A132596 -id:A132596 - cp's OEIS frontend

A173115 a(n) = -(sin(2narccos(sqrt(3))))^2.

0, 24, 2400, 235224, 23049600, 2258625624, 221322261600, 21687323011224, 2125136332838400, 208241673295152024, 20405558846592060000, 1999536525292726728024, 195934173919840627286400

Offset: 0

Artur Jasinski, Feb 10 2010

Index entries for linear recurrences with constant coefficients, signature (99,-99,1)

Cf. A001079, A108741, A132592, A146311, A146312, A146313.

Table[ -Round[N[Sin[2 n ArcCos[Sqrt[3]]]^2, 100]], {n, 0, 20}]
OR
Table[Round[N[ -1/2 + (1/4) (49 + 20 Sqrt[6])^n + (1/4) (49 - 20 Sqrt[6])^n]], {n, 0, 6}]
OR
Clear[a]; a[n_] := a[n] = 99 a[n - 1] - 99 a[n - 2] + a[n - 3]; a[0] = 0; a[1] = 24; a[2] = 2400; Table[a[n], {n, 0, 10}]

a(n)=([0,1,0;0,0,1;1,-99,99]^n*[0;24;2400])[1,1] \\ Charles R Greathouse IV, Jun 11 2015

a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3), n > 2.
Binet formula: a(n) = -1/2 + (1/4)(49 + 20*sqrt(6))^n + (1/4)(49 - 20*sqrt(6))^n.
a(n) = 24*A108741(n).
From R. J. Mathar, Aug 23 2012: (Start)
G.f.: -24*x*(1+x) / ( (x-1)*(x^2-98*x+1) ).
a(n) = A132596(2n). (End)

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 8, 2, 0, 0, 49, 24, 3, 0, 0, 288, 242, 48, 4, 0, 0, 1681, 2400, 675, 80, 5, 0, 0, 9800, 23762, 9408, 1444, 120, 6, 0, 0, 57121, 235224, 131043, 25920, 2645, 168, 7, 0, 0, 332928, 2328482, 1825200, 465124, 58080, 4374, 224, 8, 0

Offset: 0

Seiichi Manyama, Dec 23 2018

			Square array begins:
   0, 0,   0,    0,      0,       0,        0, ...
   0, 1,   8,   49,    288,    1681,     9800, ...
   0, 2,  24,  242,   2400,   23762,   235224, ...
   0, 3,  48,  675,   9408,  131043,  1825200, ...
   0, 4,  80, 1444,  25920,  465124,  8346320, ...
   0, 5, 120, 2645,  58080, 1275125, 27994680, ...
   0, 6, 168, 4374, 113568, 2948406, 76545000, ...

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

Columns 0-5 give A000004, A001477, A033996, A322675, A322677, A322745.
Rows 0-9 give A000004, A001108, A132596, A007654(n+1), A132584, A322707, A322708, A322709, A132592, A132593.
Main diagonal gives A322746.
Cf. A173175 (A(n,2*n)), A322790.

Unprotect[Power]; 0^0 := 1; Protect[Power]; Table[(-1 + Sum[Binomial[2 k, 2 j] (# + 1)^(k - j)*#^j, {j, 0, k}])/2 &[n - k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jan 01 2019 *)
nmax = 9; row[n_] := LinearRecurrence[{4n+3, -4n-3, 1}, {0, n, 4n(n+1)}, nmax+1]; T = Array[row, nmax+1, 0]; A[n_, k_] := T[[n+1, k+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 06 2019 *)

def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  (0..n).map{|i| (0..k).inject(-1){|s, j| s + ncr(2 * k, 2 * j) * (i + 1) ** (k - j) * i ** j} / 2}
end
def A322699(n)
  a = []
  (0..n).each{|i| a << A(i, n - i)}
  ary = []
  (0..n).each{|i|
    (0..i).each{|j|
      ary << a[i - j][j]
    }
  }
  ary
end
p A322699(10)

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

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Original entry on oeis.org

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055

Offset: 1

Bruno Berselli, Dec 10 2013

For n>1, partial sums of A080872 starting from A080872(1).

