A063496 -id:A063496 - cp's OEIS frontend

A063490 a(n) = (2n - 1)(7n^2 - 7n + 6)/6.

1, 10, 40, 105, 219, 396, 650, 995, 1445, 2014, 2716, 3565, 4575, 5760, 7134, 8711, 10505, 12530, 14800, 17329, 20131, 23220, 26610, 30315, 34349, 38726, 43460, 48565, 54055, 59944, 66246, 72975, 80145, 87770, 95864, 104441, 113515, 123100, 133210, 143859, 155061

Offset: 1

N. J. A. Sloane, Aug 01 2001

From Omar E. Pol, Oct 23 2019: (Start)
a(n) is also the sum of terms that are in the n-th finite row and in the n-th finite column of the square [1,n]x[1,n] of the natural number array A000027; e.g., the [1,3]x[1,3] square is
1..3..6
2..5..9
4..8..13,
so that a(1) = 1, a(2) = 2 + 3 + 5 = 10, a(3) = 4 + 6 + 8 + 9 + 13 = 40.
Hence the partial sums give A185505. (End)

Harry J. Smith, Table of n, a(n) for n = 1..1000
Hyunsoo Cho, JiSun Huh, Hayan Nam, and Jaebum Sohn, Combinatorics on bounded free Motzkin paths and its applications, arXiv:2205.15554 [math.CO], 2022. (See p. 14).
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Cf. A064788, A185505.

[(2*n-1)*(7*n^2-7*n+6)/6: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2*n-1)*(7*n^2-7*n+6)/6, {n,1,50}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,10,40,105}, 50] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(7*n^2 - 7*n + 6)/6 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-6 + 12*x + 21*x^2 + 14*x^3 )*exp(x)/6 + 1)) \\ G. C. Greubel, Dec 01 2017

G.f.: x*(1+x)*(1+5*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
a(n) = Sum_{k = n^2-2*n+2..n^2} A064788(k). - Lior Manor, Jan 13 2013
From G. C. Greubel, Dec 01 2017: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
E.g.f.: (-6 + 12*x + 21*x^2 + 14*x^3)*exp(x)/6 + 1. (End)

A063488 a(n) = (2n-1)(n^2 -n +2)/2.

Original entry on oeis.org

1, 6, 20, 49, 99, 176, 286, 435, 629, 874, 1176, 1541, 1975, 2484, 3074, 3751, 4521, 5390, 6364, 7449, 8651, 9976, 11430, 13019, 14749, 16626, 18656, 20845, 23199, 25724, 28426, 31311, 34385, 37654, 41124, 44801, 48691, 52800, 57134, 61699, 66501, 71546, 76840

Offset: 1

N. J. A. Sloane, Aug 01 2001

Sum of two consecutive terms of A006003(n) = n*(n^2+1)/2. a(n) = A006003(n-1) + A006003(n). - Alexander Adamchuk, Jun 03 2006

If a 2-set Y and a 3-set Z are disjoint subsets of an n-set X then a(n-4) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

Harry J. Smith, Table of n, a(n) for n = 1..1000
Milan Janjic, Two Enumerative Functions
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*n*(2*n^2 - 3*n + 1) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Cf. A000217, A006003.
Partial sums of A005918.

[(2*n-1)*(n^2 -n +2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2 n - 1) (n^2 - n + 2)/2, {n, 1, 40}] (* Bruno Berselli, Oct 14 2016 *)
LinearRecurrence[{4,-6,4,-1}, {1,6,20,49}, 50] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(n^2 - n + 2)/2 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2 + 4*x + 3*x^2 + 2*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

G.f.: (1 + x)*(1 + x + x^2)/(1 - x)^4. - Jaume Oliver Lafont, Aug 30 2009
a(n) = A000217(A000217(n)) - A000217(A000217(n-2)). - Bruno Berselli, Oct 14 2016
E.g.f.: (-2 + 4*x + 3*x^2 + 2*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017

A068745 Number of potential flows in 4 X 4 array with integer velocities in -n..n, i.e., number of 4 X 4 arrays with adjacent elements differing by no more than n, counting arrays differing by a constant only once.

Original entry on oeis.org

1, 690437, 1133641543, 164185416899, 6913624013061, 138190481342321, 1678843050246451, 14285299502131463, 93044501704039945, 492225938556374973, 2204710243834695807, 8617480381892283531

Offset: 0

R. H. Hardin, Feb 27 2002

nonn

2 X 2 A063496, 3 X 3 A068744, 5 X 5 A068746, 6 X 6 A068747, by velocity limit 1..14 A068748-A068761, solenoidal flows A068722-A068738.

