|
| |
|
|
A156619
|
|
Numbers n that are 7 or 18 mod 25.
|
|
1
| |
|
|
7, 18, 32, 43, 57, 68, 82, 93, 107, 118, 132, 143, 157, 168, 182, 193, 207, 218, 232, 243, 257, 268, 282, 293, 307, 318, 332, 343, 357, 368, 382, 393, 407, 418, 432, 443, 457, 468, 482, 493, 507, 518, 532, 543, 557, 568, 582, 593, 607, 618, 632, 643, 657, 668
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,1
|
|
|
COMMENTS
| Also, numbers n such that n^2+1=0 mod 25.
Also, if a(1)=7, a(2)=18, a(n)=2*a(n-1)-a(n-2)-3 (if n is even); a(n)=2*a(n-1)-a(n-2)+3 (if n is odd); example: a(3)=2*18-7+3=32; a(4)=2*32-18-3=43; a(5)=2*43-32+3=57; a(6)=2*57-43-3=68; the sequence (7,18,32,43,57,68,82,93) repeat to (107,118,..,) (207,218,..,) (307,318,..,) (407,418,...,) and so on
Numbers of the form 25*n+7 or 25*n+18. Numbers b such that 25 is a base-b Euler pseudoprime. - Karsten Meyer, Jan 05 2011,
|
|
|
LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (1,1,-1).
|
|
|
FORMULA
| n^2+1=0 mod (5^2)
a(n)=a(n-1)+a(n-2)-a(n-3) = 25*n/2-25/4-3*(-1)^n/4. G.f.: x(7+11x+7x^2)/((1+x)(1-x)^2). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 19 2009]
|
|
|
MATHEMATICA
| fQ[n_] := Mod[n^2 + 1, 25] == 0; Select[ Range@ 670, fQ]
|
|
|
CROSSREFS
| Sequence in context: A103571 A103572 A049532 * A033537 A000566 A169677
Adjacent sequences: A156616 A156617 A156618 * A156620 A156621 A156622
|
|
|
KEYWORD
| nonn,easy
|
|
|
AUTHOR
| Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Feb 11 2009
|
| |
|
|
