A054602 -id:A054602 - cp's OEIS frontend

A006527 a(n) = (n^3 + 2*n)/3.

0, 1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531, 28424, 30405

Offset: 0

N. J. A. Sloane

Number of ways to color vertices (or edges) of a triangle using <= n colors, allowing only rotations.
Also: dot_product (1,2,...,n)*(2,3,...,n,1), n >= 0. - Clark Kimberling
Start from triacid and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002
Starting with offset 1 = row sums of triangle A158822 and binomial transform of (1, 3, 4, 2, 0, 0, 0, ...). - Gary W. Adamson, Mar 28 2009
One-ninth of sum of three consecutive cubes: a(n) = ((n-1)^3 + n^3 + (n+1)^3)/9. - Zak Seidov, Jul 22 2013
For n > 2, number of different cubes, formed after splitting a cube in color C_1, by parallel planes in the colors C_2, C_3, ..., C_n in three spatial dimensions (in the order of the colors from a fixed vertex). Generally, in a large hypercube n^d is f(n,d) = C(n+d-1, d) + C(n, d) different small hypercubes. See below for my formula a(n) = f(n,3). - Thomas Ordowski, Jun 15 2014
a(n) is a square for n = 1, 2 & 24; and for no other values up to 10^7 (see M. Gardner). - Michel Marcus, Sep 06 2015
Number of unit tetrahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb. - Jason Pruski, Aug 23 2017

M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
B. Babcock and A. van Tuyl, Revisiting the spreading and covering numbers, arXiv preprint arXiv:1109.5847 [math.AC], 2011.
Richard A. Brualdi and Geir Dahl, Alternating Sign Matrices and Hypermatrices, and a Generalization of Latin Square, arXiv:1704.07752 [math.CO], 2017. See p. 8.
Peter Esser?, Guenter Stertenbrink, Triangles with Mac Mahon's pieces, digest of 14 messages in polyforms Yahoo group, Apr 14 - May 2, 2002.
Th. Grüner, A. Kerber, R. Laue and M. Meringer, Mathematics for Combinatorial Chemistry
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
polyforms list, Triangles with MacMahon's pieces.
Taskcentre, McMahon's Triangles 2
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 of triangle A094414. Row 6 of the array in A107735.
Cf. A002264, A005843, A054602, A059100, A135184, A158822, A177342.
Cf. A000292 (unoriented), A000292(n-2) (chiral), A000290 (achiral) triangle colorings.
Row 2 of A324999 (simplex vertices and facets) and A327083 (simplex edges and ridges).

a006527 n = n * (n ^ 2 + 2) `div` 3  -- Reinhard Zumkeller, Jan 06 2014

[(n^3 + 2*n)/3: n in [0..50]]; // Vincenzo Librandi, May 15 2011

A006527:=z*(1+z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
with(combinat):seq(lcm(fibonacci(4,n),fibonacci(2,n))/3,n=0..42); # Zerinvary Lajos, Apr 20 2008

Table[ (n^3 + 2*n)/3, {n, 0, 45} ]
LinearRecurrence[{4,-6,4,-1},{0,1,4,11},46] (* or *) CoefficientList[ Series[(x+x^3)/(x-1)^4,{x,0,49}],x] (* Harvey P. Dale, Jun 13 2011 *)

