|
|
A004771
|
|
a(n) = 8*n + 7. Or, numbers whose binary expansion ends in 111.
|
|
42
|
|
|
7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 119, 127, 135, 143, 151, 159, 167, 175, 183, 191, 199, 207, 215, 223, 231, 239, 247, 255, 263, 271, 279, 287, 295, 303, 311, 319, 327, 335, 343, 351, 359, 367, 375, 383, 391, 399, 407, 415, 423, 431
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
These numbers cannot be expressed as the sum of 3 squares. - Artur Jasinski, Nov 22 2006
These numbers cannot be perfect squares. - Cino Hilliard, Sep 03 2006
a(n-2), n >= 2, appears in the second column of triangle A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014
The initial terms 7, 15, 23, 31 are the generating set for the rest of the sequence in the sense that, by Lagrange's Four Square Theorem, any number n of the form 8*k+7 can always be written as a sum of no fewer than four squares, and if n = a^2 + b^2 + c^2 + d^2, then (a mod 4)^2 + (b mod 4)^2 + (c mod 4)^2 + (d mod 4)^2 must be one of 7, 15, 23, 31. - Walter Kehowski, Jul 07 2014
Define a set of consecutive positive odd numbers {1, 3, 5, ..., 12*n + 9} and skip the number 6*n + 5. Then the contraharmonic mean of that set gives this sequence. For example, ContraharmonicMean[{1, 3, 7, 9}] = 7. - Hilko Koning, Aug 27 2018
Jacobi symbol (2, a(n)) = Kronecker symbol (a(n), 2) = 1. - Jianing Song, Aug 28 2018
|
|
LINKS
|
|
|
FORMULA
|
O.g.f: (7 + x)/(1 - x)^2 = 8/(1 - x)^2 - 1/(1 - x). - R. J. Mathar, Nov 30 2007
a(n) = t(t(t(n))), where t(i) = 2*i + 1.
E.g.f.: exp(x)*(7 + 8*x).
|
|
MAPLE
|
|
|
MATHEMATICA
|
8 Range[0, 60] + 7 (* or *) Range[7, 500, 8] (* or *) Table[8 n + 7, {n, 0, 60}] (* Bruno Berselli, Dec 28 2016 *)
|
|
PROG
|
(Haskell)
a004771 = (+ 7) . (* 8)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|