Andrew Cogliano has authored 3 sequences.
A361246
a(n) is the smallest integer k > 1 that satisfies k mod j <= 1 for all integers j in 1..n.
Original entry on oeis.org
2, 2, 3, 4, 16, 25, 36, 120, 505, 721, 2520, 2520, 41041, 83161, 83161, 196560, 524161, 524161, 3160080, 3160080, 3160080, 3160080, 68468401, 68468401, 68468401, 68468401, 4724319601, 4724319601, 26702676000, 26702676000
Offset: 1
a(7)=36 since 36 mod 7 = 1, 36 mod 6 = 0, 36 mod 5 = 1, 36 mod 4 = 0, 36 mod 3 = 0, 36 mod 2 = 0, 36 mod 1 = 0 and 36 is the smallest integer greater than 1 where all of these remainders are 1 or less.
-
isok(k, n) = for (j=1, n, if ((k % j) > 1, return(0))); return(1);
a(n) = my(k=2); while(!isok(k, n), k++); k; \\ Michel Marcus, Mar 17 2023
-
final=100
k=2
for n in range(1, final+1):
j = n+1
while (j > 1):
j -= 1
if k%j>1:
k += j-(k%j)
j = n+1
print(k)
-
from math import lcm
from itertools import product
from sympy.ntheory.modular import solve_congruence
def A361246(n):
if n == 1: return 2
alist, blist, c, klist = [], [], 1, list(range(n,1,-1))
while klist:
k = klist.pop(0)
if not c%k:
blist.append(k)
else:
c = lcm(c,k)
alist.append(k)
for m in klist.copy():
if not k%m:
klist.remove(m)
for d in product([0,1],repeat=len(alist)):
x = solve_congruence(*list(zip(d,alist)))
if x is not None:
y = x[0]
if y > 1:
for b in blist:
if y%b > 1:
break
else:
if y < c:
c = y
return int(c) # Chai Wah Wu, Jun 19 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
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.
-
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)
A361247
a(n) is the smallest integer k > 2 that satisfies k mod j <= 2 for all integers j in 1..n.
Original entry on oeis.org
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
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.
-
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)