a(n)=n*(n^2+2)/3 \\ Charles R Greathouse IV, Jul 25 2011

a(0)=0, a(1)=1, a(2)=4, a(3)=11; for n > 3, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jun 13 2011
From Paul Barry, Mar 13 2003: (Start)
a(n) = 2*binomial(n+1, 3) + binomial(n, 1).
G.f.: x*(1+x^2)/(1-x)^4. (End)
a(n) = A000292(n) + A000292(n-2). - Alexander Adamchuk, May 20 2006
a(n) = n*A059100(n)/3. - Lekraj Beedassy, Feb 06 2007
a(n) = A054602(n)/3. - Zerinvary Lajos, Apr 20 2008
a(n) = ( n + Sum_{i=1..n} A177342(i) )/(n+1), with n > 0. - Bruno Berselli, May 19 2010
a(n) = A002264(A000578(n) + A005843(n)). - Reinhard Zumkeller, Jun 16 2011
a(n) = binomial(n+2, 3) + binomial(n, 3). - Thomas Ordowski, Jun 15 2014
a(n) = A000292(n) - A000292(-n). - Bruno Berselli, Sep 22 2016
E.g.f.: (x/3)*(3 + 3*x + x^2)*exp(x). - G. C. Greubel, Sep 01 2017
From Robert A. Russell, Oct 20 2020: (Start)
a(n) = 1*C(n,1) + 2*C(n,2) + 2*C(n,3), where the coefficient of C(n,k) is the number of oriented triangle colorings using exactly k colors.
a(n) = 2*A000292(n) - A000290(n) = 2*A000292(n-2) + A000290(n). (End)
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/4 = 1.45245201414472469745354677573358867... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

More terms from Alexander Adamchuk, May 20 2006
Corrected and replaced 5th formula from Harvey P. Dale, Jun 13 2011
Deleted an erroneous comment. - N. J. A. Sloane, Dec 10 2018

A073133 Table by antidiagonals of T(n,k) = n*T(n,k-1) + T(n,k-2) starting with T(n,1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 4, 10, 12, 5, 1, 5, 17, 33, 29, 8, 1, 6, 26, 72, 109, 70, 13, 1, 7, 37, 135, 305, 360, 169, 21, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89

Offset: 1

Henry Bottomley, Jul 16 2002

Columns of the array are generated from Fibonacci polynomials f(x). They are: (1), (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 +1), ... If column headings start 0, 1, 2, ... then the terms in the n-th column are generated from the n-th degree Fibonacci polynomial. For example, column 5 (8, 70, 360, ...) is generated from f(x), x = 1,2,3,...; fifth-degree polynomial x^5 + 4x^3 + 3x; e.g., f(2) = 70 = 2^5 + 4*8 + 3*2. - Gary W. Adamson, Apr 02 2006
The ratio of two consecutive entries of the sequence in the n-th row approaches (n + sqrt(n^2 + 4))/2. Example: The sequence beginning (1, 3, 10, 33, ...) tends to 3.302775... = (3 + sqrt(13))/2. - Gary W. Adamson, Aug 12 2013
As to the array sequences, (n+1)-th sequence is the INVERT transform of the n-th sequence. - Gary W. Adamson, Aug 20 2013
The array can be extended infinitely above the Fibonacci row by taking successive INVERTi transforms, resulting in:
  ...
  1,  -2,   5, -12,  29, -70, ...
  1,  -1,   2,  -3,   5,  -8, ...
  l,   0,   1,   0,   1,   0, ...
  1,   1,   2,   3,   5,   8, ...
  1,   2,   5,  12,  29,  70, ...
  ...
  This results in an infinite array in which sequences above the (1, 0, 1, 0, ...) are reflections of the sequences below, except for the alternate signs. Any sequence in the (+ sign) row starting (1, n, ...) is the (2*n-th) INVERT transform of the same sequence but with alternate signs.  Example: (1, 2, 5, 12, ...) is the (2*2) = fourth INVERT transform of (1, -2, 5, -12, ...) by inspection. Conjecture:  This "reflection" principle will result from taking successive INVERT transforms of any aerated sequence starting 1, ... and with positive signs. Likewise, the rows above the aerated sequence are successive INVERTi transforms of the aerated sequence. - Gary W. Adamson, Jul 14 2019
From Michael A. Allen, Feb 21 2023: (Start)
Row n is the n-metallonacci sequence.
T(n,k) is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			Table begins:
  1, 1,  2,  3,   5,    8,   13, ...
  1, 2,  5, 12,  29,   70,  169, ...
  1, 3, 10, 33, 109,  360, 1189, ...
  1, 4, 17, 72, 305, 1292, 5473, ... etc.

G. C. Greubel, Antidiagonals n = 1..100, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows are A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A041041, A049666, A041061, A140455, A041085, etc.
Columns are A000012, A000027, A002522, A054602, A057721, A124152, etc.
Different from A081572.

