A005900 -id:A005900 - cp's OEIS frontend

A124258 Triangle whose rows are sequences of increasing and decreasing squares: 1; 1,4,1; 1,4,9,4,1; ...

1, 1, 4, 1, 1, 4, 9, 4, 1, 1, 4, 9, 16, 9, 4, 1, 1, 4, 9, 16, 25, 16, 9, 4, 1, 1, 4, 9, 16, 25, 36, 25, 16, 9, 4, 1, 1, 4, 9, 16, 25, 36, 49, 36, 25, 16, 9, 4, 1, 1, 4, 9, 16, 25, 36, 49, 64, 49, 36, 25, 16, 9, 4, 1, 1, 4, 9, 16, 25, 36, 49, 64, 81, 64, 49, 36, 25, 16, 9, 4, 1, 1, 4, 9, 16

Offset: 1

Jonathan Vos Post, Dec 16 2006

The triangle A003983 with individual entries squared and each 2nd row skipped.
Analogous to A004737. - Peter Bala, Sep 25 2007
T(n,k) =  min(n,k)^2. The order of the list T(n,k) is by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). - Boris Putievskiy, Jan 13 2013

			Triangle starts
  1;
  1, 4, 1;
  1, 4, 9, 4, 1:
  1, 4, 9, 16, 9, 4, 1:
From _Boris Putievskiy_, Jan 13 2013: (Start)
The start of the sequence as table:
  1...1...1...1...1...1...
  1...4...4...4...4...4...
  1...4...9...9...9...9...
  1...4...9..16..16..16...
  1...4...9..16..25..25...
  1...4...9..16..25..36...
  ...
The start of the sequence as triangle array read by rows:
  1;
  1, 4, 1;
  1, 4, 9,  4,  1;
  1, 4, 9, 16,  9,  4,  1;
  1, 4, 9, 16, 25, 16,  9,  4, 1;
  1, 4, 9, 16, 25, 36, 25, 16, 9, 4, 1;
  ...
Row number k contains 2*k-1 numbers 1,4,...,(k-1)^2,k^2,(k-1)^2,...,4,1. (End)

Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A005900 (row sums), A004737, A133819, A133823, A133824, A133825, A003983.

A003983 := proc(n,k) min(n,k) ; end: A124258 := proc(n,k) A003983(n,k)^2 ; end: for d from 1 to 20 by 2 do for c from 1 to d do printf("%d, ",A124258(d+1-c,c)) ; od: od: # R. J. Mathar, Sep 21 2007
# second Maple program:
T:= n-> i^2$i=1..n, (n-i)^2$i=1..n-1:
seq(T(n), n=1..10);  # Alois P. Heinz, Feb 15 2022

Flatten[Table[Join[Range[n]^2,Range[n-1,1,-1]^2],{n,10}]] (* Harvey P. Dale, Jun 14 2015 *)

O.g.f.: (1+qx)^2/((1-x)(1-qx)^2(1-q^2x)) = 1 + x(1 + 4q + q^2) + x^2(1 + 4q + 9q^2 + 4q^3 + q^4) + ... . - Peter Bala, Sep 25 2007
From Boris Putievskiy, Jan 13 2013: (Start)
a(n) = (A004737(n))^2.
a(n) = (floor(sqrt(n-1)) - |n- floor(sqrt(n-1))^2- floor(sqrt(n-1))-1| +1)^2. (End)

More terms from R. J. Mathar, Sep 21 2007

Edited by N. J. A. Sloane, Jun 30 at the suggestion of R. J. Mathar

A322611 Numbers that are sums (of a nonempty sequence) of consecutive centered square numbers.

Original entry on oeis.org

1, 5, 6, 13, 18, 19, 25, 38, 41, 43, 44, 61, 66, 79, 84, 85, 102, 113, 127, 140, 145, 146, 181, 187, 198, 212, 221, 225, 230, 231, 258, 259, 265, 300, 313, 325, 326, 338, 343, 344, 365, 402, 404, 421, 439, 445, 470, 481, 483, 486, 488, 489, 524, 545, 547, 578, 585, 613, 626, 651, 660

Offset: 1

Ilya Gutkovskiy, Dec 20 2018

nonn

Eric Weisstein's World of Mathematics, Centered Square Number

Cf. A001844, A005900, A034705, A320728, A322479, A322610, A322638, A322640.

anmax = 1000; nmax = Floor[Sqrt[anmax/2]] + 1; Select[Union[Flatten[Table[Sum[k^2 + (k + 1)^2, {k, i, j}], {i, 0, nmax}, {j, i, nmax}]]], # <= anmax &] (* Vaclav Kotesovec, Dec 21 2018 *)

A061804 a(n) = 2n(2*n^2 + 1).

