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A125651 Numbers k such that A125650(k) is a perfect square. +0
5
1, 3, 24, 147, 864, 5043, 29400, 171363, 998784, 5821347, 33929304, 197754483, 1152597600, 6717831123, 39154389144, 228208503747, 1330096633344, 7752371296323, 45184131144600, 263352415571283, 1534930362283104 (list; graph; listen)
OFFSET

1,2

COMMENT

Corresponding numbers m such that m^2 = A125650(a(n)) are listed in A125652.

3 divides a(n) for n>1. For n>1 a(n) = 3*A001108(n-1), where A001108(k) = {0, 1, 8, 49, 288, 1681, ...}, A001108(k)-th triangular number is a square. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jan 19 2007

FORMULA

For n>1, a(n+2) = 6*a(n+1) - a(n) + 6.

For n>1, a(n) = ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))*3/4 - 3/2.

a(2k) = 3*A002315(n)^2; a(2k+1) = 6*A001542(n)^2.

a(n) = 3*A001108(n-1) for n>1. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jan 19 2007

EXAMPLE

a(2)=3 because A125650(3)=9=3^2; a(3)=24 because A125650(24)=81=9^2.

CROSSREFS

Cf. A125650, A125652.

Cf. A001108.

Sequence in context: A069515 A056350 A056344 this_sequence A043017 A003443 A119581

Adjacent sequences: A125648 A125649 A125650 this_sequence A125652 A125653 A125654

KEYWORD

nonn

AUTHOR

Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 29 2006

EXTENSIONS

Edited by Max Alekseyev (maxale(AT)gmail.com), Jan 11 2007

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Last modified December 8 08:31 EST 2009. Contains 170430 sequences.


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