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Branch term s_n(b), b > 1 of equivalence classes of prime sequences {s_n(k)} for k > 0 derived by records of first differences of Rowland-like recurrences with increasing even start values >= 4.
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%I #29 Oct 30 2017 12:54:24

%S 0,0,0,0,131,0,233,167,2381,647,0,233,0,941,263,0,0,353,0,0,797,0,0,0,

%T 941,0,0,8273,569,0,0,569,1181,0,0,22133,761,0,761,1721,839,1811,881,

%U 0,1811,929,1973,0,0,1049,1181,9323,2309,1187,0,2441,2441

%N Branch term s_n(b), b > 1 of equivalence classes of prime sequences {s_n(k)} for k > 0 derived by records of first differences of Rowland-like recurrences with increasing even start values >= 4.

%C See A291528 (leaves) for equivalence classes.

%C If the conjecture of an inverse tree of primes with the leaves in A291528 using the same index n holds, except a(1)=0 all terms a(n) == 0 indicates that the branch point is not yet found.

%C This is a k-ary tree, k > 2, such as a(7) == a(12) == 233.

%C Maybe these simple Rowland-like recurrences giving all primes are related to a simple bounded physical quantum system with a "Hamiltonian for the zeros of the Riemann zeta function" (cf. Bender et al.) having degenerated energy eigenvalues a(n).

%C [Note: the editors feel that any such connection is extremely unlikely. - _N. J. A. Sloane_, Oct 30 2017]

%H Carl M. Bender, Dorje C. Brody, Markus P. Müller, <a href="https://arxiv.org/abs/1608.03679">Hamiltonian for the zeros of the Riemann zeta function</a>, arXiv:1608.03679 [quant-ph], 2016.

%F a(n) > A291528(n) || a(n) == 0.

%e n=5: Some equivalence classes of prime sequences {s_n(k)} have the same tail for a constant C_n < k, such as {s_2(k)} = {7,13,29,59,131,...} and {s_5(k)} = {31,61,131,...} with common tail {a(5),...} = {131,...} and the branch 131 = a(5). Thus it seems that all terms != 0 are branches of a kind of an inverse prime-tree with the root at infinity.

%t For[i = 2; pl = {}; fp = {}; bp = {}, i < 350, i++,

%t ps = Union@FoldList[Max, 1, Rest@# - Most@#] &@

%t FoldList[#1 + GCD[#2, #1] &, 2 i, Range[2, 10^5]];

%t p = Select[ps, (i <= #) && ! MemberQ[pl, #] &, 1];

%t If[p != {},

%t fp = Join[fp, {p}];

%t b = Select[Drop[ps, po = Position[ps, p[[1]]][[1]][[1]]],

%t MemberQ[pl, #] &, 1];

%t If[b != {}, bp = Join[bp, {b}], bp = Join[bp, {{0}}]];

%t pl = Union[pl, Drop[ps, po - 1]]]]; Flatten@bp

%Y Cf. A134162, A134734, A167168 (equivalence classes), A291528 (leaves)

%K nonn

%O 1,5

%A _Ralf Steiner_, Aug 28 2017