			153 is in the sequence because 3*153*154/2+1 = 188^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Sequence A129444 gives n+1.
Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.
a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.

A098297 Member r=12 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 12, 121, 1200, 11881, 117612, 1164241, 11524800, 114083761, 1129312812, 11179044361, 110661130800, 1095432263641, 10843661505612, 107341182792481, 1062568166419200, 10518340481399521, 104120836647576012

Offset: 0

Wolfdieter Lang, Oct 18 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A097784, A098296.

a:=[0,1,12];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+ a[n-3]; od; a; # G. C. Greubel, May 24 2019

I:=[0,1,12]; [n le 3 select I[n] else 11*Self(n-1)-11*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, May 24 2019

LinearRecurrence[{11,-11,1}, {0,1,12}, 30] (* G. C. Greubel, May 24 2019 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+x)/((1-x)*(1-10*x+x^2)))) \\ G. C. Greubel, May 24 2019

(x*(1+x)/((1-x)*(1-10*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019

a(n) = (T(n, 5)-1)/4 with Chebyshev's polynomials of the first kind evaluated at x=5: T(n, 5) = A001079(n) = ((5 + 2*sqrt(6))^n + (5 - 2*sqrt(6))^n)/2.
a(n) = 10*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=12.
G.f.: x*(1+x)/((1-x)*(1-10*x+x^2)) = x*(1+x)/(1-11*x+11*x^2-x^3) (from the Stephan link, see A092184).
a(n) = A132596(n) / 2. - Peter Bala, Dec 31 2012

A175497 Numbers k with the property that k^2 is a product of two distinct triangular numbers.

Original entry on oeis.org

0, 6, 30, 35, 84, 180, 204, 210, 297, 330, 546, 840, 1170, 1189, 1224, 1710, 2310, 2940, 2970, 3036, 3230, 3900, 4914, 6090, 6930, 7134, 7140, 7245, 7440, 8976, 10710, 12654, 14175, 14820, 16296, 16380, 17220, 19866, 22770, 25172, 25944, 29103

Offset: 1

Zak Seidov, May 30 2010

nonn

From Robert G. Wilson v, Jul 24 2010: (Start)
Terms in the i-th row are products contributed with a factor A000217(i):
(1) 0, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, ...
(2) 30, 297, 2940, 29103, 288090, 2851797, 28229880, ...
(3) 84, 1170, 16296, 226974, 3161340, ...
(4) 180, 3230, 57960, 1040050, 18662940, ...
(5) 330, 7245, 159060, 3492075, 76666590, ...
(6) 546, 14175, 368004, 9553929, ...
(7) 840, 25172, 754320, 22604428, ...
(8) 210, 1224, 7134, 41580, 242346, 1412496, 8232630, 47983284, ...
(9) 1710, 64935, 2465820, 93636225, ...
(10) 2310, 96965, 4070220, ...
(11) 3036, 139590, 6418104, ...
(12) 3900, 194922, 9742200, ...
(13) 4914, 265265, 14319396, ...
(14) 6090, 353115, 20474580, ...
(15) 7440, 461160, 28584480, ...
(End)
Numbers m with property that m^2 is a product of two distinct triangular numbers T(i) and T(j) such that i and j are in the same row of the square array A(n, k) defined in A322699. - Onur Ozkan, Mar 17 2023

R. J. Mathar, Table of n, a(n) for n = 1..1000 replacing a b-file of _Robert G. Wilson v_ of 2010.
R. J. Mathar, OEIS A175497, Mar 16 2023
Project Euler, Problem 833. Square Triangle Products.
M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT], 2008. Theorem 5.6.