Original entry on oeis.org

0, 6, 36, 114, 264, 510, 876, 1386, 2064, 2934, 4020, 5346, 6936, 8814, 11004, 13530, 16416, 19686, 23364, 27474, 32040, 37086, 42636, 48714, 55344, 62550, 70356, 78786, 87864, 97614, 108060, 119226, 131136, 143814, 157284, 171570, 186696, 202686, 219564

Offset: 0

Amarnath Murthy, May 28 2001

Sum of n-th row of triangle of even numbers: (2, 4), (6, 8, 10, 12), (14, 16, 18, 20, 22, 24), (26, 28, 30, 32, 34, 36, 38, 40), ..., where n-th row contains 2n terms.

Also, integer values of n^3/2 + n. - Arkadiusz Wesolowski, Jul 20 2012

			a(3) = 114 = 14 + 16 + 18 + 20 + 22 + 24.

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

Cf. A002061, A005900.

A061804:=n->2*n*(2*n^2 + 1); seq(A061804(n), n=0..50); # Wesley Ivan Hurt, Mar 20 2014

Table[2*n*(2*n^2 + 1),{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Dec 04 2010 *)
LinearRecurrence[{4,-6,4,-1},{0,6,36,114},40] (* Harvey P. Dale, Feb 04 2023 *)

a(n) = { 2*n*(2*n^2 + 1) } \\ Harry J. Smith, Jul 28 2009

def A061804(n): return n*((n**2<<2) + 2) # Chai Wah Wu, Aug 29 2022

a(n) = 6*A005900(n).
G.f.: 6*x*(1 + x)^2/(1 - x)^4. - Colin Barker, Apr 20 2012
a(n) = A002061(A002061(n+1)) - A002061(A002061(n)). - Daniel Poveda Parrilla, Jun 10 2017

More terms from Larry Reeves (larryr(AT)acm.org) and Alford Arnold, May 29 2001

Better description from Dean Hickerson, Jun 05 2001

A142985 a(1) = 1, a(2) = 6, a(n+2) = 6a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 6, 42, 324, 2784, 26424, 275472, 3132576, 38629440, 513708480, 7331489280, 111798455040, 1814503057920, 31234337164800, 568451665152000, 10906950910464000, 220060558384128000, 4657890328906752000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A005900, A142983, A142984, A142986, A142987.

Cf. A002378.

a142985 n = a142985_list !! (n-1)
a142985_list = 1 : 6 : zipWith (+)
                       (map (* 6) $ tail a142985_list)
                       (zipWith (*) (drop 2 a002378_list) a142985_list)
-- Reinhard Zumkeller, Jul 17 2015

p := n -> (2*n^3+n)/3: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==6,a[n]==6a[n-1]+(n-1)n*a[n-2]},a,{n,20}] (* Harvey P. Dale, Sep 20 2013 *)

a(n) = n!*p(n+1)*Sum {k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^3 + n)/3 = A005900(n).
Recurrence: a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 6, b(2) = 38. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(6 + 1*2/(6 + 2*3/(6 + 3*4/(6 + ... + n*(n - 1)/6)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k = 1..oo} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(6 + 1*2/(6 + 2*3/(6 + 3*4/(6 + ... + n*(n - 1)/(6 + ...))))) = 6*log(2) - 4, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

A166345 Triangle T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)p(n, x)/x, p(n, x) = xD( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x(1+x)/(1-x)^3, and p(3, x) = x(1+2*x+x^2)/(1-x)^4, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 7, 7, 1, 1, 18, 42, 18, 1, 1, 41, 198, 198, 41, 1, 1, 88, 799, 1584, 799, 88, 1, 1, 183, 2925, 10331, 10331, 2925, 183, 1, 1, 374, 10056, 58874, 103310, 58874, 10056, 374, 1, 1, 757, 33160, 305888, 869794, 869794, 305888, 33160, 757, 1

