A023039 -id:A023039 - cp's OEIS frontend

A253368 a(n) = F(12*n)/(12^2) with the Fibonacci numbers F = A000045.

1, 322, 103683, 33385604, 10750060805, 3461486193606, 1114587804280327, 358893811492071688, 115562692712642803209, 37210828159659490561610, 11981771104717643318035211, 3858093084890921488916776332, 1242293991563772001787883943693

Offset: 1

Peter M. Chema, Dec 30 2014

The Fibonacci sequence with seeds 10/9, 10/9 for n = 0 and 1 has integer values precisely for every 12th entry, namely (10/9)*F(12*n), n >= 1: 160, 51520, 16589280, 5341696640, 1720009728800, ... .
Because F(12*n)/(9*2^4) = (F(6*n)/2^3)*(L(6*n)/2) = A049660(n)*A023039(n), this is indeed an integer sequence. Here L = A000032 (Lucas).
For the digital root of this sequence see A253298.
The final digits cycle a sequence of period 10 [1,2,3,4,5,6,7,8,9,0...], see A010879. - Peter M. Chema, Nov 11 2015

Michael De Vlieger, Table of n, a(n) for n = 1..399
Index entries for linear recurrences with constant coefficients, signature (322, -1).

Cf. A000045, A010879, A023039, A049660, A253298.

Table[Fibonacci[12 n]/144, {n, 11}] (* Michael De Vlieger, Apr 03 2015 *)
LinearRecurrence[{322, -1},{1, 322},11] (* Ray Chandler, Aug 12 2015 *)

Vec(x/(x^2-322*x+1) + O(x^20)) \\ Colin Barker, Dec 31 2014

vector(20, n, fibonacci(12*n)/(9*2^4)) \\ Altug Alkan, Nov 11 2015

a(n) = F(12*n)/(9*2^4) = A049660(n)*A023039(n), n >= 1.
G.f.: x / (x^2 - 322*x + 1). - Colin Barker, Dec 31 2014
From Peter Bala, Apr 03 2015: (Start)
For integer k, 1 + k*(36 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/8*Sum_{n >= 1} Fibonacci(6*n)*x^n )*( 1 + k/8*Sum_{n >= 1} Fibonacci(6*n)*(-x)^n ).
1 + 64*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Fibonacci(6*n+3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(6*n+3)*(-x)^n ).
1 + 320*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(6*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(6*n)*(-x)^n ).
(End)
a(n) = (((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^(2*n))))/(144*sqrt(5)). - Colin Barker, Jun 02 2016

Errors in name, data and formula corrected by Colin Barker, Dec 31 2014
Edited: numbers and name changed, formula and programs adjusted by Wolfdieter Lang, Jan 20 2015
Name simplified using "12" as the common number.Peter M. Chema, Mar 31 2016

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1

Offset: 0

Seiichi Manyama, Dec 26 2018

			Square array begins:
   1,  1,   1,    1,      1,       1,         1, ...
   1,  3,  17,   99,    577,    3363,     19601, ...
   1,  5,  49,  485,   4801,   47525,    470449, ...
   1,  7,  97, 1351,  18817,  262087,   3650401, ...
   1,  9, 161, 2889,  51841,  930249,  16692641, ...
   1, 11, 241, 5291, 116161, 2550251,  55989361, ...
   1, 13, 337, 8749, 227137, 5896813, 153090001, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-3 give A000012, A005408, A069129(n+1), A322830.
Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986.
Main diagonal gives A173174.
A(n-1,n) gives A173148(n).
Cf. A322699, A322747.

A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *)

a(n) = 2 * A322699(n) + 1.
A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k).
A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k).
A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1.
A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = 2*n+1.

A132584 a(0)=0, a(1)=4; for n > 1, a(n) = 18*a(n-1) - a(n-2) + 8.