Cf. 4 X 4 this sequence (degree 4*4-1) with factor 2n-1 ; 3 X 3 A068744 (degree 3*3-1) with factor (2n-1)^2 ; 2 X 2 A063496 (degree 2*2-1) with factor 2n-1.

Let y = 2*n - 1; it appears that a(n) = y*(2623243666*y^14 + 9598591135*y^12 + 17180805187*y^10 + 20342655905*y^8 + 17636121503*y^6 + 10907793260*y^4 + 3135618144*y^2 + 304819200)/81729648000. - R. H. Hardin, Jan 01 2007

A068746 Number of potential flows in 5 X 5 array with integer velocities in -n..n, i.e., number of 5 X 5 arrays with adjacent elements differing by no more than n, counting arrays differing by a constant only once.

Original entry on oeis.org

1, 1366395515, 184422574177355, 523957519578572209, 207345516734034667209, 24953087551680958151267, 1354915464537160758459123

Offset: 0

R. H. Hardin, Feb 27 2002

nonn

2 X 2 A063496, 3 X 3 A068744, 4 X 4 A068745, 6 X 6 A068747, by velocity limit 1..14 A068748-A068761, solenoidal flows A068722-A068738.

A142992 Square array, read by ascending antidiagonals, of the crystal ball sequences for the root lattices of type C_n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 19, 25, 7, 1, 1, 33, 85, 49, 9, 1, 1, 51, 225, 231, 81, 11, 1, 1, 73, 501, 833, 489, 121, 13, 1, 1, 99, 985, 2471, 2241, 891, 169, 15, 1, 1, 129, 1765, 6321, 8361, 4961, 1469, 225, 17, 1

Offset: 0

Peter Bala, Jul 18 2008

The lattice C_n consists of all integer lattice points v = (x_1,...,x_n) in Z^n such that the sum x_1 + ... + x_n is even. Let ||v|| = 1/2 * Sum_{i = 1..n} |x_i|; this defines a norm on C_n. The k-th term of the crystal ball sequence of C_n gives the number of lattice points v in C_n with ||v|| <= k [Bacher et al.]. The case n = 2 is illustrated in the Example section below.
This array has a remarkable relationship with the constant log(2). The row, column and (conjecturally) the diagonal entries of the array occur in series acceleration formulas for log(2) (see the Formula section below for some examples).
See A103884 for the table of coordination sequences of the C_n lattices. For the crystal ball sequences for the A_n and D_n lattices see A108625 and A108553 respectively. For the crystal ball sequences for the product lattices A_1 x ... x A_1(n copies) and A_n x A_n see A008288 and A143007 respectively.

			The square array begins
n\k|0...1....2.....3.....4......5
=================================
.0.|1...1....1.....1.....1......1
.1.|1...3....5.....7.....9.....11
.2.|1...9...25....49....81....121 A016754
.3.|1..19...85...231...489....891 A063496
.4.|1..33..225...833..2241...4961 A142993
.5.|1..51..501..2471..8361..22363 A142994
...
Triangular array begins
n\k|0...1...2...3...4...5
=========================
.0.|1
.1.|1...1
.2.|1...3...1
.3.|1...9...5...1
.4.|1..19..25...7...1
.5.|1..33..85..49...9...1
Case n = 2: The C_2 lattice consists of all integer lattice points v = (x,y) in Z x Z such that x + y is even, equipped with the taxicab type norm ||v|| = 1/2 * (|x| + |y|). There are 8 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 16 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the C_2 lattice (row 2 of the table) begins 1, 1+8 = 9, 1+8+16 = 25, ... .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 . 2 . . . .
. . . 2 . 1 . 2 . . .
. . 2 . 1 . 1 . 2 . .
. 2 . 1 . 0 . 1 . 2 .
. . 2 . 1 . 1 . 2 . .
. . . 2 . 1 . 2 . . .
. . . . 2 . 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .

Seiichi Manyama, Rows n = 0..139, flattened
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.
J. H. Conway and N. J. A. Sloane, Low dimensional lattices VII Coordination sequences, Proc. R. Soc. Lond., Ser. A, 453 (1997), 2369-2389.

Row 2-10 give A016754, A063496, A142993, A142994, A305549, A305721, A305722, A305723, A305724.