Offset: 1

Roger L. Bagula, Oct 12 2009

			Triangle begins as:
  1;
  1,   1;
  1,   2,     1;
  1,   7,     7,      1;
  1,  18,    42,     18,      1;
  1,  41,   198,    198,     41,      1;
  1,  88,   799,   1584,    799,     88,      1;
  1, 183,  2925,  10331,  10331,   2925,    183,     1;
  1, 374, 10056,  58874, 103310,  58874,  10056,   374,   1;
  1, 757, 33160, 305888, 869794, 869794, 305888, 33160, 757, 1;

Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, MaGraw-Hill, New York, 1976, page 91

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A166340, A166341, A166343, A166344, A166346, A166349.

Cf. A005900, A123125.

(* First program *)
p[x_, 1]:= x/(1-x)^2;
p[x_, 2]:= x*(1+x)/(1-x)^3;
p[x_, 3]:= x*(1+10*x+x^2)/(1-x)^4;
p[x_, n_]:= p[x, n]= x*D[p[x, n-1], x]
Table[CoefficientList[(1-x)^(n+1)*p[x, n]/x, x], {n,12}]//Flatten
(* Second program *)
b[n_, k_, m_]:= If[n<2, 1, If[k==0, 0, k^(n-1)*((m+3)*k^2 - m)/3]];
t[n_, k_, m_]:= t[n,k,m]= Sum[(-1)^(k-j)*Binomial[n+1, k-j]*b[n,j,m], {j,0,k}];
T[n_, k_, m_]:= T[n,k,m]= If[k==1, 1, t[n-1,k,m] - t[n-1,k-1,m]];
Table[T[n,k,-1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Mar 11 2022 *)

def b(n,k,m):
    if (n<2): return 1
    elif (k==0): return 0
    else: return k^(n-1)*((m+3)*k^2 - m)/3
@CachedFunction
def t(n,k,m): return sum( (-1)^(k-j)*binomial(n+1, k-j)*b(n,j,m) for j in (0..k) )
def A166345(n,k): return 1 if (k==1) else t(n-1,k,-1) - t(n-1,k-1,-1)
flatten([[A166345(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Mar 11 2022

T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)*p(n, x)/x, p(n, x) = x*D( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x*(1+x)/(1-x)^3, and p(3, x) = x*(1+2*x+x^2)/(1-x)^4.
From G. C. Greubel, Mar 11 2022: (Start)
T(n, k) = t(n-1, k) - t(n-1, k-1), T(n,1) = 1, where t(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1, k-j)*b(n, j), b(n, k) = k^(n-2)*A005900(k), b(n, 0) = 1, and b(1, k) = 1.
T(n, n-k) = T(n, k). (End)

Edited by G. C. Greubel, Mar 11 2022

A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.

Original entry on oeis.org

1, 3, 1, 5, 6, 1, 7, 19, 9, 1, 9, 44, 42, 12, 1, 11, 85, 138, 74, 15, 1, 13, 146, 363, 316, 115, 18, 1, 15, 231, 819, 1059, 605, 165, 21, 1, 17, 344, 1652, 2984, 2470, 1032, 224, 24, 1, 19, 489, 3060, 7380, 8378, 4974, 1624, 292, 27, 1, 21, 670, 5301, 16488

Offset: 1

Clark Kimberling, Mar 03 2012

For a discussion and guide to related arrays, see A208510.
Riordan array ((1+x)/(1-x)^2, x(1+x)/(1-x)^2) (follows from Kruchinin formula). - Ralf Stephan, Jan 02 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A099375 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

			First five rows:
1
3...1
5...6....1
7...19...9....1
9...44...42...12...1
First five polynomials v(n,x):
1
3 + x
5 + 6x + x^2
7 + 19x + 9x^2 + x^3
9 + 44x + 42x^2 + 12x^3 + x^4
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
/1        \/1        \/1        \      /1            \
|3 1      ||0 1      ||0 1      |      |3  1         |
|5 3 1    ||0 3 1    ||0 0 1    |... = |5  6  1      |
|7 5 3 1  ||0 5 3 1  ||0 0 3 1  |      |7 19  9  1   |
|9 7 5 3 1||0 7 5 3 1||0 0 5 3 1|      |9 44 42 12 1 |
|...      ||...      ||...      |      |...
(End)

