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A181437
Size of the longest increasing sequence of primes starting with 2, 3 and with second-order differences bounded by n.
1
4, 57, 57, 421, 421, 1860, 1860, 24661, 24661, 380028, 380028, 2964603, 2964603
OFFSET
1,1
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
EXAMPLE
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.
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 *)
CROSSREFS
Sequence in context: A060497 A092273 A193745 * A353001 A156873 A320977
KEYWORD
nonn,nice,more
AUTHOR
Esteban Crespi de Valldaura, Jan 31 2011, Feb 03 2011
STATUS
approved