T:= function(n,k)
    if k<0 then return 0;
    elif k=1 then return 1;
    else return n*T(n,k-1) + T(n,k-2);
    fi;
  end;
Flat(List([1..15], n-> List([1..n], k-> T(n-k+1,k) ))); # G. C. Greubel, Aug 12 2019

A073133 := proc(n,k)
    option remember;
    if k <= 1 then
        k;
    else
        n*procname(n,k-1)+procname(n,k-2) ;
    end if;
end proc:
seq(seq( A073133(d-k,k),k=1..d-1),d=2..13) ; # R. J. Mathar, Aug 16 2019

T[n_, 1]:= 1; T[n_, k_]:= T[n, k] = If[k<0, 0, n*T[n, k-1] + T[n, k-2]]; Table[T[n-k+1, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Aug 12 2019 *)

T(n,k) = if(k==1, 1, k<0, 0, n*T(n,k-1)+T(n,k-2));
for(n=1,15, for(k=1,n, print1(T(n-k+1,k), ", "))) \\ G. C. Greubel, Aug 12 2019

def T(n, k):
    if (k<0): return 0
    elif (k==1): return 1
    else: return n*T(n, k-1) + T(n, k-2)
[[T(n-k+1, k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Aug 12 2019

T(n, k) = A073134(n, k) + 2*A073135(n, k-2) = Sum_{j=0..k-1} abs(A049310(k-1, j)*n^j).
T(n,k) = [[0,1; 1,n]^{k+1}]{1,1}, n,k in {1,2,...}. - _L. Edson Jeffery, Sep 23 2012
G.f. for row n: x/(1-n*x-x^2). - L. Edson Jeffery, Aug 28 2013

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 10, 12, 5, 0, 1, 5, 17, 33, 29, 8, 0, 1, 6, 26, 72, 109, 70, 13, 0, 1, 7, 37, 135, 305, 360, 169, 21, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 0, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89, 0, 1, 11, 101, 747, 4289, 18200, 53353, 98145, 98209, 42837, 5741, 144

Offset: 1

Roger L. Bagula, Jan 29 2010

Equals A073133 with an additional column A(.,0).
If the first column and top row are deleted, antidiagonal reading yields A118243.
Adding a top row of 1's and antidiagonal reading downwards yields A157103.
Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....
From Jianing Song, Jul 14 2018: (Start)
All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.
Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.
If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.
Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.
Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:
((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4
......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*
......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*
......1.........3,7.............1...................0...................0
.....-1.........1,5.............0...................0...................1
.....-1.........3,7.............0...................1...................0
* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
(End)
From Michael A. Allen, Mar 06 2023: (Start)
Removing the first (n=0) row of A352361 gives this sequence.
Row n is the n-metallonacci sequence.
A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			The array, A(n, k), starts in row n = 1 with columns k >= 0 as
  0      1      1      2      3      5      8
  0      1      2      5     12     29     70
  0      1      3     10     33    109    360
  0      1      4     17     72    305   1292
  0      1      5     26    135    701   3640
  0      1      6     37    228   1405   8658
  0      1      7     50    357   2549  18200
  0      1      8     65    528   4289  34840
  0      1      9     82    747   6805  61992
  0      1     10    101   1020  10301 104030
  0      1     11    122   1353  15005 166408
Antidiagonal triangle, T(n, k), begins as:
  0;
  0, 1;
  0, 1, 1;
  0, 1, 2,  2;
  0, 1, 3,  5,   3;
  0, 1, 4, 10,  12,   5;
  0, 1, 5, 17,  33,  29,    8;
  0, 1, 6, 26,  72, 109,   70,   13;
  0, 1, 7, 37, 135, 305,  360,  169,  21;
  0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;

Jianing Song, Antidiagonals n = 1..100, flattened (the first 30 antidiagonals from Vincenzo Librandi)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows n include: A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9), A041041 (n=10), A049666 (n=11), A041061 (n=12), A140455 (n=13), A041085 (n=14), A154597 (n=15), A041113 (n=16), A178765 (n=17), A041145 (n=18), A243399 (n=19), A041181 (n=20). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.)
Columns k include: A000004 (k=0), A000012 (k=1), A000027 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Entry points for A(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3).
Pisano period for A(n,k) modulo m: A001175 (n=1), A175181 (n=2), A175182 (n=3), A175183 (n=4), A175184 (n=5), A175185 (n=6).
Number of zeros in a period for A(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3).
Cf. A084844, A084845, A118243, A157103, A271782, A316269.
Sums include: A304357, A304359.
Similar to: A073133.