Cf. A008288, A103884, A108625, A142979, A143007.

with combinat: T := (n,k) -> add(binomial(2n,2i)*binomial(k+i,n),i = 0..n): for n from 0 to 9 do seq(T(n,k), k = 0..9) end do;

t[n_, k_] := Sum[ Binomial[2*n, 2*i]*Binomial[k+i, n], {i, 0, n}]; Table[t[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

T(n,k) = Sum_{i = 0..n} C(2*n,2*i)*C(k+i,n).
O.g.f. for row n: 1/(1-x)^(n+1) * Sum_{k = 0..n} C(2*n,2*k)*x^k = 1/(1-x) * T(n,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind.
O.g.f. for the array: 1/(1-x) * {(1-t) - x*(1+t)}/{(1-t)^2 - x*(1+t)^2} = (1+x+x^2+x^3+...) + (1+3*x+5*x^2+7*x^3+...)*t + (1+9*x+25*x^2+49*x^3+...)*t^2 + ... .
Row n of the array has the form [p_n(0),p_n(1),p_n(2),...], where the polynomial function p_n(x) = Sum_{k = 0..n} C(2*n,2*k)*C(x+k,n). The first few are p_0(x) = 1, p_1(x) = 2*x+1, p_2(x) = (2*x+1)^2, p_3(x) = (2*x+1)*(8*x^2+8*x+3)/3 and p_4(x) = (2*x+1)^2*(4*x^2+4*x+3)/3.
Alternative expressions for p_n(x) include p_n(x) = Sum_{k = 0..n} 2^(2*k)*n/(n+k)*C(n+k,2*k)*C(x,k) and p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(2*x+1,k).
The polynomials p_n(x) satisfy the 3-term recurrence relation n*p_n(x) = 2*(2*x+1)*p_(n-1)(x)+(n-2)*p(n-2)(x) for n >= 2; their generating function is 1/2*((1+t)/(1-t))^(2*x+1) = 1/2 + (2*x+1)*t + (2*x+1)^2*t^2 + (2*x+1)*(8*x^2+8*x+3)/3*t^3 + ... . Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(2*x+1;b,c) at b = 0, c = -1. Compare with A142979.
The polynomial p_n(x) is the unique polynomial solution to the difference equation (2*x+1)*{f(x+1/2) - f(x-1/2)} = 2*n*f(x), normalized so that f(0) = 1. The function p_n(x) is also the unique polynomial solution to the difference equation (2*x+1)*{(x+1)*f(x+1) + x*f(x-1)} = ((2*x+1)^2 + 2*n^2)*f(x), normalized so that f(0) = 1.
The zeros of p_n(x) lie on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p.4 of [BUMP et al.]).
For n > 0, the entries in row n of the array occur in series acceleration formulas for log(2): 2*log(2) = 1 + (1/2 - 1/6 +...+(-1)^n/(n*(n-1))) + (-1)^(n+1)*Sum_{k >= 1} 1/(k*T(n,k-1)*T(n,k)). For example, the fourth row of the table (n = 3) gives 2*log(2) = 4/3 + 1/(1*1*19) + 1/(2*19*85) + 1/(3*85*231) + ... .
The corresponding result for column k is 2*log(2) = 1 + (1/(1*3) + 1/(2*3*5) +...+ 1/(k*(2*k-1)*(2k+1)) + (2*k+1)*Sum_{n >= 1} (-1)^(n+1)/(n*(n+1)*T(n,k)* T(n+1,k)).
For example, the third column of the table (k = 2) gives 2*log(2) = 41/30 + 5*(1/(1*2*5*25) - 1/(2*3*25*85) + 1/(3*4*85*225) - ... ).
For the main diagonal calculation suggests the result: 2*log(2) = 4/3 + Sum_{n >= 1} (-1)^(n+1)*(5*n+3)/(n*(n+1)*T(n,n)*T(n+1,n+1)).
Similar series acceleration formulas for log(2) come from the row, column and diagonal entries of the square array of Delannoy numbers, A008288 (which may viewed as the array of crystal ball sequences for the product lattices A_1 x...x A_1). For corresponding results for the constants zeta(2) and zeta(3) see A108625 and A143007 respectively.

