|
| |
|
|
A055792
|
|
a(n) and floor(a(n)/2) are both squares; i.e. squares which remain squares when written in base 2 and last digit is removed.
|
|
32
|
|
|
|
0, 1, 9, 289, 9801, 332929, 11309769, 384199201, 13051463049, 443365544449, 15061377048201, 511643454094369, 17380816062160329, 590436102659356801, 20057446674355970889, 681362750825443653409, 23146276081390728245001, 786292024016459316676609
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
a(n) > 0 is a square such that a(n) - 1 is a product of powers. - Michel Lagneau, Feb 16 2012
|
|
|
LINKS
|
Charles R Greathouse IV, Table of n, a(n) for n = 0..654
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).
|
|
|
FORMULA
|
a(n) = 34*a(n-1) - a(n-2) - 16 = A001541(n-1)^2 = 2*A001542(n-1)^2 + 1 = 8*A001110(n-1) + 1.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3. G.f.: -x*(9*x^2-26*x+1) / ((x-1)*(x^2-34*x+1)). - Colin Barker, Sep 15 2014
a(n) = c*k^n + 1/2 + o(1) with k = 17+sqrt(288) = 33.97... and c = 17/4 - sqrt(18). - Charles R Greathouse IV, May 07 2015
a(n) = (4 + 2*(17+12*sqrt(2))^(1-n) + (34-24*sqrt(2))*(17+12*sqrt(2))^n)/8 for n>0. - Colin Barker, Mar 02 2016
|
|
|
EXAMPLE
|
a(2) = 9 because 9 = 3^2 = 1001 base 2 and 100 base 2 = 4 = 2^2.
|
|
|
PROG
|
(PARI) concat(0, Vec(-x*(9*x^2-26*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100))) \\ Colin Barker, Sep 15 2014
(PARI) is(n)=issquare(n) && issquare(n\2) \\ Charles R Greathouse IV, May 07 2015
|
|
|
CROSSREFS
|
Cf. A023110, A247375.
Sequence in context: A110553 A078326 A118893 * A053935 A086699 A027834
Adjacent sequences: A055789 A055790 A055791 * A055793 A055794 A055795
|
|
|
KEYWORD
|
nonn,base,easy
|
|
|
AUTHOR
|
Henry Bottomley, Jul 14 2000
|
|
|
STATUS
|
approved
|
| |
|
|
