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A139398 a(n) = Sum_{k >= 0} binomial(n,5*k). 17
1, 1, 1, 1, 1, 2, 7, 22, 57, 127, 254, 474, 859, 1574, 3004, 6008, 12393, 25773, 53143, 107883, 215766, 427351, 843756, 1669801, 3321891, 6643782, 13333932, 26789257, 53774932, 107746282, 215492564, 430470899, 859595529, 1717012749, 3431847189, 6863694378 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,6
COMMENTS
From Gary W. Adamson, Mar 13 2009: (Start)
M^n * [1,0,0,0,0] = [a(n), A139761(n), A139748(n), A139714(n), A133476(n)]
where M = the 5 X 5 matrix [1,1,0,0,0; 0,1,1,0,0; 0,0,1,1,0; 0,0,0,1,1; 1,0,0,0,1]
Sum of terms = 2^n. Example: M^6 * [1,0,0,0,0] = [7, 15, 20, 15, 7]; sum = 2^6 = 64. (End)
{A139398, A133476, A139714, A139748, A139761} is the difference analog of the hyperbolic functions of order 5, {h_1(x), h_2(x), h_3(x), h_4(x), h_5 (x)}. For a definition see the reference "Higher Transcendental Functions" and the Shevelev link. - Vladimir Shevelev, Jun 14 2017
REFERENCES
A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Ch. 18.
LINKS
John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv preprint arXiv:1610.09361 [math.NT], 2016.
FORMULA
G.f.: -(x-1)^4/((2*x-1)*(x^4-2*x^3+4*x^2-3*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
E.g.f.: (exp(z)^2+2*exp(3/4*z+1/4*z*sqrt(5))*cos(1/4*z*sqrt(2)*sqrt(5+sqrt(5)))+ 2*exp(3/4*z-1/4*z*sqrt(5))*cos(1/4*z*sqrt(2)*sqrt(5-sqrt(5))))/5. - Peter Luschny, Jul 10 2012
a(n) = (2^n + sqrt(5)*(cos(Pi*n/5) - (-1)^n*cos(2*Pi*n/5))*A000045(n) + (cos(Pi*n/5) + (-1)^n*cos(2*Pi*n/5))*A000032(n))/5. - Vladimir Reshetnikov, Oct 04 2016
From Vladimir Shevelev, Jun 17 2017: (Start)
a(n) = round((2/5)*(2^(n-1) + phi^n*cos(Pi*n/5))), where phi is the golden ratio and round(x) is the integer nearest to x.
The formula follows from the identity a(n)=1/5*Sum_{j=1..5}((omega_5)^j + 1)^n, where omega_5=exp(2*Pi*i)/5 (cf. Theorem 1 of [Shevelev] link for i=1, n=5, m:=n). Further note that for a=cos(x)+i*sin(x), a+1 = 2*cos ^2 (x/2) + i*sin(x), and for the argument y of a+1 we have tan(y)=tan(x/2) and r^2 = 4*cos^4(x/2) + sin^2(x) = 4*cos^2(x/2). So (a+1)^n = (2*cos(x /2))^n*(cos(n*x/2) + i*sin(n*x/2)). Using this, for x=2*Pi/5, we have (omega_5+1)^n = phi^n(cos(Pi*n/5) + i*sin(Pi*n/5)). Since (omega_5)^4+1=(1+omega_5)/omega_5, we easily find that ((omega_5)^4+1)^n is conjugate to (omega_5+1)^n. So (omega_5+1)^n+((omega_5)^4+1)^n = phi^n*cos(Pi*n/5). Further, we similarly obtain that (omega_5)^2+1 is conjugate to (omega_5) ^3+1=(1+(omega_5)^2)/(omega_5)^2 and ((omega_5)^2+1)^n +((omega_5)^3+1)^n = 2*(sqrt(2-phi))^n*cos(2*Pi*n/5). The absolute value of the latter <= 2*(2-phi)^(n/2) and quickly tends to 0. Finally, ((omega_5)^5+1)^n=2^n, and the formula follows. (End)
a(n+m) = a(n)*a(m) + H_2(n)*H_5(m) + H_3(n)*H_4(m) + H_4(n)*H_3(m) + H_5(n)*H_2(m), where H_2=A133476, H_3=A139714, H_4=A139748, H_5=A139761. - Vladimir Shevelev, Jun 17 2017
MAPLE
f:=(n, r, a) -> add(binomial(n, r*k+a), k=0..n); fs:=(r, a)->[seq(f(n, r, a), n=0..40)];
A139398_list := proc(n) local i; (exp(z)^2+2*exp(3/4*z+1/4*z*sqrt(5))* cos(1/4*z*sqrt(2)*sqrt(5+sqrt(5)))+2*exp(3/4*z-1/4*z*sqrt(5))* cos(1/4*z*sqrt(2)*sqrt(5-sqrt(5))))/5; series(%, z, n+2): seq(simplify(i!*coeff(%, z, i)), i=0..n) end: A139398_list(35); # Peter Luschny, Jul 10 2012
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 2}, {1, 1, 1, 1, 1}, 40] (* Harvey P. Dale, Jun 11 2015 *)
Expand@Table[(2^n + Sqrt[5] (Cos[Pi n/5] - (-1)^n Cos[2 Pi n/5]) Fibonacci[n] + (Cos[Pi n/5] + (-1)^n Cos[2 Pi n/5]) LucasL[n])/5, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 04 2016 *)
PROG
(Magma) [n le 5 select 1 else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+2*Self(n-5): n in [1..40]]; // Vincenzo Librandi, Jun 27 2017
CROSSREFS
Sequence in context: A348290 A099132 A365798 * A365736 A364522 A226910
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jun 13 2008
STATUS
approved

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Last modified May 18 15:24 EDT 2024. Contains 372664 sequences. (Running on oeis4.)