Original entry on oeis.org

0, 4, 80, 1444, 25920, 465124, 8346320, 149768644, 2687489280, 48225038404, 865363202000, 15528312597604, 278644263554880, 5000068431390244, 89722587501469520, 1610006506595061124, 28890394531209630720, 518417095055178291844, 9302617316461999622480

Offset: 0

Mohamed Bouhamida, Nov 14 2007

The old definition given for this sequence was "Sequence allows us to find X values of the equation: X(X + 1) - 5*Y^2 = 0".

With this old definition, if X = a(n), then Y = A207832(n). Now, with u = 2X+1, this Diophantine equation becomes the Pell-Fermat equation u^2 - 20*Y^2 = 1, and then, u = A023039(n) and Y = A207832(n). - Bernard Schott, Jan 25 2023

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000032, A000045, A000217, A007654, A023039, A049683, A292443, A207832.

I:=[0,4,80]; [n le 3 select I[n] else 18*Self(n-1)-Self(n-2)+8: n in [1..30]]; // Vincenzo Librandi, Dec 24 2018

LinearRecurrence[{19, -19, 1}, {0, 4, 80}, 40] (* Vincenzo Librandi, Dec 24 2018 *)
nxt[{a_,b_}]:={b,18b-a+8}; NestList[nxt,{0,4},20][[;;,1]] (* Harvey P. Dale, Aug 25 2024 *)

a(n) = (A023039(n) - 1)/2. - Max Alekseyev, Nov 13 2009
G.f.: -4*x*(x+1)/((x-1)*(x^2-18*x+1)). - Colin Barker, Oct 24 2012
From Amiram Eldar, Jan 11 2022: (Start)
a(n) = 5*Fibonacci(3*n)^2/4 - 1 if n is odd and 5*Fibonacci(3*n)^2/4 if n is even.
A000217(a(n)) = A292443(n). (End)
a(n) = (Lucas(6*n)-2)/4. - Jeffrey Shallit, Jan 20 2023
a(n) = 4 * A049683(n). - Alois P. Heinz, Jan 20 2023

More terms from Max Alekseyev, Nov 13 2009

New definition by Antti Karttunen, Oct 24 2012

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

Original entry on oeis.org

0, 2, 36, 646, 11592, 208010, 3732588, 66978574, 1201881744, 21566892818, 387002188980, 6944472508822, 124613502969816, 2236098580947866, 40125160954091772, 720016798592704030

Offset: 0

Gary Detlefs, Feb 20 2012

Denote as {a,b,c,d} the second-order linear recurrence a(n) = c*a(n-1) + d*a(n-2) with initial terms a, b. The following sequences and recurrence formulas are related to integer solutions of k*x^2 + 1 = y^2.
.
   k             x                        y
   -  -----------------------  -----------------------
A001542 {0,2,6,-1}       A001541 {1,3,6,-1}
A001353 {0,1,4,-1}       A001075 {1,2,4,-1}
A060645 {0,4,18,-1}      A023039 {1,9,18,-1}
A001078 {0,2,10,-1}      A001079 {1,5,10,-1}
A001080 {0,3,16,-1}      A001081 {1,8,16,-1}
A001109 {0,1,6,-1}       A001541 {1,3,6,-1}
A084070 {0,1,38,-1}      A078986 {1,19,38,-1}
A001084 {0,3,20,-1}      A001085 {1,10,20,-1}
A011944 {0,2,14,-1}      A011943 {1,7,14,-1}
A075871 {0,180,1298,-1}  A114047 {1,649,1298,-1}
A068204 {0,4,30,-1}      A069203 {1,15,30,-1}
A001090 {0,1,8,-1}       A001091 {1,4,8,-1}
A121740 {0,8,66,-1}      A099370 {1,33,66,-1}
A202299 {0,4,34,-1}      A056771 {1,17,34,-1}
A174765 {0,39,340,-1}    A114048 {1,179,340,-1}
a(n)    {0,2,18,-1}      A023039 {1,9,18,-1}
A174745 {0,12,110,-1}    A114049 {1,55,110,-1}
A174766 {0,42,394,-1}    A114050 {1,197,394,-1}
A174767 {0,5,48,-1}      A114051 {1,24,48,-1}
A004189 {0,1,10,-1}      A001079 {1,5,10,-1}
A174768 {0,10,102,-1}    A099397 {1,51,102,-1}
The sequence of the c parameter is listed in A180495.

