A187749
Maximum k for which a partition of the product of the first k primes (the k-th primorial) into the sum of two products gives a sum that yields only n composites from that sum down to 1+1=2 by sequential deletion of the largest prime.
Original entry on oeis.org
17, 20, 24, 27, 31, 33, 35, 40, 41, 42, 44, 49, 50, 54
Offset: 0
a(0): Both 2*5*11*23*37*41*43*59 + 3*7*13*17*19*29*31*47*53 and 2*3*5*11*13*19*29*41*43*47*53 + 7*17*23*31*37*59 are primes such that repeatedly removing the largest prime down to the sum 1 + 1 yields a prime, and these are the only examples for the primes through prime(17)=59. That is, for either of the above cases, neither portion may be multiplied by 61 to give a prime sum. Therefore, a(0) = 17, and this means -- identically -- that none of the primes counted in A103787(18) naturally link by deletion to primes counted in A103787(k) in a chain from k equals 17 down to 1, but that at least one of the primes counted in A103787(17) -- both of the above -- do link in such a way to primes counted in A103787(k) for k equals 16 down to 1.
a(2): 2*3*5*11*13*19*29*41*43*47*53*59*71*79*89 + 7*17*23*31*37*61*67*73*83 is prime and removing the primes sequentially yields a composite only after removal of 79 and then of 61. This twosome of composites is uniquely fewest for partitions of this many of the first primes (24), and no such sum that includes the next prime (97) yields fewer than 3 composites by deletion. So, a(2) = 24.
A293756
a(n) = smallest number k with n prime factors such that d + k/d is prime for every d | k.
Original entry on oeis.org
1, 2, 6, 30, 210, 186162
Offset: 0
a(2) = 2*3 = 6 because k = 6 is the smallest number with 2 prime factors such that for d = {1, 2, 3, 6} we have 1 + 6/1 = 6 + 6/6 = 7 is prime and 2 + 6/2 = 3 + 6/3 = 5 is prime.
From _Michael De Vlieger_, Nov 14 2017: (Start)
First differences of prime indices of a(n):
n a(n) A287352(a(n))
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1 2 1
2 6 1, 1
3 30 1, 1, 1
4 210 1, 1, 1, 1
5 186162 1, 1, 6, 1, 11
(End)
Subsequence of
A080715 (d + k/d is prime for every d|k).
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with(numtheory): P:=proc(q) local a,b,j,k,n,ok; print(1);for n from 1 to q do for k from 2 to q do a:=ifactors(k)[2]; a:=add(a[j][2],j=1..nops(a)); if a=n then b:=divisors(k); ok:=1;
for j from 1 to nops(b) do if not isprime(b[j]+k/b[j]) then ok:=0; break; fi; od; if ok=1 then print(k); break; fi; fi; od; od; end: P(10^8); # Paolo P. Lava, Nov 16 2017
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isok(k, n)=if (!issquarefree(k), return (0)); if (omega(k) != n, return (0)); fordiv(k, d, if (!isprime(d+k/d), return(0))); 1;
a(n) = {my(k=1); while( !isok(k, n), k++); k;} \\ Michel Marcus, Nov 11 2017
A103788
a(n) = number of ks that make primorial P(n)/A019565(k)-A019565(k) prime.
Original entry on oeis.org
0, 1, 3, 6, 13, 28, 39, 78, 138, 207, 437, 865, 1423, 2750, 4904, 8861, 16201, 33346, 58534, 111878, 208914, 397522
Offset: 1
P(2)/A(0)-A(0)=6-1=5 is prime, so a(2)=1;
P(4)/A(k)-A(k): 210/2-2=103; 210/3-3=67; 210/6-6=29; 210/5-5=37; 210/10-10=11; 210/7-7=23; so a(4)=6;
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npd = 1; Do[npd = npd*Prime[n]; tn = 0; tt = 1; cp = npd/tt - tt; ct = 0; While[IntegerQ[cp], If[(cp > 0) && PrimeQ[cp], ct = ct + 1]; tn = tn + 1; tt = 1; k1 = tn; o = 1; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, tt = tt*Prime[o]]; k1 = ( k1 - k2)/2; o = o + 1]; cp = npd/tt - tt]; Print[ct], {n, 1, 22}]
A331665
Numbers k with a record number of divisors d < sqrt(k) such that d + k/d is prime.
Original entry on oeis.org
1, 2, 6, 30, 210, 2310, 3570, 4830, 11550, 30030, 43890, 111930, 131670, 510510, 690690, 870870, 1021020, 2459730, 9699690, 13123110, 17160990, 40750710, 146006070, 223092870, 340510170, 358888530, 688677990, 1462190730, 2445553110, 2911018110, 6469693230
Offset: 1
2 has one divisor below sqrt(2), 1, such that 1 + 2/1 = 3 is prime.
6 has 2 divisors below sqrt(6), 1 and 2, such that 1 + 6/1 = 7 and 2 + 6/2 = 5 are primes.
30 has 4 divisors below sqrt(30), 1, 2, 3, and 5 such that 1 + 30/1 = 31, 2 + 30/2 = 17, 3 + 30/3 = 13 and 5 + 30/5 = 11 are primes.
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divCount[n_] := DivisorSum[n, Boole @ PrimeQ[# + n/#] &, #^2 < n &]; seq = {}; dm = -1; Do[d1 = divCount[n]; If[d1 > dm, dm = d1; AppendTo[seq, n]], {n, 1,10^6}]; seq
Showing 1-4 of 4 results.
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