A107231 -id:A107231 - cp's OEIS frontend

A034828 a(n) = floor(n^2/4)*(n/2).

0, 0, 1, 3, 8, 15, 27, 42, 64, 90, 125, 165, 216, 273, 343, 420, 512, 612, 729, 855, 1000, 1155, 1331, 1518, 1728, 1950, 2197, 2457, 2744, 3045, 3375, 3720, 4096, 4488, 4913, 5355, 5832, 6327, 6859, 7410, 8000, 8610, 9261, 9933, 10648, 11385, 12167, 12972, 13824

Offset: 0

N. J. A. Sloane

Wiener index of cycle of length n.

a(n+1) is the sum of labeled number of boxes arranged as pyramid with base n. The sum of boxes is A002620(n+1). See the illustration in links. - Kival Ngaokrajang, Jul 02 2013

			G.f.: x^2 + 3*x^3 + 8*x^4 + 15*x^5 + 27*x^6 + 42*x^7 + 64*x^8 + 90*x^9 + ...

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Kival Ngaokrajang, Illustration for n = 1..10.
Eric Weisstein's World of Mathematics, Wiener Index.
H. J. Wiener, Structural Determination of Paraffin Boiling Points, J. Amer. Chem. Soc. 69 (1947), 17-20.
J. Zerovnik, Szeged index of symmetric graphs, J. Chem. Inf. Comput. Sci., 39 (1999), 77-80.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Equals A005996/2.
Partial sums of A001318.
Cf. A107231.
Cf. A000578, A000330, A002620.
Cf. A002620, A007310.
Cf. A062717.

[Floor(n^2/4)*(n/2): n in [0..50]]; // G. C. Greubel, Feb 23 2018

A034828:=n->n*floor(n^2/4)/2; seq(A034828(k), k=0..100); # Wesley Ivan Hurt, Nov 05 2013

Table[Floor[n^2/4] n/2, {n, 0, 50}] (* Harvey P. Dale, Jun 10 2011 *)
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {0, 0, 1, 3, 8, 15}, 50] (* Harvey P. Dale, Jun 10 2011 *)

{a(n) = (n^2 \ 4) * n / 2} /* Michael Somos, Sep 06 2008 */

{a(n) = if( n<0, -a(-n), polcoeff( x^2 * (1 + x + x^2) / ((1 - x)^2 * (1 - x^2)^2) + x * O(x^n), n))} /* Michael Somos, Sep 06 2008 */

a(n) = (n^2-1)*n/8 if n is odd, otherwise n^3/8.
From Paul Barry, May 13 2005: (Start)
G.f.: x^2*(1+x+x^2)/((1-x)^2*(1-x^2)^2).
a(n) = 2*a(n-1) +a(n-2) -4*a(n-3) +a(n-4) +2*a(n-5) -a(n-6).
a(n) = (2*n^3 +12*n^2 +23*n +14)/16 +(n+2)*(-1)^n/16.
a(n) = Sum_{k=0..floor((n+2)/2)} ((n+2)/(n+2-k))(-1)^k*C(n+2-k, k)* C(n-2*k+2, 2)*C(n-2*k, floor((n-2*k)/2)). [Typo corrected by R. J. Mathar, Aug 18 2008] (End)
a(n) = (2*n^2 - 1 + (-1)^n) * n / 16. - Michael Somos, Sep 06 2008
Euler transform of length 3 sequence [3, 2, -1]. - Michael Somos, Sep 06 2008
a(-n) = -a(n). - Michael Somos, Sep 06 2008
a(2n) = A000578(n). a(2n+1) = 3*A000330(n). a(n) = n*A002620(n)/2. - Michael Somos, Sep 06 2008
a(n) = (-n + Sum_{k=1..n} A007310(k)^2)/24. - Jesko Matthes, Feb 19 2021
Sum_{n>=2} 1/a(n) = 6 - 8*log(2) + zeta(3). - Amiram Eldar, Apr 16 2022
a(n) = Sum_{k=1..n} A062717(k)/4. - Sela Fried, Jun 27 2022

Definition reworded by Michael Somos, Sep 06 2008

A107239 Sum of squares of tribonacci numbers (A000073).

Original entry on oeis.org

0, 0, 1, 2, 6, 22, 71, 240, 816, 2752, 9313, 31514, 106590, 360606, 1219935, 4126960, 13961456, 47231280, 159782161, 540539330, 1828631430, 6186215574, 20927817799, 70798300288, 239508933824, 810252920400, 2741065994769, 9272959837818, 31370198430718

Offset: 0

Jonathan Vos Post, May 17 2005

			a(7) = 71 = 0^2 + 0^2 + 1^2 + 1^2 + 2^2 + 4^2 + 7^2

R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.

G. C. Greubel, Table of n, a(n) for n = 0..1000
M. Feinberg, Fibonacci-Tribonacci, Fib. Quart. 1(3) (1963), 71-74.
Z. Jakubczyk, Advanced Problems and Solutions, Fib. Quart. 51 (3) (2013) 185, H-715.
Eric Weisstein's World of Mathematics, Tribonacci Number
Index entries for linear recurrences with constant coefficients, signature (3,1,3,-7,1,-1,1).