Bruno Berselli, Table of n, a(n) for n = 0..500
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039, A049660, A079962.

m:=16; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2*x/(1-18*x+x^2))); // Bruno Berselli, Jun 19 2019

readlib(issqr):for x from 1 to 720016798592704030 do if issqr(20*x^2+1) then print(x) fi od;

LinearRecurrence[{18, -1}, {0, 2}, 16] (* Bruno Berselli, Feb 21 2012 *)
Table[2 ChebyshevU[-1 + n, 9], {n, 0, 16}]  (* Herbert Kociemba, Jun 05 2022 *)

makelist(expand(((2+sqrt(5))^(2*n)-(2-sqrt(5))^(2*n))/(4*sqrt(5))), n, 0, 15); /* Bruno Berselli, Jun 19 2019 */

a(n) = 18*a(n-1) - a(n-2).
From Bruno Berselli, Feb 21 2012: (Start)
G.f.: 2*x/(1-18*x+x^2).
a(n) = -a(-n) = 2*A049660(n) = ((2 + sqrt(5))^(2*n)-(2 - sqrt(5))^(2*n))/(4*sqrt(5)). (End)
a(n) = Fibonacci(6*n)/4. - Bruno Berselli, Jun 19 2019
For n>=1, a(n) = A079962(6n-3). - Christopher Hohl, Aug 22 2021

A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).

Original entry on oeis.org

0, 1, 324, 104329, 33593616, 10817040025, 3483053294436, 1121532343768369, 361129931640120384, 116282716455774995281, 37442673568827908360100, 12056424606446130716956921, 3882131280602085262951768464, 1250034215929265008539752488489

Offset: 0

Bruno Berselli, Jan 12 2018

16*k*a(n) provides infinitely many x-values solutions (x,y) of x*(5*x + k) = y^2.
This follows from the fact that 5*16*a(n) + 1 is a perfect square: more precisely, 80*a(n) + 1 = A023039(n)^2.
This is a divisibility sequence, that is a(n) divides a(m) if n divides m. It is the case P1 = 324, P2 = 644, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Jan 19 2018

Colin Barker, Table of n, a(n) for n = 0..300
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A023039, A049660, A225786, A253368, A298271.

P:=proc(n) trunc(evalf(((2+sqrt(5))^(4*n)+(2-sqrt(5))^(4*n)-2)/320,1000));
end: seq(P(i),i=0..13); # Paolo P. Lava, Jan 18 2018

CoefficientList[x (1 + x)/((1 - x) (1 - 322 x + x^2)) + O[x]^20, x]

makelist(coeff(taylor(x*(1+x)/((1-x)*(1-322*x+x^2)), x, 0, n), x, n), n, 0, 20);

first(n) = Vec(x*(1 + x)/((1 - x)*(1 - 322*x + x^2)) + O(x^n), -n) \\ Iain Fox, Jan 12 2018

gf = x*(1+x)/((1-x)*(1-322*x+x^2))
print(taylor(gf, x, 0, 20).list())

G.f.: x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).
a(n) = a(-n) = ((2 + sqrt(5))^(4*n) + (2 - sqrt(5))^(4*n) - 2)/320.
a(n) = A225786(n)/48. This is the case k=3 of the first comment. Example: for n = 2, 16*3*a(2) = A225786(2) = 15552 and 15552*(5*15552+3) = 34776^2.
a(n) = A049660(n)^2.
a(n)*(80*a(n) + 1) = 81*A253368(n)^2 for n>0.
a(n)*a(n-2) = (a(n-1) - 1)^2.
a(n) = 322*a(n-1) - a(n-2) + 2.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3). - Iain Fox, Jan 12 2018
a(n) = A298271(n)+A298271(n-1). - R. J. Mathar, Nov 20 2020

A074076 One-sixth of the area of some primitive Heronian triangles with a distance of 2n+1 between the median and altitude points on the longest side.