A172236:= func< n,k | k le 1 select k else Evaluate(DicksonSecond(k-1,-1), n-k) >;
[A172236(n,k): k in [0..n-1], n in [1..13]]; // G. C. Greubel, Sep 29 2024

A172236[n_,k_]:=Fibonacci[k, n-k];
Table[A172236[n, k], {n,15}, {k,0,n-1}]//Flatten

A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));
tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), ", ")); print); \\ Jianing Song, Jul 14 2018 (program from Michel Marcus; see also A316269)

A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018

def A172236(n,k): return sum(binomial(k-j-1,j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))
flatten([[A172236(n,k) for k in range(n)] for n in range(1,14)]) # G. C. Greubel, Sep 29 2024

A(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. - Jianing Song, Jun 27 2018
For n >= 1, Sum_{i=1..k} 1/A(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/A(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/A(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). - Jianing Song, Apr 21 2019
From G. C. Greubel, Sep 29 2024: (Start)
A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).
T(n, k) = F_{k}(n-k) (antidiagonal triangle).
Sum_{k=0..n-1} T(n, k) = A304357(n) - (1-(-1)^n)/2.
Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.
T(2*n, n) = A084844(n).
T(2*n+1, n+1) = A084845(n). (End)

More terms from Jianing Song, Jul 14 2018

A352361 Array read by ascending antidiagonals. A(n, k) = Fibonacci(k, n), where Fibonacci(n, x) are the Fibonacci polynomials.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 2, 2, 0, 0, 1, 3, 5, 3, 1, 0, 1, 4, 10, 12, 5, 0, 0, 1, 5, 17, 33, 29, 8, 1, 0, 1, 6, 26, 72, 109, 70, 13, 0, 0, 1, 7, 37, 135, 305, 360, 169, 21, 1, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1

Offset: 0

Peter Luschny, Mar 18 2022

From Michael A. Allen, Mar 26 2023: (Start)
Row n is the n-metallonacci sequence for n>0.
A(n,k), for n > 0 and k > 0, is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			Array, A(n,k), starts:
  n\k 0, 1, 2,  3,   4,    5,     6,      7,       8,        9, ...
  -------------------------------------------------------------------------
  [0] 0, 1, 0,  1,   0,    1,     0,      1,       0,        1, ... A000035;
  [1] 0, 1, 1,  2,   3,    5,     8,     13,      21,       34, ... A000045;
  [2] 0, 1, 2,  5,  12,   29,    70,    169,     408,      985, ... A000129;
  [3] 0, 1, 3, 10,  33,  109,   360,   1189,    3927,    12970, ... A006190;
  [4] 0, 1, 4, 17,  72,  305,  1292,   5473,   23184,    98209, ... A001076;
  [5] 0, 1, 5, 26, 135,  701,  3640,  18901,   98145,   509626, ... A052918;
  [6] 0, 1, 6, 37, 228, 1405,  8658,  53353,  328776,  2026009, ... A005668;
  [7] 0, 1, 7, 50, 357, 2549, 18200, 129949,  927843,  6624850, ... A054413;
  [8] 0, 1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, ... A041025;
  [9] 0, 1, 9, 82, 747, 6805, 61992, 564733, 5144589, 46866034, ... A099371;
      |  |  |  | A054602 |   A124152;
      |  |  |  A002522   A057721;
      |  |  A001477;
      |  A000012;
      A000004;
Antidiagonals, T(n, k), begin as:
  0;
  0, 1;
  0, 1, 0;
  0, 1, 1,  1;
  0, 1, 2,  2,   0;
  0, 1, 3,  5,   3,   1;
  0, 1, 4, 10,  12,   5,   0;
  0, 1, 5, 17,  33,  29,   8,   1;
  0, 1, 6, 26,  72, 109,  70,  13,  0;
  0, 1, 7, 37, 135, 305, 360, 169, 21, 1;

G. C. Greubel, Antidiagonals n = 0..50, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Other versions of this array are A073133, A157103, A172236.
Rows n: A000035 (n=0), A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9).
Columns k: A000004 (k=0), A000012 (k=1), A001477 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Cf. A084844 (main diagonal), A352362 (Lucas polynomials), A350470 (Jacobsthal polynomials).
Sums include: A304357 (row sums), A304359.
Cf. A084845.

