|
| |
|
|
A031150
|
|
Appending a digit to n^2 gives another perfect square.
|
|
10
|
|
|
|
1, 2, 4, 5, 6, 12, 18, 43, 80, 154, 191, 228, 456, 684, 1633, 3038, 5848, 7253, 8658, 17316, 25974, 62011, 115364, 222070, 275423, 328776, 657552, 986328, 2354785, 4380794, 8432812, 10458821, 12484830, 24969660, 37454490
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Square root of 'Squares from A023110 with last digit removed'.
One could include an initial '0', and even list it with multiplicity 3 or 4, since 00, 01, 04 and 09 are all perfect squares: In analogy to corresponding sequences for other bases, this sequence could be defined as sqrt(floor[A023110/10]), see A204512 [base 8], A204517 (base 7), A204519 (base 6), A204521 [base 5], A001353 [base 3], A001542 [base 2]. (For bases 4 and 9, the corresponding sequence contains all integers.) - M. F. Hasler, Jan 16 2012
|
|
|
REFERENCES
|
R. K. Guy, Neg and Reg, preprint, Jan 2012.
|
|
|
LINKS
|
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,38,0,0,0,0,0,0,-1).
|
|
|
FORMULA
|
G.f.: x*(x^10+2*x^9+4*x^8+5*x^7+18*x^6+12*x^5+6*x^4+5*x^3+4*x^2+2*x+1) / (x^14-38*x^7+1). - Colin Barker, Jan 30 2013
|
|
|
EXAMPLE
|
5^2 = 25 and 16^2 = 256, so 5 is in the sequence.
115364^2 = 13308852496, 364813^2 = 133088524969.
|
|
|
MAPLE
|
for i from 1 to 150000 do if (floor(sqrt(10 * i^2 + 9)) > floor(sqrt(10 * i^2))) then print(i) end if end do;
|
|
|
MATHEMATICA
|
CoefficientList[Series[(x^10 + 2 x^9 + 4 x^8 + 5 x^7 + 18 x^6 + 12 x^5 + 6 x^4 + 5 x^3 + 4 x^2 + 2 x + 1)/(x^14 - 38 x^7 + 1), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 19 2013 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 38, 0, 0, 0, 0, 0, 0, -1}, {1, 2, 4, 5, 6, 12, 18, 43, 80, 154, 191, 228, 456, 684}, 40] (* Harvey P. Dale, Jun 09 2017 *)
|
|
|
CROSSREFS
|
See A202303 for the resulting squares.
|
|
|
KEYWORD
|
nonn,base,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
