A283425 Difference between A002110(n) and the largest semiprime b*c < A002110(n) where b is prime(n+1).
1, 61, 127, 113, 199, 191, 701, 233, 457, 241, 3701, 557, 3673, 421, 499, 947, 2437, 4349, 8513, 4951, 3229, 937, 4813, 881, 6863, 1499, 2803, 12497, 2029, 88493, 5857, 10853, 28627, 9551, 43691, 85049, 15973, 75209, 4933, 5009, 22613, 14731, 74489, 16993, 90887, 307, 3581, 15083, 12893, 71317, 3583, 1907
Offset: 4
Keywords
Examples
Sequence starts at n=4. For n=5, a(n)=61. Pn(5): a=2310, b=13, c=173, d=61. I.e., d = a - (b*c) = 2310 - (13*173) = 2310 - 2249 = 61. Pn(4): a=210, b=11, c=19, d=1, Pn(5): a=2310, b=13, c=173, d=61, Pn(6): a=30030, b=17, c=1759, d=127, Pn(7): a=510510, b=19, c=26863, d=113, Pn(8): a=9699690, b=23, c=421717, d=199, Pn(9): a=223092870, b=29, c=7692851, d=191. a(n) = a - (b*c) where a(n) has a high probability of being prime, and b*c is the largest semiprime below A002110(n) where b is prime (n+1).
Programs
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Mathematica
Table[Function[{P, q}, P - NextPrime[P/q, -1] q] @@ {Product[Prime@ i, {i, n}], Prime[n + 1]}, {n, 4, 55}] (* Michael De Vlieger, May 15 2017 *)
Formula
a(n) = A002110(n) - A000040(n+1)*prevprime(A002110(n)/A000040(n+1)) for n >= 4. - Michael De Vlieger, May 15 2017
Comments