A072562
Smallest of exactly n consecutive integers divisible respectively by the first n primes.
Original entry on oeis.org
4, 2, 8, 158, 3098, 788, 210998, 5316098, 34415168, 703693778, 194794490678, 5208806743928, 138782093170508, 5006786309605868, 253579251611336438, 12551374903381164638, 142908008812141343558, 77053322014980646906358
Offset: 1
a(4)=158 because 158 is the least number such that 158, 159, 160 and 161 are divisible by 4 consecutive primes, namely 2, 3, 5 and 7 respectively.
a(5) does not equal A069561(5)=788 because 788 is the smallest integer in a run of 6 (not 5) consecutive integers that are divisible respectively by the first 6 consecutive primes. - _Geoffrey Critzer_, Oct 29 2014
-
A:= proc(n)
local r;
if n = 1 then return 4 fi;
r:= chrem([seq(-i,i=0..n-1)],[seq(ithprime(i),i=1..n)]);
if r + n mod ithprime(n+1) = 0 then r + mul(ithprime(i),i=1..n) else r fi
end proc:
seq(A(n),n=1..50); # Robert Israel, Oct 29 2014
-
f[n_] := Block[{p = Prime@ Range@ n}, r = ChineseRemainder[-Range@ n + 1, p]; If[ Mod[r + n, Prime[n + 1]] == 0, r + Times @@ p, r]]; f[1] = 4; Array[f, 20] (* Robert G. Wilson v, Oct 30 2014 *)
-
a(n)=if(n==1,return(4)); my(m=chinese(vector(n, k, Mod(1-k, prime(k)))), p=prime(n+1), t=lift(m)); if((t+n)%p, t, t+m.mod) \\ Charles R Greathouse IV, Jun 20 2015
A360478
Least k such that the first n primes divide k and the next n primes divide k+1.
Original entry on oeis.org
2, 174, 11010, 877590, 3576536040, 7300395162060, 8122095802580760, 15497422946114018910, 6949903578918639188850, 482875127106370562524140180, 2448313281623289989477792853630, 20024982066721727911275778517919720, 29200503600421680216708710172770859570
Offset: 1
a(3) = 11010 because the first 3 primes {2, 3, 5} divide 11010 and the next 3 primes {7, 11, 13} divide 11011.
-
K = {}; For[n = 1, n <= 13, n++, R = Prime[Range[2 n]];
A = R[[1 ;; n]]; t = Times @@ A; B = R[[n + 1 ;; 2 n]]; p = t + 1;
For[b = 1, b <= n, b++, While[Mod[p, Part[B, b]] != 0, p += t];
t *= Part[B, b]]; AppendTo[K, p - 1]]; Print[K]
-
a(n)={my(p=primes(2*n)); lift(chinese(Mod(0, vecprod(p[1..n])), Mod(-1, vecprod(p[n+1..2*n]))))} \\ Andrew Howroyd, Feb 08 2023
A185681
a(n) = start of n consecutive numbers divisible respectively by prime(k)^n, for k=1..n.
Original entry on oeis.org
2, 8, 21248, 1197741248, 16414088688381248, 579329868703698452660781248, 20182773361697812807734811854155781248, 28343998868273668587268878406666355122557128720825781248
Offset: 1
a(1) = 2 as 2 is divisible by 2^1 ;
a(2) = 8 as 8 and 9 are divisible by 2^2 and 3^2 respectively ;
a(3) = 21248 as 21248, 21249 and 21250 are divisible by 2^3, 3^3 and 5^3 respectively.
-
Table[ ChineseRemainder[ Table[ -i, {i, 0, n - 1}], Table[ Prime[i]^n, {i, 1, n}]], {n, 2, 10}]
A212313
Start of a run of n consecutive even numbers divisible respectively by first n primes.
Original entry on oeis.org
0, 4, 16, 106, 1576, 1576, 421996, 932506, 68830336, 1407387556, 189028491226, 2996875353046, 277564186341016, 10013572619211736, 507158503222672876, 25102749806762329276, 285816017624282687116, 36818262670554322829446, 5783948949281495901009676
Offset: 1
106 = 2*53, 108 = 3*36, 110 = 5*22, 112 = 7*16.
-
Table[ChineseRemainder[Table[2 k - 2, {k, n}], Table[Prime[k], {k, n}]], {n, 20}]
A248331
a(n) is the first term in a length n sequence of consecutive integers that are divisible respectively by the square of the first n primes.
Original entry on oeis.org
0, 8, 548, 29348, 1308248, 652312448, 180110691548, 65335225716548, 38733853511213648, 4368761145612023948, 1804216772228848838648, 14884872991210984993091648, 9816873967836575781598117448, 143397994078495393809327283088348
Offset: 1
a(4)=29348. 29348 is divisible by 4, 29349 is divisible by 9, 29350 is divisible by 25, 29351 is divisible by 49.
The first few rows of the triangle of quotients are:
0;
2, 1;
137, 61, 22;
7337, 3261, 1174, 599;
327062, 145361, 52330, 26699, 10812;
163078112, 72479161, 26092498, 13312499, 5391012, 3859837;
- _Michel Marcus_, Oct 27 2014
- K. H. Rosen, Elementary Number Theory and its Applications, Addison-Wesley, 1984, page 113.
-
Table[ChineseRemainder[Reverse[Range[-k, 0]], Table[Prime[n]^2, {n, 1, k + 1}]], {k, 0, 13}]
Showing 1-5 of 5 results.
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