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Numbers that are congruent to {1, 4} mod 5.
54

%I #112 Aug 10 2024 01:30:21

%S 1,4,6,9,11,14,16,19,21,24,26,29,31,34,36,39,41,44,46,49,51,54,56,59,

%T 61,64,66,69,71,74,76,79,81,84,86,89,91,94,96,99,101,104,106,109,111,

%U 114,116,119,121,124,126,129,131,134,136,139,141,144,146,149,151,154

%N Numbers that are congruent to {1, 4} mod 5.

%C Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 72 ).

%C Cf. property described by _Gary Detlefs_ in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 5). - _Bruno Berselli_, Nov 17 2010

%C The sum of the alternating series (-1)^(n+1)/a(n) from n=1 to infinity is (Pi/5)*cot(Pi/5), that is (1/5)*sqrt(1 + 2/sqrt(5))*Pi. - _Jean-François Alcover_, May 03 2013

%C These numbers appear in the product of a Rogers-Ramanujan identity. See A003114 also for references. - _Wolfdieter Lang_, Oct 29 2016

%C Let m be a product of any number of terms of this sequence. Then m - 1 or m + 1 is divisible by 5. Closed under multiplication. - _David A. Corneth_, May 11 2018

%H Reinhard Zumkeller, <a href="/A047209/b047209.txt">Table of n, a(n) for n = 1..1000</a>

%H William A. Stein, <a href="http://wstein.org/Tables/">The modular forms database</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DeterminedbySpectrum.html">Determined by Spectrum</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F G.f.: (1+3x+x^2)/((1-x)(1-x^2)).

%F a(n) = floor((5n-2)/2). [corrected by _Reinhard Zumkeller_, Jul 19 2013]

%F a(1) = 1, a(n) = 5(n-1) - a(n-1). - _Benoit Cloitre_, Apr 12 2003

%F From _Bruno Berselli_, Nov 17 2010: (Start)

%F a(n) = (10*n + (-1)^n - 5)/4.

%F a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.

%F a(n) = a(n-2) + 5 for n > 2.

%F a(n) = 5*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.

%F a(n)^2 = 5*A036666(n) + 1 (cf. also Comments). (End)

%F a(n) = 5*floor(n/2) + (-1)^(n+1). - _Gary Detlefs_, Dec 29 2011

%F E.g.f.: 1 + ((10*x - 5)*exp(x) + exp(-x))/4. - _David Lovler_, Aug 23 2022

%p seq(floor(5*k/2)-1, k=1..100); # _Wesley Ivan Hurt_, Sep 27 2013

%t Select[Range[0, 200], MemberQ[{1, 4}, Mod[#, 5]] &] (* _Vladimir Joseph Stephan Orlovsky_, Feb 12 2012 *)

%t LinearRecurrence[{1,1,-1},{1,4,6},70] (* _Harvey P. Dale_, Jul 19 2024 *)

%o (Haskell)

%o a047209 = (flip div 2) . (subtract 2) . (* 5)

%o a047209_list = 1 : 4 : (map (+ 5) a047209_list)

%o -- _Reinhard Zumkeller_, Jul 19 2013, Jan 05 2011

%o (PARI) a(n)=(10*n+(-1)^n-5)/4 \\ _Charles R Greathouse IV_, Sep 24 2015

%Y Cf. A000566, A036666, A003114, A203776, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887.

%Y Cf. A005408 (n=1 or 3 mod 4), A007310 (n=1 or 5 mod 6).

%Y Cf. A045468 (primes), A032527 (partial sums).

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_

%E Edited by _Michael Somos_, Sep 22 2002