A361246 -id:A361246 - cp's OEIS frontend

A361247 a(n) is the smallest integer k > 2 that satisfies k mod j <= 2 for all integers j in 1..n.

3, 3, 3, 4, 5, 6, 30, 42, 56, 72, 792, 792, 1080, 1080, 1080, 30240, 246961, 246961, 636482, 636482, 1360801, 2162162, 2162162, 2162162, 39412802, 39412802, 107881202, 107881202, 3625549202, 3625549202, 3625549202, 170918748001, 170918748001, 170918748001, 170918748001, 170918748001

Offset: 1

Andrew Cogliano, Mar 05 2023

nonn

			a(7)=30 since 30 mod 7 = 2, 30 mod 6 = 0, 30 mod 5 = 0, 30 mod 4 = 2, 30 mod 3 = 0, 30 mod 2 = 0 and 30 is the smallest integer greater than 2 where all of these remainders are 2 or less.

Chai Wah Wu, Table of n, a(n) for n = 1..40

Cf. A003418 (all remainders 0).
Cf. also A361246, A361248.
Equals {A056697}+1.  - Hugo Pfoertner, May 11 2023

isok(k, n) = for (j=1, n, if ((k % j) > 2, return(0))); return(1);
a(n) = my(k=3); while(!isok(k, n), k++); k; \\ Michel Marcus, Mar 17 2023

final=100
k=3
for n in range(1, final+1):
    j = n+1
    while (j > 2):
        j -= 1
        if k%j>2:
              k += j-(k%j)
              j = n+1
    print(k)

a(32)-a(36) from Chai Wah Wu, Apr 24 2023

A361248 a(n) is the smallest integer k > 3 that satisfies k mod j <= 3 for all integers j in 1..n.

Original entry on oeis.org

4, 4, 4, 4, 5, 6, 7, 8, 56, 72, 91, 651, 651, 1080, 1080, 1443, 20163, 20163, 246962, 246962, 246962, 609843, 2162162, 2162162, 29055601, 29055601, 107881202, 107881202, 205405203, 205405203, 3625549202, 5675443203, 8374212002, 8374212002, 8374212002, 8374212002, 131668891200, 131668891200

Offset: 1

Andrew Cogliano, Mar 05 2023

nonn

			a(11)=91 since 91 mod 11 = 3, 91 mod 10 = 1, 91 mod 9 = 1, 91 mod 8 = 3, 91 mod 7 = 0, 91 mod 6 = 1, 91 mod 5 = 1, 91 mod 4 = 3, 91 mod 3 = 1, 91 mod 2 = 1, 91 mod 1 = 0 and 91 is the smallest integer greater than 3 where all of these remainders are 3 or less.

Chai Wah Wu, Table of n, a(n) for n = 1..48

Cf. A003418 (all remainders 0).

Cf. A361246, A361247.

isok(k, n) = for (j=5, n, if ((k % j) > 3, return(0))); return(1);
a(n) = my(k=4); while(!isok(k, n), k++); k; \\ Michel Marcus, Mar 17 2023

final=100
k=4
for n in range(1, final+1):
    j = n+1
    while (j > 3):
        j -= 1
        if k%j>3:
              k += j-(k%j)
              j = n+1
    print(k)

For n > 2, n <= a(n) < A003418(n). - Charles R Greathouse IV, Apr 27 2023

cp's OEIS Frontend

A361247 a(n) is the smallest integer k > 2 that satisfies k mod j <= 2 for all integers j in 1..n.

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