A352361:= func< n, k | k le 1 select k else Evaluate(DicksonSecond(k-1, -1), n-k) >;
[A352361(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Sep 29 2024

seq(seq(combinat:-fibonacci(k, n - k), k = 0..n), n = 0..11);

Table[Fibonacci[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten
(* or *)
A[n_, k_] := With[{s = Sqrt[n^2 + 4]}, ((n + s)^k - (n - s)^k) / (2^k*s)];
Table[Simplify[A[n, k]], {n, 0, 9}, {k, 0, 9}] // TableForm

A(n, k) = ([1, k; 1, k-1]^n)[2, 1] ;
export(A)
for(k = 0, 9, print(parvector(10, n, A(n - 1, k))))

def A352361(n, k): return lucas_number1(k,n-k,-1)
flatten([[A352361(n, k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Sep 29 2024

A(n, k) = Sum_{j=0..floor((k-1)/2)} binomial(k-j-1, j)*n^(k-2*j-1).
A(n, k) = ((n + s)^k - (n - s)^k) / (2^k*s) where s = sqrt(n^2 + 4).
A(n, k) = [x^k] (x / (1 - n*x - x^2)).
A(n, k) = n^(k-1)*hypergeom([1 - k/2, 1/2 - k/2], [1 - k], -4/n^2) for n,k >= 1.
A(n, n) = T(2*n, n) = A084844(n).
From G. C. Greubel, Sep 29 2024: (Start)
T(n, k) = A(n-k, k) (antidiagonal triangle).
T(2*n+1, n+1) = A084845(n).
Sum_{k=0..n} T(n, k) = A304357(n) (row sums).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)*A304359(n). (End)

A185651 A(n,k) = Sum_{d|n} phi(d)*k^(n/d); square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 6, 3, 0, 0, 4, 12, 12, 4, 0, 0, 5, 20, 33, 24, 5, 0, 0, 6, 30, 72, 96, 40, 6, 0, 0, 7, 42, 135, 280, 255, 84, 7, 0, 0, 8, 56, 228, 660, 1040, 780, 140, 8, 0, 0, 9, 72, 357, 1344, 3145, 4200, 2205, 288, 9, 0

Offset: 0

Alois P. Heinz, Aug 29 2013

Dirichlet convolution of phi(n) and k^n. - Richard L. Ollerton, May 07 2021

			Square array A(n,k) begins:
  0, 0,  0,   0,    0,     0,     0, ...
  0, 1,  2,   3,    4,     5,     6, ...
  0, 2,  6,  12,   20,    30,    42, ...
  0, 3, 12,  33,   72,   135,   228, ...
  0, 4, 24,  96,  280,   660,  1344, ...
  0, 5, 40, 255, 1040,  3145,  7800, ...
  0, 6, 84, 780, 4200, 15810, 46956, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0..10 give A000004, A001477, A053635, A054610, A054611, A054612, A054613, A054614, A054615, A054616, A054617.
Rows n=0..10 give A000004, A001477, A002378, A054602, A054603, A054604, A054605, A054606, A054607, A054608, A054609.
Main diagonal gives A228640.
Cf. A000010, A075195, A258170.

with(numtheory):
A:= (n, k)-> add(phi(d)*k^(n/d), d=divisors(n)):
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[, 0] = a[0, ] = 0; a[n_, k_] := Sum[EulerPhi[d]*k^(n/d), {d, Divisors[n]}]; Table[a[n - k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 06 2013 *)