Cf. A208660, A208510. A099375.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A208660 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A208904 *)

u(n,x)=u(n-1,x)+2x*v(n-1,x),
v(n,x)=u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Vladimir Kruchinin, Mar 11 2013: (Start)
T(n,k) = sum(i=0..n, binomial(i+k-1,2*k-1)*binomial(k,n-i))
((x+x^2)/(1-x)^2)^k = sum(n>=k, T(n,k)*x^n).
T(n,2)=A005900(n).
T(2*n-1,n) / n = A003169(n).
T(2*n,n) = A156894(n), n>1.
sum(k=1..n, T(n,k)) = A003946(n).
sum(k=1..n, T(n,k)*(-1)^(n+k)) = A078050(n).
n*sum(k=1..n, T(n,k)/k) = A058481(n). (End)
Recurrence: T(n+1,k+1) = sum {i = 0..n-k} (2*i + 1)*T(n-i,k). - Peter Bala, Jul 21 2014

A213752 Rectangular array: (row n) = b**c, where b(h) = 2*h-1, c(h) = b(n-1+h), n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 6, 3, 19, 14, 5, 44, 37, 22, 7, 85, 76, 55, 30, 9, 146, 135, 108, 73, 38, 11, 231, 218, 185, 140, 91, 46, 13, 344, 329, 290, 235, 172, 109, 54, 15, 489, 472, 427, 362, 285, 204, 127, 62, 17, 670, 651, 600, 525, 434, 335, 236, 145, 70, 19, 891, 870, 813

Offset: 1

Clark Kimberling, Jun 20 2012

Principal diagonal:  A100157
Antidiagonal sums:  A071238
row 1,  (1,3,5,7,9,...)**(1,3,5,7,9,...): A005900
row 2,  (1,3,5,7,9,...)**(3,5,7,9,11,...): A143941
row 3,  (1,3,5,7,9,...)**(5,7,9,11,13,...): (2*k^3 + 12*k^2 + k)/6
row 4,  (1,3,5,7,9,...)**(7,9,11,13,15,,...): (2*k^3 + 18*k^2 + k)/6
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1...6....19...44....85....146
3...14...37...76....135...218
5...22...55...108...185...290
7...30...73...140...235...362
9...38...91...172...285...434

Cf. A213500.

b[n_] := 2 n - 1; c[n_] := 2 n - 1;
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213752 *)
Table[t[n, n], {n, 1, 40}] (* A100157 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A071238 *)

T(n,k) = 4*T(n,k-1)-6*T(n,k-2)+4*T(n,k-3)-T(n,k-4).

G.f. for row n: f(x)/g(x), where f(x) = 2*n - 1 + 2*x - (2*n - 3)*x^2 and g(x) = (1 - x )^4.

A279512 Sierpinski octahedron numbers a(n) = 26^n + 32^n + 1.

Original entry on oeis.org

6, 19, 85, 457, 2641, 15649, 93505, 560257, 3360001, 20156929, 120935425, 725600257, 4353576961, 26121412609, 156728377345, 940370067457, 5642220011521, 33853319282689, 203119914123265, 1218719481593857, 7312316883271681, 43873901287047169, 263243407697117185

Offset: 0

Steven Beard, Dec 14 2016

Sierpinski recursion applied to octahedron. Cf. A279511 for square pyramids.

Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, Octahedron flake and Sierpinski tetrahedron.
Index entries for linear recurrences with constant coefficients, signature (9,-20,12).

Cf. A005900, A047999, A279511.

LinearRecurrence[{9, -20, 12}, {6, 19, 85}, 50] (* or *) Table[2*6^n + 3*2^n + 1, {n,0,50}] (* G. C. Greubel, Dec 22 2016 *)