From Robert G. Wilson v, Jul 24 2010: (Start)
A001109 (with the exception of 1), A011945, A075848 and A055112 are all proper subsets.
Many terms are in common with A147779.
Cf. A001108, A132596, A007654, A001108, A132593. (End)
Cf. A152005 (two distinct tetrahedral numbers).

isA175497 := proc(n)
    local i,Ti,Tj;
    if n = 0 then
        return true;
    end if;
    for i from 1 do
        Ti := i*(i+1)/2 ;
        if Ti > n^2 then
            return false;
        else
            Tj := n^2/Ti ;
            if Tj <> Ti and type(Tj,'integer') then
                if isA000217(Tj) then  # code in A000217
                    return true;
                end if;
            end if;
        end if;
    end do:
end proc:
for n from 0 do
    if isA175497(n) then
        printf("%d,\n",n);
    end if;
end do: # R. J. Mathar, May 26 2016

triangularQ[n_] := IntegerQ[Sqrt[8n + 1]];
okQ[n_] := Module[{i, Ti, Tj}, If[n == 0, Return[True]]; For[i = 1, True, i++, Ti = i(i+1)/2; If[Ti > n^2, Return[False], Tj = n^2/Ti; If[Tj != Ti && IntegerQ[Tj], If[ triangularQ[Tj], Return[True]]]]]];
Reap[For[k = 0, k < 30000, k++, If[okQ[k], Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Jun 13 2023, after R. J. Mathar *)

from itertools import count, islice, takewhile
from sympy import divisors
from sympy.ntheory.primetest import is_square
def A175497_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda k:not k or any(map(lambda d: is_square((d<<3)+1) and is_square((k**2//d<<3)+1), takewhile(lambda d:d**2A175497_list = list(islice(A175497_gen(),20)) # Chai Wah Wu, Mar 13 2023

def A175497_list(n):
    def A322699_A(k, n):
        p, q, r, m = 0, k, 4*k*(k+1), 0
        while m < n:
            p, q, r = q, r, (4*k+3)*(r-q) + p
            m += 1
        return p
    def a(k, n, j):
        if n == 0: return 0
        p = A322699_A(k, n)*(A322699_A(k, n)+1)*(2*k+1) - a(k, n-1, 1)
        q = (4*k+2)*p - A322699_A(k, n)*(A322699_A(k, n)+1)//2
        m = 1
        while m < j: p, q = q, (4*k+2)*q - p; m += 1
        return p
    A = set([a(k, 1, 1) for k in range(n+1)])
    k, l, m = 1, 1, 2
    while True:
        x = a(k, l, m)
        if x < max(A):
            A |= {x}
            A  = set(sorted(A)[:n+1])
            m += 1
        else:
            if m == 1 and l == 1:
                if k > n:
                    return sorted(A)
                k += 1
            elif m > 1:
                l += 1; m = 1
            elif l > 1:
                k += 1; l, m = 1, 1
# Onur Ozkan, Mar 15 2023

a(n)^2 = A169836(n). - R. J. Mathar, Mar 12 2023

A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+3)}/{1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+1)} at m = 3.

Original entry on oeis.org

2, 8, 1, 16, 1, 96, 1, 176, 1, 968, 1, 1760, 1, 9600, 1, 17440, 1, 95048, 1, 172656, 1, 940896, 1, 1709136, 1, 9313928, 1, 16918720, 1, 92198400, 1, 167478080, 1, 912670088, 1, 1657862096, 1

Offset: 0

Peter Bala, Jan 06 2013

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 3. For other cases see A221073 (m = 2), A221075 (m = 4) and A221076 (m = 5).

			Product {n >= 0} {1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+1)} = 2.11180 16361 44098 52896 ...
= 2 + 1/(8 + 1/(1 + 1/(16 + 1/(1 + 1/(96 + 1/(1 + 1/(176 + ...))))))).
Since (sqrt(3) - sqrt(2))^3 = 9*sqrt(3) - 11*sqrt(2) we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+1)} = 1 + 1/(16 + 1/(1 + 1/(968 + 1/(1 + 1/(17440 + 1/(1 + 1/(940896 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,10,0,-10,0,-1,0,1).