A(n,k) = Sum_{d|n} phi(d)*k^(n/d).
A(n,k) = Sum_{i=0..min(n,k)} C(k,i) * i! * A258170(n,i). - Alois P. Heinz, May 22 2015
G.f. for column k: Sum_{n>=1} phi(n)*k*x^n/(1-k*x^n) for k >= 0. - Petros Hadjicostas, Nov 06 2017
From Richard L. Ollerton, May 07 2021: (Start)
A(n,k) = Sum_{i=1..n} k^gcd(n,i).
A(n,k) = Sum_{i=1..n} k^(n/gcd(n,i))*phi(gcd(n,i))/phi(n/gcd(n,i)).
A(n,k) = A075195(n,k)*n for n >= 1, k >= 1.  (End)

A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

Original entry on oeis.org

1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531

Offset: 0

Klaus Brockhaus, Nov 14 2009

a(n) = ((n*(n+1)*(n+2))+(n+(n+1)+(n+2)))/3, n >= 0.
Equals A006527 without initial term 0: a(n) = A006527(n+1).
Binomial transform of A167876.
Inverse binomial transform of A080930.
a(n) = A007290(n+2)+n+1.
a(n) = A014820(n)/(n+1) for n > 0.
a(n) = A116731(n+2)-1.
a(n) = A033547(n+1)-n.
a(n) = A054602(n)/3.
a(n) = A086514(n+3)-2.
a(n) = A002061(n+1)+a(n-1) for n > 0.
a(n) = A005894(n)-a(n-1) for n > 0.
First bisection is A057813.
Second differences are in A004277.
a(n) = A177342(n)*(-1)+a(n-1)*5 with n>0. For n=8, a(8)=-A177342(8)+a(7)*5=-631+176*5=249. - Bruno Berselli, May 18 2010

			a(0) = (0*1*2+0+1+2)/3 = (0+3)/3 = 1.
a(1) = (1*2*3+1+2+3)/3 = (6+6)/3 = 4.
a(6)-4*a(5)+6*a(4)-4*a(3)+a(2) = 119-4*76+6*45-4*24+11 = 0. - _Bruno Berselli_, May 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001477 (nonnegative integers),
A006527 ((n^3+2*n)/3),
A167876 (1, 3, 4, 2, 0, 0, 0, 0, ...),
A080930,
A007290 (2*C(n, 3)),
A014820 ((1/3)*(n^2+2*n+3)*(n+1)^2),
A116731,
A033547 (n*(n^2+5)/3),
A054602 (Sum_{d|3} phi(d)*n^(3/d)),
A086514 ((n^3-6*n^2+14*n-6)/3),
A002061 (n^2-n+1),
A005894 (centered tetrahedral numbers),
A057813 ((2*n+1)*(4*n^2+4*n+3)/3),
A004277 (1 and the positive even numbers),
A028387 (n+(n+1)^2),
A166941, A166942, A166943.

[ (&*s + &+s)/3 where s is [n..n+2]: n in [0..42] ];

Select[Table[(n*(n+1)*(n+2)+n+(n+1)+(n+2))/3,{n,0,5!}],IntegerQ[#]&] (* Vladimir Joseph Stephan Orlovsky, Dec 04 2010 *)
(Times@@#+Total[#])/3&/@Partition[Range[0,65],3,1]  (* Harvey P. Dale, Mar 14 2011 *)

a(n)=(n+1)*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (n^3+3*n^2+5*n+3)/3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+2 for n > 3; a(0)=1, a(1)=4, a(2)=11, a(3)=24.
G.f.: (1+x^2)/(1-x)^4.
a(n) = SUM(A109613(k)*A005408(n-k): 0<=k<=n). - Reinhard Zumkeller, Dec 05 2009
a(n)-4*a(n-1)+6*a(n-2)-4*a(n-3)+a(n-4)=0 for n>3. - Bruno Berselli, May 26 2010

A360850 Array read by antidiagonals: T(m,n) is the number of (undirected) paths in the complete bipartite graph K_{m,n}.

