A004771 - OEIS

A004771

a(n) = 8*n + 7. Or, numbers whose binary expansion ends in 111.

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

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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

Vincenzo Librandi, Table of n, a(n) for n = 0..5000

Tanya Khovanova, Recursive Sequences

Leo Tavares, Illustration: Twin Square Frames

Leo Tavares, Illustration: Mid-line Hexagons

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 962

Index entries for linear recurrences with constant coefficients, signature (2,-1).

FORMULA

O.g.f: (7 + x)/(1 - x)^2 = 8/(1 - x)^2 - 1/(1 - x). - R. J. Mathar, Nov 30 2007

a(n) = 2*a(n-1) - a(n-2) for n >= 2. - Vincenzo Librandi, May 28 2011

A056753(a(n)) = 7. - Reinhard Zumkeller, Aug 23 2009