Original entry on oeis.org

60, 4620, 2024, 5984, 11480, 22960, 41580, 8096, 45920, 521640, 226884, 392920, 438944, 803320, 6725544, 207900, 37966500, 1544620, 6846840, 2295200, 2785484, 9009000, 4016600, 188375200, 3383500, 149240, 5738000, 875124, 12013456, 8848840

Offset: 1

Lekraj Beedassy, Aug 28 2002

nonn

With N=2n+1, such a triangle has sides N*u +/- M, 2*M*u (the latter being cut into M*u +/- N by the corresponding altitude) and inradius M*(N - M)*v. The first entry, in particular, is associated with sequence A023039.

Ray Chandler, Table of n, a(n) for n = 1..499
Eric Weisstein's World of Mathematics, Heronian Triangle.
Wikipedia, Heronian Triangle.

Cf. A074074, A074075, A023039, A002349, A002350.

A033313 := proc(Dcap) local c,i,fr,nu,de ; if issqr(Dcap) then -1; else c := numtheory[cfrac](sqrt(Dcap)) ; for i from 1 do try fr := numtheory[nthconver](c,i) ; nu := numer(fr) ; de := denom(fr) ; if nu^2-Dcap*de^2=1 then RETURN(nu) ; fi; catch: RETURN(-1) ; end try; od: fi: end:
A074076 := proc(n) local Dmin,xmin,Dcap ; Dmin := -1; xmin := -1; mmin := -1; ymin := -1; for m from 1 to n do Dcap := (2*n+1+2*m)*(2*n+1-2*m) ; x := A033313(Dcap) ; if xmin = -1 or (x >0 and xA074076(n),n=1..80) ; # R. J. Mathar, Sep 21 2009

a(n) = M(n)*D(n)*u(n)*v(n)/6, where (u, v) is the fundamental solution to x^2 - D*y^2 = 1, with M = 2*A074075(n); D = A074074(n) = N^2 - M^2.

Removed assertion that these are the minimum areas - R. J. Mathar, Sep 21 2009

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A180498 a(n) = n^2 - 5*floor(n/sqrt(5))^2.

Original entry on oeis.org

1, 4, 4, 11, 5, 16, 4, 19, 1, 20, 41, 19, 44, 16, 45, 11, 44, 4, 41, 80, 36, 79, 29, 76, 20, 71, 9, 64, 121, 55, 116, 44, 109, 31, 100, 16, 89, 164, 76, 155, 61, 144, 44, 131, 25, 116, 4, 99, 196, 80, 181, 59, 164, 36, 145, 11, 124, 239, 101, 220, 76, 199, 49, 176, 20, 151

Offset: 1

Carmine Suriano, Sep 08 2010

nonn

a(n)=1 for n=A023039.

			a(4)=11 since 4^2-5*floor(4/sqrt(5))^2=16-5=11.

Zak Seidov, Table of n, a(n) for n = 1..1000
Zak Seidov, Graphic of A180498.

Cf. A023039, A082532.

cp's OEIS Frontend

A253368 a(n) = F(12*n)/(12^2) with the Fibonacci numbers F = A000045.

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A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j.

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A132584 a(0)=0, a(1)=4; for n > 1, a(n) = 18*a(n-1) - a(n-2) + 8.

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Magma

Mathematica

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A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

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Maple

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Maxima

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A298101 Expansion of x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).

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A074076 One-sixth of the area of some primitive Heronian triangles with a distance of 2n+1 between the median and altitude points on the longest side.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).