A091522 -id:A091522 - cp's OEIS frontend

A091523 Graham-Pollak sequence with initial term 8.

8, 12, 17, 24, 34, 48, 68, 96, 136, 193, 273, 386, 546, 772, 1092, 1545, 2185, 3090, 4370, 6180, 8740, 12360, 17480, 24721, 34961, 49443, 69923, 98886, 139846, 197772, 279692, 395544, 559384, 791089, 1118769, 1582179, 2237539, 3164358

Offset: 1

Eric W. Weisstein, Jan 18 2004

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..500
Th. Stoll, On Families of Nonlinear Recurrences Related to Digits, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.2.
Eric Weisstein's World of Mathematics, Graham-Pollak Sequence

Cf. A001521, A091522.

See A001521.

A100673 A Graham-Pollak-like sequence with cube root instead of square root.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 15, 20, 26, 34, 44, 56, 71, 90, 114, 144, 182, 230, 291, 367, 463, 584, 737, 929, 1171, 1476, 1860, 2344, 2954, 3723, 4691, 5911, 7448, 9385, 11825, 14899, 18772, 23652, 29800, 37546, 47306, 59603, 75096, 94616, 119209, 150195, 189235

Offset: 0

Jonathan Vos Post, Dec 06 2004

When the multiplier in the recurrence is 2 and the recurrence has two terms inside a square root, 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) = a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3)) have corresponding properties." This sequence is the given one, having the multiplier in the recurrence as 2 and three terms inside a cube root and with a(0) = 1. Through n=50, the primes are when n = 1, 2, 3, 6, 13, 20, 21, 24, 25, 31. Through n=50, the semiprimes are when n = 4, 7, 9, 10, 19, 23, 32, 34, 36, 40, 42, 47, 49.

			a(6) = 11 because a(5) = 8; so a(6) = Floor((2*8*(8+1)*(8+2))^(1/3))
= floor(1440^(1/3)) = 11, which happens to be prime.
a(45) = 119209 because a(44) = 94616, so a(45) = Floor((2*94616*(94616+1)*(94616+2))^(1/3)) = floor(1694094050176992^(1/3)) = 119209 = 23 * 71 * 73, which happens to be a 3-brilliant number.

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

Harvey P. Dale, Table of n, a(n) for n = 0..1000
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

Cf. A001521, A091522, A091523.

NestList[Floor[Surd[2#(#+1)(#+2),3]]&,1,50] (* Harvey P. Dale, Feb 24 2016 *)

a(0) = 1, a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3))

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

Original entry on oeis.org

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

Jonathan Vos Post, Dec 06 2004

nonn

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.

			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.

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

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

Cf. A001521, A091522, A091523.

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 *)

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

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

cp's OEIS Frontend

A091523 Graham-Pollak sequence with initial term 8.

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A100673 A Graham-Pollak-like sequence with cube root instead of square root.

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

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