Original entry on oeis.org

1, 3, 3, 6, 12, 6, 10, 33, 33, 10, 15, 72, 135, 72, 15, 21, 135, 438, 438, 135, 21, 28, 228, 1140, 2224, 1140, 228, 28, 36, 357, 2511, 8850, 8850, 2511, 357, 36, 45, 528, 4893, 27480, 55725, 27480, 4893, 528, 45, 55, 747, 8700, 70462, 265665, 265665, 70462, 8700, 747, 55

Offset: 1

Andrew Howroyd, Feb 23 2023

T(m,n) is the number of induced paths including zero length paths in the m X n rook graph. This is also the number of induced trees in these graphs since these are the only induced trees.

			Array begins:
===================================================
m\n|  1   2    3     4      5        6        7 ...
---+-----------------------------------------------
1  |  1   3    6    10     15       21       28 ...
2  |  3  12   33    72    135      228      357 ...
3  |  6  33  135   438   1140     2511     4893 ...
4  | 10  72  438  2224   8850    27480    70462 ...
5  | 15 135 1140  8850  55725   265665   962010 ...
6  | 21 228 2511 27480 265665  2006316 11158203 ...
7  | 28 357 4893 70462 962010 11158203 98309827 ...
   ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Eric Weisstein's World of Mathematics, Complete Bipartite Graph.
Eric Weisstein's World of Mathematics, Graph Path.
Eric Weisstein's World of Mathematics, Rook Graph.

Main diagonal is A288035.
Rows 1..2 are A000217, A054602.
Cf. A360849 (cycles), A360851.

T(m,n) = sum(j=1, min(m,n), j!^2*binomial(m,j)*binomial(n,j)*(1 + (m+n)/2 - j))

T(m,n) = Sum_{j=1..min(m,n)} j!^2*binomial(m,j)*binomial(n,j)*(1 + (m+n)/2 - j).

T(m,n) = T(n,m).

A206334 Numbers n such that there is a triangle with area n, side n, and the other two sides rational.

Original entry on oeis.org

3, 5, 7, 10, 12, 15, 16, 18, 19, 23, 25, 26, 27, 28, 29, 30, 33, 34, 36, 38, 39, 40, 41, 42, 43, 44, 46, 47, 51, 52, 55, 57, 58, 59, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 80, 83, 84, 85, 86, 87, 88, 89, 90, 91, 93, 95, 96, 97, 103, 104, 105, 106, 107, 109, 115, 119, 122, 123, 124, 125, 126

Offset: 1

James R. Buddenhagen, Feb 06 2012

nonn

n>3 is in the sequence just in case the elliptic curve y^2 = 4*x^4 + (n^2+8)*x^2 + 4 has positive rank.  Note that (0,2) is on that curve.
n is in the sequence just in case there are positive rational numbers x,y such that x*y>1 and x - 1/x + y - 1/y = n.
The triangle whose sides are [(4*k^6+8*k^5+8*k^4+4*k^3+2*k^2+2*k+1)/((k+1)*k*(2*k^2+2*k+1)), (4*k^6+16*k^5+28*k^4+28*k^3+18*k^2+6*k+1)/((k+1)*k*(2*k^2+2*k+1)), 4*k^2+4*k+4] has area equal to its third side.  Hence, starting with the second term, A112087 is a subsequence of the present sequence.
The triangle whose sides are [(k^6+2*k^4+k^2+1)/(k*(k^2+1)), (k^4+3*k^2+1)/(k*(k^2+1)), (k^2+2)*k] has area equal to its third side. Hence, starting with the first positive term, A054602 is a subsequence of the present sequence. [This subsequence found by Dragan K, see second link, below.]
The triangle whose sides are [(k^8+6*k^6+13*k^4+13*k^2+4)/(k*(k^2+2)*(k^2+1)), (k^6+3*k^4+5*k^2+4)/(k*(k^2+2)*(k^2+1)), k*(k^2+4)] has area equal to its third side. Hence A155965 is a subsequence of the present sequence.

			5 is in the sequence because the triangle with sides (37/6, 13/6, 5) has area 5, one side 5, and the other two sides rational.

James R. Buddenhagen, Table of triangles up to n = 145
Dragan K and Rita the dog, Question and answer [broken link]
Ian Connell, APECS elliptic curve software (which runs under old versions of Maple).
Eric Weisstein's World of Mathematics, Heronian Triangle.

