A047209 Numbers that are congruent to {1, 4} mod 5.

1, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 109, 111, 114, 116, 119, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

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

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

These numbers appear in the product of a Rogers-Ramanujan identity. See A003114 also for references. - Wolfdieter Lang, Oct 29 2016

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

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
• William A. Stein, The modular forms database.
• Eric Weisstein's World of Mathematics, Determined by Spectrum.
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A000566, A036666, A001622, A003114, A203776, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887, A352324.

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

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

Programs

Haskell

a047209 = (flip div 2) . (subtract 2) . (* 5)
a047209_list = 1 : 4 : (map (+ 5) a047209_list)
-- Reinhard Zumkeller, Jul 19 2013, Jan 05 2011

Maple

seq(floor(5*k/2)-1, k=1..100); # Wesley Ivan Hurt, Sep 27 2013

Mathematica

Select[Range[0, 200], MemberQ[{1, 4}, Mod[#, 5]] &] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
LinearRecurrence[{1,1,-1},{1,4,6},70] (* Harvey P. Dale, Jul 19 2024 *)

PARI

a(n)=(10*n+(-1)^n-5)/4 \\ Charles R Greathouse IV, Sep 24 2015

Formula

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

a(n) = floor((5n-2)/2). [corrected by Reinhard Zumkeller, Jul 19 2013]

a(1) = 1, a(n) = 5(n-1) - a(n-1). - Benoit Cloitre, Apr 12 2003

From Bruno Berselli, Nov 17 2010: (Start)

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

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

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

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

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

a(n) = 5*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011

E.g.f.: 1 + ((10*x - 5)*exp(x) + exp(-x))/4. - David Lovler, Aug 23 2022

From Amiram Eldar, Nov 22 2024: (Start)

Product_{n>=1} (1 - (-1)^n/a(n)) = phi (A001622).

Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/5) * cosec(Pi/5) (A352324). (End)

Extensions

Edited by Michael Somos, Sep 22 2002