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 A004771 a(n) = 8*n + 7. Or, numbers whose binary expansion ends in 111. 28
 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. [Proof. Assume x^2 = 8*k + 7. Then x is odd of the form 2*m + 1. So (2*m + 1)^2 - 7 = 8*k, 4*m^2 + 4*m - 6 = 8*k, 2*m^2 + 2*m - 3 = 4*k or odd = even a contradiction. So the assumption that x^2 = 8*k + 7 is false. - Cino Hilliard, Sep 03 2006 a(n) is the set of numbers congruent to {7, 15, 23} mod 24. - Gary Detlefs, Mar 07 2010 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, ContraharmonicMean[{1, 3, 5, 7, 9, 13, 15, 17, 19, 21}] = 15, ContraharmonicMean[{1, 3, 5, 7, 9, 11, 13, 15, 19, 21, 23, 25, 27, 29, 31, 33}] = 23. - 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 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) = floor((24*n - 2)/3) with offset 1, a(1) = 7. - Gary Detlefs, Mar 07 2010 a(n) = 2*a(n-1) - a(n-2) for n > 1, a(0) = 7, a(1) = 15.  - Vincenzo Librandi, May 28 2011 A056753(a(n)) = 7. - Reinhard Zumkeller, Aug 23 2009 a(n) = t(t(t(n))), where t(i) = 2*i + 1. MAPLE A004771:=n->8*n+7; seq(A004771(n), n=0..100); # Wesley Ivan Hurt, Dec 22 2013 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 (MAGMA) [8*n+7: n in [0..60]]; // Vincenzo Librandi, May 28 2011 (PARI) a(n)=8*n+7 \\ Charles R Greathouse IV, Sep 23 2012 (Haskell) a004771 = (+ 7) . (* 8) a004771_list = [7, 15 ..]  -- Reinhard Zumkeller, Jan 29 2013 (GAP) List([0..60], n->8*n+7); # Muniru A Asiru, Aug 28 2018 CROSSREFS Cf. A008590, A017077, A017137. Cf. A007522 (primes), subsequence of A047522. Cf. A227144, A227146. Sequence in context: A059562 A017149 A133655 * A029724 A194400 A177768 Adjacent sequences:  A004768 A004769 A004770 * A004772 A004773 A004774 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified May 23 05:25 EDT 2019. Contains 323508 sequences. (Running on oeis4.)