Cf. A112087, A054602, A155965, and A206351 (subsequences, see comments).

A120573 a(n) = n^5 + 3n^3 + 2n = n(n^2+1)(n^2+2).

Original entry on oeis.org

6, 60, 330, 1224, 3510, 8436, 17850, 34320, 61254, 103020, 165066, 254040, 377910, 546084, 769530, 1060896, 1434630, 1907100, 2496714, 3224040, 4111926, 5185620, 6472890, 8004144, 9812550, 11934156, 14408010, 17276280, 20584374, 24381060

Offset: 1

David W. Wilson, Jun 17 2006

Largest area of any triangle with integer sides a <= b <= c and inradius n. Triangle has sides (n^2+2, n^4+2n^2+1, n^4+3n^2+1).

a(n) = A002522(n)*A054602(n). - Zerinvary Lajos, Apr 20 2008

David W. Wilson, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (6, -15, 20, -15, 6, -1).

See A120062 for sequences related to integer-sided triangles with integer inradius n.

Cf. A002522, A054602.

with(combinat):seq(lcm(fibonacci(4,n),fibonacci(3,n)),n=1..30); # Zerinvary Lajos, Apr 20 2008

LinearRecurrence[{6,-15,20,-15,6,-1},{6,60,330,1224,3510,8436},30] (* Harvey P. Dale, Aug 14 2023 *)

A085151 Numbers generated by the Fibonacci polynomial x^4 + 3x^2 + 1.

Original entry on oeis.org

5, 29, 109, 305, 701, 1405, 2549, 4289, 6805, 10301, 15005, 21169, 29069, 39005, 51301, 66305, 84389, 105949, 131405, 161201, 195805, 235709, 281429, 333505, 392501, 459005, 533629, 617009, 709805, 812701, 926405, 1051649, 1189189

Offset: 1

Gary W. Adamson, Jun 21 2003

nonn

Start with the Fibonacci polynomials of A011973 (see "examples") and put in appropriate exponents, e.g. {1,1} = x^2 + 1, the generator of A002522; {1,2} = x^3 + 2x, the generator of A054602; and to get the next polynomial, multiply by x and add the previous polynomial, such that the generator for A085151 = x^4 + 3x^2 + 1 = (x)(x^3+2x) + (x^2+1).

			a(2) = f(2) of x^4 + 3x^2 + 1 = 29.
a(2) = 29 = (2)*A054602(2) + A002522(2) = (2)(12) + 5.
[2,2,2,2] = 12/29; a(2) = 29, & 12 = A054602(2). Thus [n,n,n,n] = A054602(n)/A085151(n).

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A002522, A011973, A054602, A057721, A085151.

Mathematica
```
a[n_] := n^4 + 3n^2 + 1; Array[a, 33]
```

def A085151(n): return (m:=n**2)*(m+3)+1 # Chai Wah Wu, May 20 2025

a(n) = n^4 + 3*n^2 + 1.
a(n) = n*A054602(n) + A002522(n).
a(n) = denominator of [n, n, n, n]; with numerator = A054602(n).
a(n) = A057721(n). - R. J. Mathar, Sep 12 2008
From Chai Wah Wu, May 20 2025: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
G.f.: x*(-x^4 - 14*x^2 - 4*x - 5)/(x - 1)^5. (End)
Sum_{n>=1} 1/a(n) = - 1/2 - (Pi/10)*((5*sinh(Pi)+sqrt(5)*sinh(sqrt(5)*Pi))/(cosh(Pi)-cosh(sqrt(5)*Pi))). - Amiram Eldar, Jun 04 2025

More terms from Robert G. Wilson v, Aug 06 2006

cp's OEIS Frontend

A006527 a(n) = (n^3 + 2*n)/3.

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A073133 Table by antidiagonals of T(n,k) = n*T(n,k-1) + T(n,k-2) starting with T(n,1) = 1.

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A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

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A352361 Array read by ascending antidiagonals. A(n, k) = Fibonacci(k, n), where Fibonacci(n, x) are the Fibonacci polynomials.

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A185651 A(n,k) = Sum_{d|n} phi(d)*k^(n/d); square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

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