E.g.f.: exp(x)*(x+x^2/2).
a(n) = a(-1-n).
a(n+1) = ((n+2)/n)*a(n), Sum_{n>=1} 1/a(n) = 2. -
Jon Perry, Jul 13 2003
For n > 0, a(n) =
A001109(n) - Sum_{k=0..n-1} (2*k+1)*
A001652(n-1-k); e.g., 10 = 204 - (1*119 + 3*20 + 5*3 + 7*0). -
Charlie Marion, Jul 18 2003
With interpolated zeros, this is n*(n+2)*(1+(-1)^n)/16. -
Benoit Cloitre, Aug 19 2003
a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j) = binomial(i+j+1, i). -
Benoit Cloitre, Aug 19 2003
a(n) = ((n+1)^3 - n^3 - 1)/6. - Xavier Acloque, Oct 24 2003
a(n) = a(n-1) + (1 + sqrt(1 + 8*a(n-1)))/2. This recursive relation is inverted when taking the negative branch of the square root, i.e., a(n) is transformed into a(n-1) rather than a(n+1). -
Carl R. White, Nov 04 2003
a(n) = a(n-2) + 2*n - 1. -
Paul Barry, Jul 17 2004
a(n) = sqrt(sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)^3)) = (Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i*j*k)^3)^(1/6). -
Alexander Adamchuk, Oct 26 2004
a(n) == 1 (mod n+2) if n is odd and a(n) == n/2+2 (mod n+2) if n is even. -
Jon Perry, Dec 16 2004
a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 1. -
Miklos Kristof, Mar 09 2005
a(n) = floor((2*n+1)^2/8). -
Paul Barry, May 29 2006
a(n)^2 + a(n+1)^2 = a((n+1)^2) [R B Nelsen, Math Mag 70 (2) (1997), p. 130]. -
R. J. Mathar, Nov 22 2006
a(n)*a(n+k)+a(n+1)*a(n+1+k) = a((n+1)*(n+1+k)). Generalizes previous formula dated Nov 22 2006 [and comments by
J. M. Bergot dated May 22 2012]. -
Charlie Marion, Feb 04 2011
(sqrt(8*a(n)+1)-1)/2 = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
A general formula for polygonal numbers is P(k,n) = (k-2)*(n-1)n/2 + n = n + (k-2)*
A000217(n-1), for n >= 1, k >= 3. -
Omar E. Pol, Apr 28 2008 and Mar 31 2013
If we define f(n,i,a) = Sum_{j=0..k-1} (binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,1), for n >= 1. -
Milan Janjic, Dec 20 2008
An exponential generating function for the inverse of this sequence is given by Sum_{m>=0} ((Pochhammer(1, m)*Pochhammer(1, m))*x^m/(Pochhammer(3, m)*factorial(m))) = ((2-2*x)*log(1-x)+2*x)/x^2, the n-th derivative of which has a closed form which must be evaluated by taking the limit as x->0.
A000217(n+1) = (lim_{x->0} d^n/dx^n (((2-2*x)*log(1-x)+2*x)/x^2))^-1 = (lim_{x->0} (2*Gamma(n)*(-1/x)^n*(n*(x/(-1+x))^n*(-x+1+n)*LerchPhi(x/(-1+x), 1, n) + (-1+x)*(n+1)*(x/(-1+x))^n + n*(log(1-x)+log(-1/(-1+x)))*(-x+1+n))/x^2))^-1. -
Stephen Crowley, Jun 28 2009
With offset 1, a(n) = floor(n^3/(n+1))/2. -
Gary Detlefs, Feb 14 2010
a(n) = 4*a(floor(n/2)) + (-1)^(n+1)*floor((n+1)/2). -
Bruno Berselli, May 23 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=1. -
Mark Dols, Aug 20 2010
a(n) + 2*a(n-1) + a(n-2) = n^2 + (n-1)^2; and
a(n) + 3*a(n-1) + 3*a(n-2) + a(n-3) = n^2 + 2*(n-1)^2 + (n-2)^2.
In general, for n >= m > 2, Sum_{k=0..m} binomial(m,m-k)*a(n-k) = Sum_{k=0..m-1} binomial(m-1,m-1-k)*(n-k)^2.
a(n) - 2*a(n-1) + a(n-2) = 1, a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 and a(n) - 4*a(n-1) + 6*a(n-2) - 4*(a-3) + a(n-4) = 0.
In general, for n >= m > 2, Sum_{k=0..m} (-1)^k*binomial(m,m-k)*a(n-k) = 0.
