A081618
Numbers n such that (product of first n primes)+1 is divisible by the (n+1)-th prime. Also n such that A075306(n)-1 is equal to A002110(n). Positions of 1 in A081617.
Original entry on oeis.org
The 8th prime, 19, divides 2*3*5*7*11*13*17+1=510511, thus 7 is a member.
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With[{nn=500},Flatten[Position[Thread[{Rest[FoldList[Times,1,Prime[ Range[ nn]]]]+ 1, Prime[ Range[2,nn+1]]}],?(Divisible[#[[1]],#[[2]]]&),{1},Heads->False]]] (* _Harvey P. Dale, Apr 18 2015 *)
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p=1; for(n=1, 10^5, p=p*prime(n); if((p+1)%prime(n+1)==0, print1(n", ")))
A081617
Smallest k such that (product of first n primes)*k+1 is divisible by the (n+1)-th prime. Also (A075306(n)-1)/A002110(n).
Original entry on oeis.org
1, 4, 3, 10, 10, 2, 1, 3, 17, 13, 10, 34, 38, 4, 51, 55, 51, 29, 68, 13, 59, 30, 27, 45, 18, 92, 77, 82, 64, 14, 68, 58, 114, 68, 8, 77, 42, 114, 31, 98, 129, 110, 43, 61, 159, 35, 109, 60, 91, 149, 193, 2, 38, 120, 259, 147, 135, 22, 140, 10, 263, 285, 286, 134, 308
Offset: 1
The 8th prime, 19, divides 2*3*5*7*11*13*17+1=510511, thus a(7)=1.
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for(n=1, 100, p=1; forprime(k=2, prime(n), p=p*k); pn=prime(n+1); s=0; while((s+1)%pn>0, s=s+p); print1(s/p", "))
A286947
Triangle read by rows in which row(n) = {T(n, k)} is the lexicographically earliest list of n numbers such that adding 1 to some T(n, k) gives a row of numbers each divisible by prime(k).
Original entry on oeis.org
1, 3, 2, 15, 20, 24, 105, 140, 84, 90, 1155, 770, 924, 1980, 2100, 15015, 10010, 24024, 4290, 13650, 23100, 255255, 340340, 204204, 364650, 464100, 353430, 60060, 4849845, 6466460, 5819814, 1385670, 3527160, 5969040, 570570, 510510, 111546435, 74364290, 44618574, 127481640, 81124680, 102965940, 39369330, 58708650, 29099070
Offset: 1
Row(1): [1]
Row(2): [3, 2]
Row(3): [15, 20, 24]
Row(4): [105, 140, 84, 90]
Row(5): [1155, 770, 924, 1980, 2100]
Row(6): [15015, 10010, 24024, 4290, 13650, 23100]
Row(7): [255255, 340340, 204204, 364650, 464100, 353430, 60060]
Row(8): [4849845, 6466460, 5819814, 1385670, 3527160, 5969040, 570570, 510510]
Row(4) = [105, 140, 84, 90].
Adding 1 to T(4, 1) gives [106,140,84,90], all elements divisible by prime(1) = 2.
Adding 1 to T(4, 2) gives [105,141,84,90], all elements divisible by prime(2) = 3.
Adding 1 to T(4, 3) gives [105,140,85,90], all elements divisible by prime(3) = 5.
Adding 1 to T(4, 4) gives [105,140,84,91], all elements divisible by prime(4) = 7.
The sum of elements in row 3 is 15 + 20 + 24 = 59. 59 + 1 = 60, a multiple of A002110(3) = 30.
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row(n) = my(pr=primes(n), p = prod(i=1, #pr, pr[i]), res=vector(n, i, lift(chinese(Mod(-1, pr[i]), Mod(0, p/pr[i]))))); res
Showing 1-3 of 3 results.
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