A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 19, 19, 1, 9, 37, 85, 51, 1, 11, 61, 231, 381, 141, 1, 13, 91, 489, 1451, 1751, 393, 1, 15, 127, 891, 3951, 9331, 8135, 1107, 1, 17, 169, 1469, 8801, 32661, 60691, 38165, 3139, 1, 19, 217, 2255, 17151, 88913, 273127, 398567, 180325, 8953, 1

Offset: 1

R. H. Hardin, Dec 02 2011

Equivalently, the number of compositions of n*(k + 1) into n parts with maximum part size 2*k+1. - Andrew Howroyd, Oct 14 2017

			Some solutions for n=7, k=3:
..1...-2....1...-1....1...-3....0....0....1....2....3...-3....0....2....1....0
.-1....2...-2....2....2....2...-1....0....2....2...-2...-1...-2...-1....2...-1
.-3...-1....1...-3....2....1....0....1....3....0....2....0...-1....2...-2...-1
..0....3....3....3...-2...-2....3....3...-3...-3....0...-1...-1...-1....0....3
..2...-1...-1...-1...-3....0...-3...-2....1...-1...-1....1....1....0....3...-1
..2...-1...-3....0....2....3....0....1...-2....1....1....1....3...-2...-3...-3
.-1....0....1....0...-2...-1....1...-3...-2...-1...-3....3....0....0...-1....3
Table starts:
.   1,      1,       1,        1,        1,         1,...
.   3,      5,       7,        9,       11,        13,...
.   7,     19,      37,       61,       91,       127,...
.  19,     85,     231,      489,      891,      1469,...
.  51,    381,    1451,     3951,     8801,     17151,...
. 141,   1751,    9331,    32661,    88913,    204763,...
. 393,   8135,   60691,   273127,   908755,   2473325,...
.1107,  38165,  398567,  2306025,  9377467,  30162301,...
.3139, 180325, 2636263, 19610233, 97464799, 370487485,...

R. H. Hardin, Table of n, a(n) for n = 1..9999
Michelle Rudolph-Lilith and Lyle E. Muller, On a link between Dirichlet kernels and central multinomial coefficients, Discrete Mathematics 338.9 (2015): 1567-1572.
Wikipedia, Dirichlet kernel.

Columns 1-10: A002426, A005191, A025012, A025014, A201549, A201550, A201551, A322538, A322539, A322540.
Rows 3-10: A003215, A063496(n+1), A083669, A201553, A201554, A322535, A322536, A322537.
Cf. A286928.

seq(print(seq(add((-1)^i*binomial(n, i)*binomial((k+1)*n-(2*k+1)*i-1, n-1), i = 0..floor((1/2)*n)), k = 1..10)), n = 1..10); # Peter Bala, Oct 16 2024

comps[r_, m_, k_] := Sum[(-1)^i*Binomial[r - 1 - i*m, k - 1]*Binomial[k, i], {i, 0, Floor[(r - k)/m]}];  T[n_, k_] := comps[n*(k + 1), 2*k + 1, n]; Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 31 2017, after Andrew Howroyd *)

comps(r, m, k)=sum(i=0, floor((r-k)/m), (-1)^i*binomial(r-1-i*m, k-1)*binomial(k, i));
T(n,k) = comps(n*(k+1), 2*k+1, n); \\ Andrew Howroyd, Oct 14 2017

Empirical: T(n,k) = Sum_{i=0..floor(k*n/(2*k+1))} (-1)^i*binomial(n,i)* binomial((k+1)*n-(2*k+1)*i-1,n-1).
The above empirical formula is true and can be derived from the formula for the number of compositions with given number of parts and maximum part size. - Andrew Howroyd, Oct 14 2017
Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k + 1
T(3,k) = 3*k^2 + 3*k + 1
T(4,k) = (16/3)*k^3 + 8*k^2 + (14/3)*k + 1
T(5,k) = (115/12)*k^4 + (115/6)*k^3 + (185/12)*k^2 + (35/6)*k + 1
T(6,k) = (88/5)*k^5 + 44*k^4 + 46*k^3 + 25*k^2 + (37/5)*k + 1
T(7,k) = (5887/180)*k^6 + (5887/60)*k^5 + (2275/18)*k^4 + (357/4)*k^3 + (6643/180)*k^2 + (259/30)*k + 1
T(m,k) = (1/Pi)*integral_{x=0..Pi} (sin((k+1/2)x)/sin(x/2))^m dx; for the proof see Dirichlet Kernel link; so f(m,n) = (1/Pi)*integral_{x=0..Pi} (Sum_{k=-n..n} exp(I*k*x))^m dx = sum(integral(exp(I(k_1+...+k_m).x),x=0..Pi)/Pi,{k_1,...,k_m=-n..n}) = sum(delta_0(k1+...+k_m),{k_1,...,k_m=-n..n}) = number of arrays of m integers in -n..n with sum zero. - Yalcin Aktar, Dec 03 2011
T(n, k) = the constant term in the expansion of (x^(-k) + ... + x^(-1) + 1 + x + ... + x^k)^n = the coefficient of x^(k*n) (i.e., the central coefficient) in the expansion of (1 + x + ... + x^(2*k))^n = the coefficient of x^(k*n) in the expansion of ( (1 - x^(2*k+1))/(1 - x) )^n. Expanding the binomials and collecting terms gives the empirical formula above. - Peter Bala, Oct 16 2024

