A181437 Size of the longest increasing sequence of primes starting with 2, 3 and with second-order differences bounded by n.

4, 57, 57, 421, 421, 1860, 1860, 24661, 24661, 380028, 380028, 2964603, 2964603

Offset: 1

Author

Esteban Crespi de Valldaura, Jan 31 2011, Feb 03 2011

Keywords

• nonn
• nice
• more

Comments

For n > 1, a(2n + 1) = a(2n) because after the primes (2,3,5), all second differences are even numbers. It is not known if the sequence remains finite for all n.

To give an idea of the size of the sequences, the largest prime in the sequence corresponding to a(12) is 1280522207.

For a given n, the Mathematica program uses recursion to find the longest list of primes with second differences bounded by n. This is not the best language to use for this problem. Even n=4 takes quite a while. The program prints longer lists as they are found. - T. D. Noe, Feb 03 2011

Examples

			For n = 1, the sequence is (2, 3, 5, 7) which has 4 members, so a(1) = 4.
For n = 2, a sequence is (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 37, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 173, 197, 223, 251, 281, 313, 347, 383, 421, 461, 503, 547, 593, 641, 691, 743, 797, 853, 911, 971, 1033, 1097, 1163, 1231, 1301, 1373, 1447, 1523, 1597, 1669, 1741, 1811, 1879, 1949, 2017, 2083) which has 57 members, so a(2) = 57.
		

Links

• MathOverflow, Is there an infinite increasing sequence of primes with bounded second or larger differences
• Esteban Crespi de Valldaura, C++ program used in computations.

Programs

Mathematica

Valid[lst_List]:=Module[{r1,r2}, r1=-n+2*lst[[-1]]-lst[[-2]]; r2=r1+2*n; r1=Max[lst[[-1]]+1,r1]; Select[Range[r1,r2],PrimeQ]]; FindPath[lst_List]:=Module[{p,ln}, p=Valid[lst]; If[p=={}, ln=Length[lst]; If[ln>len, len=ln; Print[{len,lst}]], Do[FindPath[Append[lst,i]], {i,p}]]]; n=2; len=0; t={2,3}; FindPath[t]; len (* T. D. Noe, Feb 03 2011 *)