cp's OEIS Frontend

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.

O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019

a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020

From Klaus Purath, Feb 02 2021: (Start)

a(n) = (2^(4*n+2)+1)/5.

a(n) = (A061654(n) + A001025(n))/2.

a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.

(End)

E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021

We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),

(53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),

(107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),

(173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),

(733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),

(997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),

(41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),

(509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),

(1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),

(773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).

G.f.: x/((1 - x)*(1 + 10*x)). - Vincenzo Librandi, Oct 22 2012

a(n) = -9*a(n-1) + 10*a(n-2). - Vincenzo Librandi, Oct 22 2012

a(n) = (-1)^(n+1)*A015585(n). - R. J. Mathar, Oct 26 2015

E.g.f.: (exp(x) - exp(-10*x))/11. - G. C. Greubel, May 26 2018

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).

E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).

O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).

Associated operator identities:

With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),

A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!

= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)

B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)

= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).

Associated binomial identities:

A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)

= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)

B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)

= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)

= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).

In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.

From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.

With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):

Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.

From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.

U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014

From Tom Copeland, Dec 26 2015: (Start)

With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.

With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)

As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016

From Tom Copeland, Sep 05 2016: (Start)

Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.

The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)

T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019

a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).

a(0)=1, a(1)=-8, a(n) = -8*a(n-1) + 9*a(n-2). - Harvey P. Dale, Aug 08 2011

G.f.: x/((1 - x)*(1 + 9*x)). - Vincenzo Librandi, Oct 22 2012

E.g.f.: (exp(x) - exp(-9*x))/10. - G. C. Greubel, May 26 2018

a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}.

G.f.: x/((1 - x)*(1 + 11*x)). - Vincenzo Librandi, Oct 22 2012

a(n) = -10*a(n-1) + 11*a(n-2). - Vincenzo Librandi, Oct 22 2012

E.g.f.: (exp(x) - exp(-11*x))/12. - G. C. Greubel, May 26 2018

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1), with q=-13.

a(n) = Sum_{j=0..n-1} (-13)^j. - Zerinvary Lajos, Dec 16 2008

G.f.: x/((1 - x)*(1 + 13*x)). - Vincenzo Librandi, Oct 22 2012

a(n) = -12*a(n-1) + 13*a(n-2). - Vincenzo Librandi, Oct 22 2012

From G. C. Greubel, May 26 2018: (Start)

a(n) = (1 - (-13)^n)/14.

E.g.f.: (exp(x) - exp(-13*x))/14. (End)

G.f.: F(q) = Sum_{n>=1} A049347(n-1) * n^2 * q^n / (1 - q^n).

G.f.: F(q) = -q * G'(q) / (9*G(q)), with G(q) = Product_{n>=1} (1 - q^n)^(9*n * A049347(n-1)).

a(n) is multiplicative with a(3^e) = 1, a(p^e) = a(p) * a(p^(e-1)) - z * a(p^(e-2)) where z = Kronecker(-3, p) * p^2 and a(p) = z + 1.

a(3*n) = a(n).

G.f.: Sum_{k>0} x^k * (1 - x^k - 6*x^(2*k) - x^(3*k) + x^(4*k)) / (1 + x^k + x^(2*k))^3. - Michael Somos, Oct 21 2007

G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = u^2 - v + w + 3*v^2 - 8*w^2 + 6*v*w - 8*u*w + 6*u*v - 9*v^3 - 54*u*v*w + 72*u*w^2 - 9*u^2*w. - Michael Somos, Dec 23 2007

a(4*n) = |(-4)^n - 1| otherwise a(n) = 1.

a(4*n) = 5*A015521(n).

O.g.f.: 1/(1 - 4*x^4) - 1/(1 + x^4) + 1/(1 - x) - 1/(1 - x^4) = x*(-1 -x -x^2 -5*x^3 +3*x^4 +3*x^5 +3*x^6 +5*x^7 +4*x^8 +4*x^9 +4*x^10) / ( (1-x) *(1+x) *(2*x^2+1) *(2*x^2-1) *(x^2+1) *(x^4+1) ).

Recurrence equation: a(n) = 4*a(n-4) + a(n-8) - 4*a(n-12).

G.f.: 1/(1 + 13*x - 14*x^2).

a(n) = ((-1)^n*14^(n + 1) + 1)/15.

a(n) = 1 - 14*a(n - 1) for n>0 and a(0)=1.

a(n) = Sum_{k = 0..n} A033999(k)*A001023(k).

Lim_{n -> infinity} a(n)/a(n + 1) = - 1/14.