A229194 Integers nearest to (2^((n-3)/2) + 3^((n-3)/2)).

1, 1, 1, 2, 3, 5, 8, 13, 21, 35, 58, 97, 163, 275, 466, 793, 1353, 2315, 3969, 6817, 11726, 20195, 34816, 60073, 103724, 179195, 309724, 535537, 926275, 1602515, 2773034, 4799353, 8307516, 14381675, 24899377, 43112257, 74651790, 129271235, 223862687, 387682633, 671402698, 1162785755, 2013837368, 3487832977, 6040770648, 10462450355, 18120829034, 31385253913, 54359521280, 94151567435, 163072632198

Offset: 0

Author

Vladimir Pletser, Sep 15 2013

Keywords

• nonn

Comments

This sequence illustrates the second law of small numbers because it is a coincidence that the terms for 1 <= n <= 8 are the same as the Fibonacci numbers F(n) (A000045): a(n) = F(n) for 1 <= n <= 8.

Furthermore, the following terms are the sum of two Fibonacci numbers: a(9) = F(9) + F(2), a(10) = F(10) + F(4), a(11) = F(11) + F(6), a(14) = F(14) + F(11); or the algebraic sum of three Fibonacci numbers: a(12) = F(12) + F(8) - F(3), a(13) = F(13) + F(10) - F(7), a(14) = F(14) + F(12) - F(10), a(18) = F(19) - F(13) - F(8), a(19) = F(20) + F(10) - F(4); or the algebraic sum of four Fibonacci numbers: a(15) = F(15) + F(12) + F(9) + F(5), a(16) = F(16) + F(14) - F(6) - F(4), a(17) = F(18) - F(13) - F(9) - F(3), a(18) = F(18) + F(16) + F(14) + F(8), a(19) = F(19) + F(18) + F(10) - F(3).

Note that, for following values of n, a(n) > F(n+1) for n >= 20.

Remark as well that (2^(1/2) + 3^(1/2)) = 3.14626437... ~= Pi (see A135611).

References

• T. Koshy, Fibonacci and Lucas Numbers with Applications, New York, Wiley-Interscience, 2001
• I. Stewart, L'univers des nombres, Belin-Pour La Science, Paris 2000.

Links

• Vladimir Pletser, Table of n, a(n) for n = 0..1000
• R. K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 3-20.
• I. Stewart, Fibonacci Forgeries
• Eric Weisstein's World of Mathematics, Fibonacci Number.
• Eric Weisstein's World of Mathematics, Strong Law of Small Numbers.

Crossrefs

Cf. A000045, A005181.

Programs

Magma

[Round(2^((n-3)/2) + 3^((n-3)/2)): n in [0..50]]; // Vincenzo Librandi, Sep 20 2013

Maple

seq(round(2^((n-3)/2)+3^((n-3)/2)), n=0..50);

Mathematica

Table[Round[2^((n - 3)/2) + 3^((n - 3)/2)], {n, 0, 50}] (* Vincenzo Librandi, Sep 20 2013 *)

Formula

a(n) = round(2^((n-3)/2) + 3^((n-3)/2)).