

A169600


Numbers that are congruent to {4, 25} mod 31.


1



4, 25, 35, 56, 66, 87, 97, 118, 128, 149, 159, 180, 190, 211, 221, 242, 252, 273, 283, 304, 314, 335, 345, 366, 376, 397, 407, 428, 438, 459, 469, 490, 500, 521, 531, 552, 562, 583, 593, 614, 624, 645, 655, 676, 686, 707, 717, 738, 748, 769, 779, 800, 810, 831, 841, 862, 872, 893, 903, 924, 934
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OFFSET

1,1


COMMENTS

Conjecture: For no n >4 in the sequence, 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is of the form p*(p+2), where p and p+2 are primes.
The conjecture is evident, it can be proved as in A169599. [Bruno Berselli, Jan 07 2013]


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,1).


FORMULA

a(n) = (62*n + 11*(1)^n  35)/4.  Vincenzo Librandi, Jan 06 2013, modified Jul 07 2015
a(n) = a(n1) + a(n2)  a(n3).  Vincenzo Librandi, Jan 06 2013
G.f.: x*(4+21*x+6*x^2) / ( (1+x)*(x1)^2 ).  R. J. Mathar, Jul 07 2015


MATHEMATICA

Select[Range[934], MemberQ[{4, 25}, Mod[#, 31]]&] (* Ray Chandler, Jul 08 2015 *)
LinearRecurrence[{1, 1, 1}, {4, 25, 35}, 61] (* Ray Chandler, Jul 08 2015 *)
Rest[CoefficientList[Series[x*(4+21*x+6*x^2)/((1+x)*(x1)^2), {x, 0, 61}], x]] (* Ray Chandler, Jul 08 2015 *)


CROSSREFS

Sequence in context: A303183 A175052 A240164 * A267765 A231176 A199772
Adjacent sequences: A169597 A169598 A169599 * A169601 A169602 A169603


KEYWORD

nonn,easy


AUTHOR

Vincenzo Librandi, Dec 03 2009


EXTENSIONS

Added missing leading terms. Clarified the comment.  R. J. Mathar, Jul 07 2015


STATUS

approved