(End)
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} 4*(sin(x))^(2*n-1)*(cos(x))^3). -
Francesco Daddi, Aug 02 2011
a(n) = -s(n+1,n), where s(n,k) are the Stirling numbers of the first kind,
A048994. -
Mircea Merca, May 03 2012
a(n)*a(n+1) = a(Sum_{m=1..n}
A005408(m))/2, for n >= 1. For example, if n=8, then a(8)*a(9) = a(80)/2 = 1620. -
Ivan N. Ianakiev, May 27 2012
G.f.: x * (1 + 3x + 6x^2 + ...) = x * Product_{j>=0} (1+x^(2^j))^3 = x * A(x) * A(x^2) * A(x^4) * ..., where A(x) = (1 + 3x + 3x^2 + x^3). -
Gary W. Adamson, Jun 26 2012
G.f.: G(0) where G(k) = 1 + (2*k+3)*x/(2*k+1 - x*(k+2)*(2*k+1)/(x*(k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). -
Sergei N. Gladkovskii, Nov 23 2012
G.f.: x + 3*x^2/(Q(0)-3*x) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Mar 14 2013
a(n) + a(n+1) + ... + a(n+8) + 6*n = a(3*n+15). -
Charlie Marion, Mar 18 2013
a(n) + a(n+1) + ... + a(n+20) + 2*n^2 + 57*n = a(5*n+55). -
Charlie Marion, Mar 18 2013
In general, a(k*n) = (2*k-1)*a(n) + a((k-1)*n-1). -
Charlie Marion, Apr 20 2015
Also, a(k*n) = a(k)*a(n) + a(k-1)*a(n-1). -
Robert Israel, Apr 20 2015
a(n+1) = det(binomial(i+2,j+1), 1 <= i,j <= n). -
Mircea Merca, Apr 06 2013
a(n) = floor(n/2) + ceiling(n^2/2) = n - floor(n/2) + floor(n^2/2). -
Wesley Ivan Hurt, Jun 15 2013
Sum_{n>=1} a(n)/n! = 3*exp(1)/2 by the e.g.f. Also see
A067764 regarding ratios calculated this way for binomial coefficients in general. -
Richard R. Forberg, Jul 15 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 2 = 0.7725887... . -
Richard R. Forberg, Aug 11 2014
Let O(n) be the oblong number n(n+1) =
A002378(n) and S(n) the square number n^2 =
A000290(n). Then a(n) + a(n+2k) = O(n+k) + S(k) and a(n) + a(n+2k+1) = S(n+k+1) + O(k). -
Charlie Marion, Jul 16 2015
A generalization of the Nov 22 2006 formula, a(n)^2 + a(n+1)^2 = a((n+1)^2), follows. Let T(k,n) = a(n) + k. Then for all k, T(k,n)^2 + T(k,n+1)^2 = T(k,(n+1)^2 + 2*k) - 2*k. -
Charlie Marion, Dec 10 2015
a(n)^2 + a(n+1)^2 = a(a(n) + a(n+1)). Deducible from
N. J. A. Sloane's a(n) + a(n+1) = (n+1)^2 and R. B. Nelson's a(n)^2 + a(n+1)^2 = a((n+1)^2). -
Ben Paul Thurston, Dec 28 2015
a(n)^2 + a(n+3)^2 + 19 = a(n^2 + 4*n + 10). -
Charlie Marion, Nov 23 2016
G.f.: x/(1-x)^3 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^3 = (1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6). -
Gary W. Adamson, Dec 03 2016
a(n) = sum of the elements of inverse of matrix Q(n), where Q(n) has elements q_i,j = 1/(1-4*(i-j)^2). So if e = appropriately sized vector consisting of 1's, then a(n) = e'.Q(n)^-1.e. -
Michael Yukish, Mar 20 2017
a(n) = Sum_{k=1..n} ((2*k-1)!!*(2*n-2*k-1)!!)/((2*k-2)!!*(2*n-2*k)!!). -
Michael Yukish, Mar 20 2017
Sum_{i=0..k-1} a(n+i) = (3*k*n^2 + 3*n*k^2 + k^3 - k)/6. -
Christopher Hohl, Feb 23 2019
a(n) == 0 (mod n) iff n is odd (see De Koninck reference). -
Bernard Schott, Jan 10 2020
8*a(k)*a(n) + ((a(k)-1)*n + a(k))^2 = ((a(k)+1)*n + a(k))^2. This formula reduces to the well-known formula, 8*a(n) + 1 = (2*n+1)^2, when k = 1. -
Charlie Marion, Jul 23 2020
a(k)*a(n) = Sum_{i = 0..k-1} (-1)^i*a((k-i)*(n-i)). -
Charlie Marion, Dec 04 2020
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)/(2*Pi).
Product_{n>=2} (1 - 1/a(n)) = 1/3. (End)
a(n) = Sum_{k=1..2*n-1} (-1)^(k+1)*a(k)*a(2*n-k). For example, for n = 4, 1*28 - 3*21 + 6*15 - 10*10 + 15*6 - 21*3 + 28*1 = 10. -
Charlie Marion, Mar 23 2022
2*a(n) =
A000384(n) - n^2 + 2*n. In general, if P(k,n) = the n-th k-gonal number, then (j+1)*a(n) = P(5 + j, n) - n^2 + (j+1)*n. More generally, (j+1)*P(k,n) = P(2*k + (k-2)*(j-1),n) - n^2 + (j+1)*n. -
Charlie Marion, Mar 14 2023
a(n) = (1/6)* Sum_{k = 0..3*n} (-1)^(n+k+1) * k*(k + 1) * binomial(3*n+k, 2*k). -
Peter Bala, Nov 03 2024
The following series telescope: for k >= 0,
Sum_{n >= 1} a(n)*a(n+2)*...*a(n+2*k)/(a(n+1)*a(n+3)*...*a(n+2*k+3)) = 1/(2*k + 3);
Sum_{n >= 1} a(n+1)*a(n+3)*...*a(n+2*k+1)/(a(n)*a(n+2)*...*a(n+2*k+2)) = 2/(2*k + 3) * Sum_{i = 1..2*k+3} 1/i. (End)
Comments