A063494 a(n) = (2n - 1)(7n^2 - 7n + 3)/3.

Original entry on oeis.org

1, 17, 75, 203, 429, 781, 1287, 1975, 2873, 4009, 5411, 7107, 9125, 11493, 14239, 17391, 20977, 25025, 29563, 34619, 40221, 46397, 53175, 60583, 68649, 77401, 86867, 97075, 108053, 119829, 132431, 145887, 160225, 175473, 191659, 208811, 226957, 246125, 266343, 287639

Offset: 1

N. J. A. Sloane, Aug 01 2001

Interpret A176271 as an infinite square array read by antidiagonals, with rows 1,5,11,19,...; 3,9,17,27,... and so on. The sum of the terms in the n X n upper submatrix are s(n) = 1, 18, 93, 296, ... = n^2*(7*n^2-1)/6, and a(n) = s(n) - s(n-1) are the first differences. - J. M. Bergot, Jun 27 2013

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n - 1)*(7*n^2 - 7*n + 3)/3: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2*n - 1)*(7*n^2 - 7*n + 3)/3, {n,1,30}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,17,75,203}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(7*n^2 - 7*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+21*x^2+14*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017

G.f.: x*(1+x)*(1+12*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
E.g.f.: (-3 + 6*x + 21*x^2 + 14*x^3)*exp(x)/3 + 1. - G. C. Greubel, Dec 01 2017
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, May 11 2023

A063491 a(n) = (2n - 1)(3n^2 - 3n + 2)/2.

Original entry on oeis.org

1, 12, 50, 133, 279, 506, 832, 1275, 1853, 2584, 3486, 4577, 5875, 7398, 9164, 11191, 13497, 16100, 19018, 22269, 25871, 29842, 34200, 38963, 44149, 49776, 55862, 62425, 69483, 77054, 85156, 93807, 103025, 112828, 123234, 134261, 145927, 158250, 171248, 184939

Offset: 1

N. J. A. Sloane, Aug 01 2001

A triangle has sides of lengths 6*n-3, 6*n^2-6*n+4, and 6*n^2-6*n+7; for n>2 its area is 6*sqrt(a(n)^2 - 1). - J. M. Bergot, Aug 30 2013

[The source of this is using (n,n+1), (n+1,n+2), and (n+2,n+3) as (a,b) in the creation of three Pythagorean triangles with sides b^2-a^2, 2*a*b, and a^2+b^2. Combine the three respective sides to create a new larger triangle, then find its area. It is not simply working backwards from the sequence. As well, the sequence has this as its first comment to show that the numbers are actually doing something to find a solution.]

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Cf. A005448, A004188.

[(2*n-1)*(3*n^2 -3*n +2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

LinearRecurrence[{4,-6,4,-1},{1,12,50,133},40] (* Harvey P. Dale, Jun 05 2016 *)
Table[(2*n-1)*(3*n^2 -3*n +2)/2, {n,1,30}] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(3*n^2 - 3*n + 2)/2 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

a <- c(0, 1, 9, 38, 110)
for(n in (length(a)+1):40)
  a[n] <- +4*a[n-1]-6*a[n-2]+4*a[n-3]-a[n-4]
a [Yosu Yurramendi, Sep 04 2013]

G.f.: x*(1+x)*(1+7*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = +4*a(n-1) -6*a(n-2) +4*a(n-3) -1*a(n-4) n > 3, a(1)=1, a(2)=12, a(3)=50, a(4)=133. - Yosu Yurramendi, Sep 04 2013
E.g.f.: (-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017
From Bruce J. Nicholson, Jun 17 2020: (Start)
a(n) = A005448(n) * A005408(n-1).
a(n) = A004188(n) + A004188(n-1). (End)

A063492 a(n) = (2n - 1)(11n^2 - 11n + 6)/6.