Cf. A000073, A000213, A085697.

Cf. A107240, A107241, A107242, A107243, A107244, A107245, A107246, A107247.

R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( x^2*(1-x-x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)*(1-x)) )); // G. C. Greubel, Nov 20 2021

b:= proc(n) option remember; `if`(n<3, [n*(n-1)/2$2],
     (t-> [t, t^2+b(n-1)[2]])(add(b(n-j)[1], j=1..3)))
    end:
a:= n-> b(n)[2]:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 22 2021

Accumulate[LinearRecurrence[{1,1,1},{0,0,1},30]^2] (* Harvey P. Dale, Sep 11 2011 *)
LinearRecurrence[{3,1,3,-7,1,-1,1}, {0,0,1,2,6,22,71}, 30] (* Ray Chandler, Aug 02 2015 *)

@CachedFunction
def T(n): # A000073
    if (n<2): return 0
    elif (n==2): return 1
    else: return T(n-1) +T(n-2) +T(n-3)
def A107231(n): return sum(T(j)^2 for j in (0..n))
[A107239(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021

a(n) = T(0)^2 + T(1)^2 + ... + T(n)^2 where T(n) = A000073(n).
From R. J. Mathar, Aug 19 2008: (Start)
a(n) = Sum_{i=0..n} A085697(i).
G.f.: x^2*(1-x-x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)*(1-x)). (End)
a(n) = 1/4 - (1/11)*Sum_{R = RootOf(_Z^3-_Z^2-_Z-1)} ((3 + 7*_R + 5*_R^2)/(3*_R^2 - 2*_R - 1)*_R^(-n) - (1/44)*Sum{_R = RootOf(Z^3+_Z^2+3*_Z-1)} ((-1 - 2*_R - 9*_R^2)/(3*_R^2 + 2*_R + 3)*_R^(-n). - _Robert Israel, Mar 26 2010
a(n+1) = A000073(n)*A000073(n+1) + ( (A000073(n+1) - A000073(n-1))^2 - 1 )/4 for n>0 [Jakubczyk]. - R. J. Mathar, Dec 19 2013

A107230 A number triangle of inverse Chebyshev transforms.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 6, 3, 1, 6, 12, 12, 4, 1, 10, 30, 30, 20, 5, 1, 20, 60, 90, 60, 30, 6, 1, 35, 140, 210, 210, 105, 42, 7, 1, 70, 280, 560, 560, 420, 168, 56, 8, 1, 126, 630, 1260, 1680, 1260, 756, 252, 72, 9, 1, 252, 1260, 3150, 4200, 4200, 2520, 1260, 360, 90, 10, 1

Offset: 0

Paul Barry, May 13 2005

First column is A001405, second column is A100071, third column is A107231. Row sums are A005773(n+1), diagonal sums are A026003. The inverse Chebyshev transform concerned takes a g.f. g(x)->(1/sqrt(1-4x^2))g(xc(x^2)) where c(x) is the g.f. of A000108. It transforms a(n) to b(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*a(n-2k). Then a(n) = Sum_{k=0..floor(n/2)} (n/(n-k))*(-1)^k*binomial(n-k,k) *b(n-2k).

Triangle read by rows: T(n,k) is the number of paths of length n with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having k H steps. Example: T(3,1)=6 because we have HUD. HUU, UDH, UHD, UHU and UUH. Sum_{k=0..n} k*T(n,k) = A132894(n). - Emeric Deutsch, Oct 07 2007

			Triangle begins
   1;
   1,  1;
   2,  2,  1;
   3,  6,  3,  1;
   6, 12, 12,  4,  1;
  10, 30, 30, 20,  5,  1;

Jinyuan Wang, Rows n=0..200 of triangle, flattened
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.

Cf. A132894.

[[Binomial(n, k)*Binomial(n-k, Floor((n-k)/2)): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 11 2019

T:=proc(n,k) options operator, arrow: binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k)) end proc: for n from 0 to 11 do seq(T(n,k),k=0..n) end do; # yields sequence in triangular form - Emeric Deutsch, Oct 07 2007

Table[Binomial[n, k]*Binomial[n-k, Floor[(n-k)/2]], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 11 2019 *)

T(n, k) = binomial(n, k)*binomial(n-k, (n-k)\2); \\ Michel Marcus, Feb 10 2019

[[binomial(n, k)*binomial(n-k, floor((n-k)/2)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 11 2019

T(n,k) = binomial(n,k)*binomial(n-k, floor((n-k)/2)).
G.f.: G=G(t,z) satisfies z*(1-2*z-t*z)*G^2+(1-2*z-t*z)*G-1=0. - Emeric Deutsch, Oct 07 2007
E.g.f.: exp(x*y)*(BesselI(0,2*x)+BesselI(1,2*x)). - Vladeta Jovovic, Dec 02 2008

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A034828 a(n) = floor(n^2/4)*(n/2).

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A107239 Sum of squares of tribonacci numbers (A000073).

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A107230 A number triangle of inverse Chebyshev transforms.

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