Cf. A098297, A105438, A132596, A174500, A221073 (m = 2), A221075 (m = 4), A221076 (m = 5).

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (sqrt(3) + sqrt(2))^(2*n) + (sqrt(3) - sqrt(2))^(2*n) - 2;
a(4*n - 1) = 1/sqrt(3)*{(sqrt(3) + sqrt(2))^(2*n + 1) + (sqrt(3) - sqrt(2))^(2*n + 1)} - 2.
a(4*n - 3) = 8*A098297(n) = 4*A132596(n); a(4*n - 1) = 4*A105038(n).
O.g.f.: 2 + x^2/(1 - x^2) + 8*x*(1 + x^2)^2/(1 - 11*x^4 + 11*x^8 - x^12) = 2 + 8*x + x^2 + 16*x^3 + x^4 + 96*x^5 + ....
If we denote the present sequence by [2; 8, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+3)]/[1 - sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+1)]. An example is given below.
O.g.f.: (x^10-2*x^8-10*x^6+20*x^4-8*x^3+x^2-8*x-2) / ((x-1)*(x+1)*(x^8-10*x^4+1)). - Colin Barker, Jan 10 2014

A169835 Perfect squares that are a product of two triangular numbers.

Original entry on oeis.org

1, 9, 36, 100, 225, 441, 784, 900, 1225, 1296, 2025, 3025, 4356, 6084, 7056, 8281, 11025, 14400, 18496, 23409, 29241, 32400, 36100, 41616, 44100, 53361, 64009, 76176, 88209, 90000, 105625, 108900, 123201, 142884, 164836, 189225, 216225, 246016, 278784, 298116

Offset: 1

R. J. Mathar, May 30 2010

nonn

Includes (except for 0) A000537 and 3/2*x*(x+1) for x in A132596. - Robert Israel, Jan 16 2015

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Erich Friedman, What's Special About This Number? (See entry 7056.)

Superset of A000537. Cf. A000217, A132596, A169836.

Cf. A000290, A010054.

a169835 n = a169835_list !! (n-1)
a169835_list = f [] (tail a000217_list) (tail a000290_list) where
   f ts us'@(u:us) vs'@(v:vs)
     | u <= v = f (u : ts) us vs'
     | any p $ map (divMod v) ts = v : f ts us' vs
     | otherwise = f ts us' vs
     where p (q, r) = r == 0 && a010054 q == 1
-- Reinhard Zumkeller, Mar 03 2015

N:= 10^6: # to get all terms <= N
A:= select(issqr,{seq(seq(a*(a+1)*b*(b+1)/4,
    b = a .. floor(sqrt(1/4+4*N/a/(a+1))-1/2)),a=1..floor(sqrt(4*N)))});
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(A, list)); # Robert Israel, Jan 16 2015

M = 10^6; (* to get all terms <= M *)
A = Union[Select[Flatten[Table[Table[(1/4) a (a+1) b (b+1), {b, a, Floor[ Sqrt[1/4 + 4M/(a (a+1))] - 1/2]}], {a, 1, Floor[Sqrt[4M]]}]], IntegerQ[ Sqrt[#]]&]] (* Jean-François Alcover, Mar 09 2019, after Robert Israel *)

istriangular(n)=issquare(8*n+1) \\ now one can use ispolygonal(n, 3)
isok(n) = {if (issquare(n), fordiv(n, d, if (d > sqrtint(n), break); if (istriangular(d) && istriangular(n/d), return (1)););); return (0);} \\ Michel Marcus, Jul 24 2013

from itertools import count, islice, takewhile
from sympy import divisors
from sympy.ntheory.primetest import is_square
def A169835_gen(): # generator of terms
    return filter(lambda k:any(map(lambda d: is_square((d<<3)+1) and is_square((k//d<<3)+1), takewhile(lambda d:d**2<=k,divisors(k)))),(m**2 for m in count(0)))
A169835_list = list(islice(A169835_gen(),20)) # Chai Wah Wu, Mar 13 2023

Corrected (missing terms inserted) by R. J. Mathar, Jun 04 2010

A171640 a(n) = 10*a(n-1)-a(n-2)-4 with a(1)=1 and a(2)=3.