Original entry on oeis.org

1, 14, 60, 161, 339, 616, 1014, 1555, 2261, 3154, 4256, 5589, 7175, 9036, 11194, 13671, 16489, 19670, 23236, 27209, 31611, 36464, 41790, 47611, 53949, 60826, 68264, 76285, 84911, 94164, 104066, 114639, 125905, 137886, 150604, 164081, 178339, 193400, 209286, 226019

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n-1)*(11*n^2-11*n+6)/6: n in [1..40]]; // Vincenzo Librandi, Dec 16 2015

A063492:=n->(2*n - 1)*(11*n^2 - 11*n + 6)/6: seq(A063492(n), n=1..50); # Wesley Ivan Hurt, Dec 16 2015

Table[(2*n-1)*(11*n^2-11*n+6)/6, {n, 5!}] (* Vladimir Joseph Stephan Orlovsky, Sep 18 2008 *)
LinearRecurrence[{4, -6, 4, -1}, {1, 14, 60, 161}, 40] (* Vincenzo Librandi, Dec 16 2015 *)

a(n) = { (2*n - 1)*(11*n^2 - 11*n + 6)/6 } \\ Harry J. Smith, Aug 23 2009

Vec(x*(1+x)*(1+9*x+x^2)/(1-x)^4 + O(x^100)) \\ Altug Alkan, Dec 16 2015

A063492_list, m = [], [22, -11, 2, 1]
for _ in range(10**2):
    A063492_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

G.f.: x*(1+x)*(1 + 9*x + x^2)/(1-x)^4. - Colin Barker, Apr 24 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4. - Wesley Ivan Hurt, Dec 16 2015
E.g.f.: (-6 + 12*x + 33*x^2 + 22*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

A063493 a(n) = (2n-1)(13n^2-13n+6)/6.

Original entry on oeis.org

1, 16, 70, 189, 399, 726, 1196, 1835, 2669, 3724, 5026, 6601, 8475, 10674, 13224, 16151, 19481, 23240, 27454, 32149, 37351, 43086, 49380, 56259, 63749, 71876, 80666, 90145, 100339, 111274, 122976, 135471, 148785, 162944, 177974, 193901, 210751, 228550, 247324, 267099

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

[(2*n-1)*(13*n^2-13*n+6)/6: n in [1..40]]; // Vincenzo Librandi, Dec 16 2015

Table[(2 n - 1) (13 n^2 - 13 n + 6)/6, {n, 1, 40}] (* Bruno Berselli, Dec 16 2015 *)
LinearRecurrence[{4,-6,4,-1}, {1,16,70,189}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(13*n^2 - 13*n + 6)/6 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-6+12*x+39*x^2+26*x^3)*exp(x)/6 + 1)) \\ G. C. Greubel, Dec 01 2017

A063493_list, m = [], [26, -13, 2, 1]
for _ in range(10**2):
    A063493_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

G.f.: x*(1+x)*(1+11*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Dec 16 2015
E.g.f.: (-6 + 12*x + 39*x^2 + 26*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

cp's OEIS Frontend

A063490 a(n) = (2*n - 1)*(7*n^2 - 7*n + 6)/6.

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A063488 a(n) = (2*n-1)*(n^2 -n +2)/2.

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A068745 Number of potential flows in 4 X 4 array with integer velocities in -n..n, i.e., number of 4 X 4 arrays with adjacent elements differing by no more than n, counting arrays differing by a constant only once.

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A068746 Number of potential flows in 5 X 5 array with integer velocities in -n..n, i.e., number of 5 X 5 arrays with adjacent elements differing by no more than n, counting arrays differing by a constant only once.

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A142992 Square array, read by ascending antidiagonals, of the crystal ball sequences for the root lattices of type C_n.

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A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

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A063494 a(n) = (2*n - 1)*(7*n^2 - 7*n + 3)/3.

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A063491 a(n) = (2*n - 1)*(3*n^2 - 3*n + 2)/2.

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A063490 a(n) = (2n - 1)(7n^2 - 7n + 6)/6.

A063488 a(n) = (2n-1)(n^2 -n +2)/2.

A063494 a(n) = (2n - 1)(7n^2 - 7n + 3)/3.

A063491 a(n) = (2n - 1)(3n^2 - 3n + 2)/2.

A063492 a(n) = (2n - 1)(11n^2 - 11n + 6)/6.

A063493 a(n) = (2n-1)(13n^2-13n+6)/6.