A077814 a(n) = #{0<=k<=n: mod(k*n,4)=2}.

0, 0, 1, 1, 0, 1, 3, 2, 0, 2, 5, 3, 0, 3, 7, 4, 0, 4, 9, 5, 0, 5, 11, 6, 0, 6, 13, 7, 0, 7, 15, 8, 0, 8, 17, 9, 0, 9, 19, 10, 0, 10, 21, 11, 0, 11, 23, 12, 0, 12, 25, 13, 0, 13, 27, 14, 0, 14, 29, 15, 0, 15, 31, 16, 0, 16, 33, 17, 0, 17, 35, 18, 0, 18, 37, 19, 0, 19, 39, 20, 0, 20, 41, 21, 0

Offset: 0

Author

John W. Layman, Dec 03 2002

Keywords

• nonn
• easy

Comments

Coefficients in the unique expansion of e/4 = Sum_{n>=1} a(n)/n!, where a(n) satisfies 0<=a(n)

Examples

			a(6) = #{1, 3, 5} = 3.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (2,-3,4,-3,2,-1).

Crossrefs

Cf. A009947, A087509, A087620.

Programs

Mathematica

a = Table[0, {i, 1, 50}]; x = Exp[1]/4; For[n = 2, n <= 51, n++, { an = 0; While [(x >= (1/n!)) && (an < (n - 1)), {an++, x = x - (1/n!)} ]}; a[[n - 1]] = an;]; a
LinearRecurrence[{2,-3,4,-3,2,-1},{0,0,1,1,0,1},90] (* Harvey P. Dale, Apr 07 2025 *)

Formula

a(n)=0 if n=4k, a(n)=k if n=4k+1, a(n)=2k+1 if n=4k+2 and a(n)=k+1 if n=4k+3.

a(n) = floor(n!*e/4) - n*floor((n-1)!*e/4). - Benoit Cloitre, Dec 07 2002

a(n) = Sum_{k=0..n} if(mod(k*n, 4)=2, 1, 0). - Paul Barry, Sep 10 2003

O.g.f.: x^2*(1-x+x^2)/((x-1)^2*(1+x^2)^2). - R. J. Mathar, Jun 13 2008

From Wesley Ivan Hurt, May 30 2015: (Start)

a(n) = 2*a(n-1)-3*a(n-2)+4*a(n-3)-3*a(n-4)+2*a(n-5)-a(n-6), n>6.

a(n) = (-1)^((1-2*n-(-1)^n)/4)*((-1)^n-2*n*(-1)^((2*n+3+(-1)^n)/4)+n*(-1)^((1+(-1)^n)/2)+n*(-1)^((2*n+1+(-1)^n)/2)-1)/8. (End)

Extensions

More terms from Benoit Cloitre, Dec 07 2002

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