A100671 A Graham-Pollak-like sequence with multiplier 3 instead of 2.

1, 2, 4, 7, 12, 21, 37, 64, 111, 193, 335, 581, 1007, 1745, 3023, 5236, 9069, 15708, 27207, 47124, 81622, 141374, 244867, 424122, 734601, 1272367, 2203805, 3817103, 6611417, 11451311, 19834253, 34353934, 59502759, 103061802, 178508278, 309185407, 535524834

Offset

0,2

Comments

When the multiplier in the recurrence is 2, we have the Graham-Pollak sequence, where there is a remarkable exact explicit formula for a(n) in terms of the union of the set of integers and the set of integer multiples of Sqrt(2). As Weisstein summarizes Borwein & Bailey: "It is not known if sequences such as a(n) = Floor(Sqrt(3*a(n-1)*(a(n-1)+1))) have corresponding properties." This sequence is the given one, with a(0) = 1. Through n=40, the primes are when n = 1, 3, 6, 9, 14, 25, 28, 29. Through n=40, the semiprimes are when n = 2, 5, 8, 10, 11, 12, 13, 16, 21, 22, 26, 27, 30, 31, 38.

References

Borwein, J. and Bailey, D., Mathematics by Experiment: Plausible Reasoning in the 21st Century. Natick, MA: A. K. Peters, 2003.

Links

Table of n, a(n) for n=0..36.

R. L. Graham and H. O. Pollak, Note on a nonlinear recurrence related to sqrt(2), Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705.

Eric Weisstein's World of Mathematics, Graham-Pollak sequence

Formula

a(0) = 1, a(n) = Floor(Sqrt(3*a(n-1)*(a(n-1)+1))).

Example

a(9) = 193 because a(8) = 111; so a(9) = Floor(Sqrt(3*111*(111+1))) = floor(sqrt(37296)) = 193, which happens to be prime.

Mathematica

RecurrenceTable[{a[0]==1, a[n]==Floor[Sqrt[3a[n-1](a[n-1]+1)]]}, a[n], {n, 40}] (* Harvey P. Dale, Sep 10 2011 *)

Crossrefs

Cf. A001521, A091522, A091523.

Sequence in context: A227376 A245531 A189593 * A189600 A005251 A014167

Adjacent sequences: A100668 A100669 A100670 * A100672 A100673 A100674

Keyword

nonn

Author

Jonathan Vos Post, Dec 06 2004

Extensions

Corrected and extended by Harvey P. Dale, Sep 10 2011

Status

approved