Vec((6 - 35*x + 34*x^2) / ((1 - x)*(1 - 2*x)*(1 - 6*x)) + O(x^30)) \\ Colin Barker, Dec 15 2016

def a(n): return 2*6**n + 3*2**n + 1
print([a(n) for n in range(23)]) # Michael S. Branicky, Jun 19 2021

a(n) = 3*2^n + 2^(n+1)*3^n + 1.
a(n) = 6a(n-1) - 6(2^n+1) + 1.
a(n) = 6a(n-1) - (3*2^(n+1) + 5).
a(n) = 2*6^n + 3*2^n + 1.
From Colin Barker, Dec 15 2016: (Start)
a(n) = 9*a(n-1) - 20*a(n-2) + 12*a(n-3) for n>2.
G.f.: (6 - 35*x + 34*x^2) / ((1 - x)*(1 - 2*x)*(1 - 6*x)).
(End)

Incorrect terms corrected by Colin Barker, Dec 15 2016

A057884 A square array based on tetrahedral numbers (A000292) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 4, 1, 1, 0, 4, 2, 1, 10, 4, 5, 3, 1, 0, 10, 8, 7, 4, 1, 20, 10, 14, 13, 10, 5, 1, 0, 20, 20, 22, 20, 14, 6, 1, 35, 20, 30, 34, 35, 30, 19, 7, 1, 0, 35, 40, 50, 56, 55, 44, 25, 8, 1, 56, 35, 55, 70, 84, 91, 85, 63, 32, 9, 1, 0, 56, 70, 95, 120, 140, 146, 129, 88, 40, 10, 1

Offset: 0

Henry Bottomley, Nov 20 2000

			Rows are (1,0,4,0,10,0,20,...), (1,1,4,4,10,10,20,...), (1,2,5,8,14,20,30,...), (1,3,7,13,22,34,50,...), (1,4,10,20,35,56,84,...) etc.

Rows are A000292 with zeros, A058187 (A000292 with terms duplicated), A006918, A002623, A000292, A000330, A005900, A001845, A008412.

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(4, 1)=4, T(0, 2n)=T(4, n) and T(0, 2n+1)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2)^4.

A130713 a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.

Original entry on oeis.org

1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Paul Curtz and Tanya Khovanova, Jul 01 2007

Self-convolution of A019590. Up to a sign the convolutional inverse of the natural numbers sequence. - Tanya Khovanova, Jul 14 2007

Iterated partial sums give the chain A130713 -> A113311 -> A008574 -> A001844 -> A005900 -> A006325 -> A033455 -> A259181, up to index. The k-th term of the n-th partial sums is (n^2-7n+14 + 4k(k+n-4))(k+n-4)!/(k-1)!/(n-1)!, for k > 3-n. Iterating partial sums in reverse (n-th differences with n zeros prepended) gives row (n+3) of A182533, modulo signs and trailing zeros. - Travis Scott, Feb 19 2023

Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.

A130713:=n->binomial(2*n, n^2); seq(A130713(n), n=0..100); # Wesley Ivan Hurt, Mar 08 2014

Table[Binomial[2 n, n^2], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 08 2014 *)

G.f.: 1 + 2*x + x^2.

a(n) = binomial(2n, n^2). - Wesley Ivan Hurt, Mar 08 2014

cp's OEIS Frontend

A124258 Triangle whose rows are sequences of increasing and decreasing squares: 1; 1,4,1; 1,4,9,4,1; ...

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A322611 Numbers that are sums (of a nonempty sequence) of consecutive centered square numbers.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A061804 a(n) = 2*n*(2*n^2 + 1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

Extensions

A142985 a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

A166345 Triangle T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)*p(n, x)/x, p(n, x) = x*D( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x*(1+x)/(1-x)^3, and p(3, x) = x*(1+2*x+x^2)/(1-x)^4, read by rows.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Mathematica

Sage

Formula

Extensions

A208904 Triangle of coefficients of polynomials v(n,x) jointly generated with A208660; see the Formula section.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A213752 Rectangular array: (row n) = b**c, where b(h) = 2*h-1, c(h) = b(n-1+h), n>=1, h>=1, and ** = convolution.

Views

Author

Keywords

Comments

Examples

Crossrefs

A061804 a(n) = 2n(2*n^2 + 1).

A142985 a(1) = 1, a(2) = 6, a(n+2) = 6a(n+1) + (n + 1)(n + 2)*a(n).

A166345 Triangle T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)p(n, x)/x, p(n, x) = xD( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x(1+x)/(1-x)^3, and p(3, x) = x(1+2*x+x^2)/(1-x)^4, read by rows.

A279512 Sierpinski octahedron numbers a(n) = 26^n + 32^n + 1.