A218129 2^(((c - 2)^2 + (c - 2))/2) + n = a(n), where c are the positive solutions to {y in N | 2*cos(2*Pi/y) is in Z}; c = {1,2,3,4,6}.

1, 2, 4, 11, 1028

Offset

0,2

Comments

The set {c} consists of the complete set of positive solutions to the short proof of the Crystallographic Restriction Theorem {1, 2, 3, 4, 6} (see A217290).

Let {V} = prime(a(n)) = {2, 3, 7, 31, 8191}. Then all elements of {V} follow form x^2 + x + 1 for some x in R; x = {(sqrt(5) - 1)/2, 1, 2, 5, 90}. (V + V(mod 2) - 2)/2 gives the complete set of Ramanujan-Nagell triangular numbers (A076046) = {0, 1, 3, 15, 4095} == (2^F_(c + 1) - 2)/2 (see A215929); F_n the n-th Fibonacci number (A000045).

Additionally, 2*V - 1 = {3, 5, 13, 61, 16381} is prime and, therefore, all elements of {V} are links in a Cunningham chain of the 2nd kind (see A005382).

Links

Table of n, a(n) for n=0..4.

Eric W. Weisstein, MathWorld: Ramanujan's Square Equation

Wikipedia, Crystallographic Restriction Theorem

Formula

to n = 4, then a(n) = 2^(a(n - 1) - 1) + n; a(-1) = 1.

Example

2^(((1 - 2)^2 + (1 - 2))/2) + 0 = 2^(a(-1) - 1) + 0 = 1 = a(0).
2^(((2 - 2)^2 + (2 - 2))/2) + 1 = 2^(a(0) - 1) + 1 = 2 = a(1).
2^(((3 - 2)^2 + (3 - 2))/2) + 2 = 2^(a(1) - 1) + 2 = 4 = a(2).
2^(((4 - 2)^2 + (4 - 2))/2) + 3 = 2^(a(2) - 1) + 3 = 11 = a(3).
2^(((6 - 2)^2 + (6 - 2))/2) + 4 = 2^(a(3) - 1) + 4 = 1028 = a(4).

Crossrefs

Cf. A217290, A215929, A076046.

Sequence in context: A057284 A156463 A227530 * A156519 A215795 A070314

Adjacent sequences: A218126 A218127 A218128 * A218130 A218131 A218132

Keyword

nonn,fini,full

Author

Raphie Frank, Oct 21 2012

Status

approved