Original entry on oeis.org

1, 3, 25, 243, 2401, 23763, 235225, 2328483, 23049601, 228167523, 2258625625, 22358088723, 221322261601, 2190864527283, 21687323011225, 214682365584963, 2125136332838401, 21036680962799043, 208241673295152025, 2061380051988721203, 20405558846592060001

Offset: 1

Mark Dols, Dec 13 2009

			a(2+1) = [5-sqrt(24)+5+sqrt(24)]^2/4 = 100/4 = 25.

Colin Barker, Table of n, a(n) for n = 1..1000
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A132596, A087799, A001263.

RecurrenceTable[{a[n] == 10 a[n - 1] - a[n - 2] - 4, a[1] == 1, a[2] == 3}, a, {n, 1, 21}] (* Michael De Vlieger, Oct 02 2015 *)
LinearRecurrence[{11,-11,1},{1,3,25},30] (* Harvey P. Dale, May 05 2018 *)

Vec(-x*(3*x^2-8*x+1)/((x-1)*(x^2-10*x+1)) + O(x^30)) \\ Colin Barker, Oct 02 2015

a(n) = A132596(n-1)+1.
2*a(n) + 2*A132596(n-1) = A087799(n-1).
a(n+1) = [(sqrt(3)-sqrt(2))^n +(sqrt(3)+ sqrt(2))^n]^2 / 4.
From Colin Barker, Oct 02 2015: (Start)
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n>3.
G.f.: -x*(3*x^2-8*x+1) / ((x-1)*(x^2-10*x+1)).
(End)
6*a(n)*(a(n)-1) = [A122653(n-1)]^2. - Jean-Luc Manguin, Jun 02 2020

A174906 a(n) is the index of the first triangular number T_i exceeding T_n such that the product of T_i*T_n is a perfect square.

Original entry on oeis.org

24, 48, 80, 120, 168, 224, 49, 360, 440, 528, 624, 728, 840, 960, 1088, 1224, 1368, 1520, 1680, 1848, 2024, 2208, 242, 2600, 2808, 3024, 3248, 3480, 3720, 3968, 4224, 4488, 4760, 5040, 5328, 5624, 5928, 6240, 6560, 6888, 7224, 7568, 7920, 8280, 8648

Offset: 2

Robert G. Wilson v, Apr 01 2010

nonn

"You can find an infinite number of [different] triangular numbers such that when multipled together form a square number. For example, for every triangular number, T_n, there are an infinite number of other triangular numbers, T_m, such that T_n*T_m is a square. For example, T_2 * T_24 = 30^2."

Clifford A. Pickover, The Loom of God, Tapestries of Mathematics and Mysticism, Sterling, NY, 2009, page 33.

Cf. A132596, A007654, A132584.

tri[n_] := n (n + 1)/2; f[n_] := Block[{k = n + 1, t = tri@n}, While[ !IntegerQ@ Sqrt[ t*tri@k], k++ ]; k]; Table[ f@n, {n, 2, 46}]

cp's OEIS Frontend

A173115 a(n) = -(sin(2*n*arccos(sqrt(3))))^2.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Ruby

Formula

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A098297 Member r=12 of the family of Chebyshev sequences S_r(n) defined in A092184.

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

A175497 Numbers k with the property that k^2 is a product of two distinct triangular numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Python

Python

Formula

A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A169835 Perfect squares that are a product of two triangular numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

A173115 a(n) = -(sin(2narccos(sqrt(3))))^2.

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).

A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+3)}/{1 - sqrt(m)[sqrt(m) - sqrt(m-1)]^(4n+1)} at m = 3.