A139398 - OEIS

%I #78 May 10 2023 11:58:12

%S 1,1,1,1,1,2,7,22,57,127,254,474,859,1574,3004,6008,12393,25773,53143,

%T 107883,215766,427351,843756,1669801,3321891,6643782,13333932,

%U 26789257,53774932,107746282,215492564,430470899,859595529,1717012749,3431847189,6863694378

%N a(n) = Sum_{k >= 0} binomial(n,5*k).

%C From _Gary W. Adamson_, Mar 13 2009: (Start)

%C M^n * [1,0,0,0,0] = [a(n), A139761(n), A139748(n), A139714(n), A133476(n)]

%C 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]

%C Sum of terms = 2^n. Example: M^6 * [1,0,0,0,0] = [7, 15, 20, 15, 7]; sum = 2^6 = 64. (End)

%C {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

%D A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Ch. 18.

%H Vincenzo Librandi, <a href="/A139398/b139398.txt">Table of n, a(n) for n = 0..1000</a>

%H John B. Dobson, <a href="http://arxiv.org/abs/1610.09361">A matrix variation on Ramus's identity for lacunary sums of binomial coefficients</a>, arXiv preprint arXiv:1610.09361 [math.NT], 2016.

%H Vladimir Shevelev, <a href="https://arxiv.org/abs/1706.01454">Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n</a>, arXiv:1706.01454 [math.CO], 2017.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,2).

%F 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

%F 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

%F 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

%F From _Vladimir Shevelev_, Jun 17 2017: (Start)

%F 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.

%F 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)

%F 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

%p f:=(n,r,a) -> add(binomial(n,r*k+a),k=0..n); fs:=(r,a)->[seq(f(n,r,a),n=0..40)];

%p 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

%t LinearRecurrence[{5,-10,10,-5,2},{1,1,1,1,1},40] (* _Harvey P. Dale_, Jun 11 2015 *)

%t 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 *)

%o (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

%Y Cf. A000749, A024493, A024494, A024495, A038503, A038504, A038505, A133476, A139714, A139748, A139761.

%K nonn,easy

%O 0,6

%A _N. J. A. Sloane_, Jun 13 2008