equal to the product plus the maximum that we may need to search through:
The concatenation \$a|b\$ of two positive numbers \$a\$ and \$b\$ is equal to \$a\times c + b\$, where \$c\$ is the first power of \10ドル\$ larger than \$b\$ (i.e. \$c = 10^{\lfloor 1 + \log_{10}{b} \rfloor}\$).
Since \$c \ge b + 1\$ we get \$a|b \ge a \times (b + 1) + b = a \times b + a + b\$.
As such the concatenation of two numbers is always at least as large as their product plus their sum. This then extends to more terms as each of the operations, concatenation and taking a product, may be applied in succession.
Jelly, 11 bytes
V+€J%ÞfÞẒƇḢ
A monadic Link that accepts a list of pairwise coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
V+€J%ÞfÞẒƇḢ - Link: Coprimes
V - Digit-smash {Coprimes}
(e.g. [2, 11, 17] -> 21117 -- N.B. V >= Product+Max so is sufficient*)
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f - keep those which are in:
Ƈ - {Comprimes} for which:
Ẓ - is prime?
Ḣ - head
* Thanks to Leo for this proof that the digit concatenation performed by V is guaranteed to be greater than or equal to the product plus the maximum that we may need to search through:
The concatenation \$a|b\$ of two positive numbers \$a\$ and \$b\$ is equal to \$a\times c + b\$, where \$c\$ is the first power of \10ドル\$ larger than \$b\$ (i.e. \$c = 10^{\lfloor 1 + \log_{10}{b} \rfloor}\$).
Since \$c \ge b + 1\$ we get \$a|b \ge a \times (b + 1) + b = a \times b + a + b\$.
As such the concatenation of two numbers is always at least as large as their product plus their sum. This then extends to more terms as each of the operations, concatenation and taking a product, may be applied in succession.
Jelly, (10?)* 1110 bytes
PḤ+€J%ÞfÞ8ḢV+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of pairwise coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
PḤ+€J%ÞfÞ8ḢV+€J%ÞfÞ8Ḣ - Link: Coprimes
P V - productDigit-smash {Coprimes}
Ḥ - double (e.g. [2, 11, 17] -> 2P21117 (-- N.B. 2PV >= P+max(Coprimes)Product+Max so is sufficient*)
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
* Thanks to (10?) IfLeo for this proof that the digit concatenation of the decimal digits of any coprime list (e.g. [3, 5, 16] ->performed by 3516V) is guaranteed to be at leastgreater than or equal to the product of the coprimes plus itsthe maximum that we may need to search through:
The concatenation (e.g.\$a|b\$ of two positive numbers [3, 5, 16]\$a\$ and ->\$b\$ is equal to 3*5*16+16 = 256)\$a\times c + b\$, thenwhere PḤ could be replaced with\$c\$ is the first power of V\10ドル\$ larger than \$b\$ (i. Ie. think it\$c = 10^{\lfloor 1 + \log_{10}{b} \rfloor}\$).
Since \$c \ge b + 1\$ we get \$a|b \ge a \times (b + 1) + b = a \times b + a + b\$.
As such the concatenation of two numbers is always big enough but I'm not 100% sureat least as large as their product plus their sum. This then extends to more terms as each of the operations, concatenation and taking a product, may be applied in succession.
Jelly, (10?)* 11 bytes
PḤ+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
PḤ+€J%ÞfÞ8Ḣ - Link: Coprimes
P - product {Coprimes}
Ḥ - double -> 2P (N.B. 2P >= P+max(Coprimes))
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
* (10?) If the concatenation of the decimal digits of any coprime list (e.g. [3, 5, 16] -> 3516) is guaranteed to be at least the product of the coprimes plus its maximum (e.g. [3, 5, 16] -> 3*5*16+16 = 256), then PḤ could be replaced with V. I think it is always big enough but I'm not 100% sure.
Jelly, 10 bytes
V+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of pairwise coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
V+€J%ÞfÞ8Ḣ - Link: Coprimes
V - Digit-smash {Coprimes}
(e.g. [2, 11, 17] -> 21117 -- N.B. V >= Product+Max so is sufficient*)
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
* Thanks to Leo for this proof that the digit concatenation performed by V is guaranteed to be greater than or equal to the product plus the maximum that we may need to search through:
The concatenation \$a|b\$ of two positive numbers \$a\$ and \$b\$ is equal to \$a\times c + b\$, where \$c\$ is the first power of \10ドル\$ larger than \$b\$ (i.e. \$c = 10^{\lfloor 1 + \log_{10}{b} \rfloor}\$).
Since \$c \ge b + 1\$ we get \$a|b \ge a \times (b + 1) + b = a \times b + a + b\$.
As such the concatenation of two numbers is always at least as large as their product plus their sum. This then extends to more terms as each of the operations, concatenation and taking a product, may be applied in succession.
Jelly, (10?)* 11 bytes
PḤ+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
PḤ+€J%ÞfÞ8Ḣ - Link: Coprimes
P - product {Coprimes}
Ḥ - double -> 2P (N.B. 2P >= P+max(Coprimes))
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
Hmm maybe* (10?) If the concatenation of the decimal digits of any coprime list (e.g. PḤ[3, 5, 16] could be -> V3516) is guaranteed to be at least the product of the coprimes plus its maximum (effectively concatenate all digits of Coprimes to make an integere.g. [3, 5, 16] -> 3*5*16+16 = 256), then PḤ could be replaced with V. I think it is always big enough.. but I'm not 100% sure.
Jelly, 11 bytes
PḤ+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
PḤ+€J%ÞfÞ8Ḣ - Link: Coprimes
P - product {Coprimes}
Ḥ - double -> 2P (N.B. 2P >= P+max(Coprimes))
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
Hmm maybe PḤ could be V (effectively concatenate all digits of Coprimes to make an integer), I think it is always big enough...
Jelly, (10?)* 11 bytes
PḤ+€J%ÞfÞ8Ḣ
A monadic Link that accepts a list of coprime integers and yields a run of the same number of consecutive composites.
How?
Uses the Chinese Remainder theorem in the most naive way possible, systematic search.
PḤ+€J%ÞfÞ8Ḣ - Link: Coprimes
P - product {Coprimes}
Ḥ - double -> 2P (N.B. 2P >= P+max(Coprimes))
J - 1-indexed indices {Coprimes} -> [1..n=length(Coprimes)]
€ - for each {v in [1..2P]}:
+ - {v} add {[1..n]} (vectorises)
Þ - sort by:
% - modulo {Coprimes} (vectorises)
Þ - sort by:
f 8 - keep those which are in Coprimes
Ḣ - head
* (10?) If the concatenation of the decimal digits of any coprime list (e.g. [3, 5, 16] -> 3516) is guaranteed to be at least the product of the coprimes plus its maximum (e.g. [3, 5, 16] -> 3*5*16+16 = 256), then PḤ could be replaced with V. I think it is always big enough but I'm not 100% sure.