A001108 -id:A001108 - cp's OEIS frontend

A076114 a(n) = start of the smallest string of n consecutive positive integers with a square sum, or 0 if no such number exists.

1, 4, 2, 0, 3, 11, 4, 1, 5, 18, 6, 0, 7, 25, 8, 0, 9, 4, 10, 0, 11, 39, 12, 2, 4, 46, 14, 0, 15, 53, 16, 9, 17, 60, 18, 0, 19, 67, 20, 3, 21, 74, 22, 0, 23, 81, 24, 0, 1, 16, 26, 0, 27, 11, 28, 4, 29, 102, 30, 0, 31, 109, 32, 0, 33, 116, 34, 0, 35, 123, 36, 5, 37, 130, 11, 0, 39

Offset: 1

Amarnath Murthy, Oct 09 2002

nonn

Equivalently, smallest k such that n(n+2k-1)/2 is a square, 0 if there is no such square. - Ralf Stephan, Mar 23 2003
a(n)=0 if and only if n has the form 4^e*m with e > 0 and m odd. - Dean Hickerson, Mar 25 2003
a(k) = 1 if k is the index of a square triangular number, i.e., k(k+1)/2 is a square, or if k belongs to A001108. If n is odd then a(n) <= (n+1)/2.

			a(2) = 4, 4+5 = 9 = 3^2. a(8) = 1, 1+2+3+4+5+6+7+8 = 36 = 6^2.

Clement Martin, Table of n, a(n) for n = 1..10000

Cf. A001108, A001109, A076115.

a(n,limit=oo) = { my(o=n*(n+1)/2, k=0); while(k

More terms from Ralf Stephan, Mar 23 2003

A076708 Numbers n such that triangular numbers T(n) and T(n+1) sum to another triangular number (that is also a perfect square).

Original entry on oeis.org

0, 5, 34, 203, 1188, 6929, 40390, 235415, 1372104, 7997213, 46611178, 271669859, 1583407980, 9228778025, 53789260174, 313506783023, 1827251437968, 10650001844789, 62072759630770, 361786555939835, 2108646576008244, 12290092900109633, 71631910824649558

Offset: 1

Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002

From T(k)+T(k+1) = (k*(k+1)+(k+1)*(k+2))/2 = (k+1)^2 any two consecutive triangular numbers sum to a square, the above sequence gives the sums that are also triangular. The units digit cycles through 0, 5, 4, 3, 8, 9, 0, 5, ...

Let P(b,e) be the polynomial 1+4*b+4*b^2+4*e+4*e^2. It appears that sequences A076708 and A076049 are special cases of the sequence of integers b such that P(b,b+n) is a perfect square. A076708 and A076049 for example are respectively the sequences of b's such that P(b,b+1) and P(b,b+2) are perfect squares. In fact it appears to be true that the sequence of integers b such that P(b,b+n) is a perfect square has the property that t(b)+t(b+n) is a triangular number. I have not had time to prove this but I do have evidence produced by Mathematica to support the assertion. - Robert Phillips (bobanne(AT)bellsouth.net), Sep 04 2009; corrected Sep 08 2009

			a(1) = (sqrt(2)*((3+2*sqrt(2))^2-(3-2*sqrt(2))^2)-8)/8 = (sqrt(2)*(9+12*sqrt(2)+8-9+12*sqrt(2)-8)-8)/8 = (sqrt(2)*24*sqrt(2)-8)/8 = (48-8)/8 = 40/8 = 5.
T(5) + T(6) = 15 + 21 = 36 = T(8).

Colin Barker, Table of n, a(n) for n = 1..1000
B. Polster, M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658, 2015
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001108, A001109, A001110.

Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/(4 Sqrt[2]) - 1, {n, 1, 20}] (* Zerinvary Lajos, Jul 14 2009 *)

concat(0, Vec(x^2*(x-5)/((x-1)*(x^2-6*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Recursion: a(n+2) = 6*a(n+1)-a(n)+4, with a(0)=0 and a(1)=5.
G.f.: (5*x^2-x^3)/((1-x)*(1-6*x+x^2)).
Closed form: a(n)= ( sqrt(2)*( (3+2*sqrt(2))^(n+1) - (3-2*sqrt(2))^(n+1) )-8 )/8.
Also, if the entries in A001109 are denoted by b(n) then a(n) = b(n+1)-1.
a(n) = sqrt(A001110(n)) - 1. - Ivan N. Ianakiev, May 03 2014

A108051 a(n+1) = 4*(a(n)+a(n-1)) for n>1, a(1)=1, a(2)=6.

Original entry on oeis.org

0, 1, 6, 28, 136, 656, 3168, 15296, 73856, 356608, 1721856, 8313856, 40142848, 193826816, 935878656, 4518821888, 21818802176, 105350496256, 508677193728, 2456110759936, 11859151814656, 57261050298368, 276480808452096

Offset: 0

Creighton Dement, Jun 01 2005

Let (a_n) be the sequence and (a_(n+1)) the sequence beginning at 1. Let B and iB be the binomial and inverse binomial transforms, respectively. Then B((a_n)) = A001108(n) (a(n)-th triangular number is a square); B((a_(n+1))) = A002315(n) (NSW Numbers); iB((a_(n+1))) = A096980(n). Note: a 2nd sequence generated by the same floretion is A057087 (Scaled Chebyshev U-polynomials evaluated at i. Generalized Fibonacci sequence.). As is often the case with two sequences corresponding to a single floretion, both satisfy the same recurrence relation.

Floretion Algebra Multiplication Program, FAMP Code: (a_n) = 2ibasekseq[A*B] (with initial term zero), (a_(n+1)) = 1tesseq[A*B], A = + .5'i - .5'j + .5'k + .5i' - .5j' + .5k' - .5'ij' - .5'ik' - .5'ji' - .5'ki'; B = - .5'i + .5'j + .5'k - .5i' + .5j' + .5k' - .5'ik' - .5'jk' - .5'ki' - .5'kj'

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (4,4).

Cf. A057087, A001108, A002315, A096980.

I:=[0, 1, 6]; [n le 3 select I[n] else 4*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

CoefficientList[Series[x*(1+2*x)/(1-4*x-4*x^2),{x,0,40}],x] (* Vincenzo Librandi, Jun 26 2012 *)

a(n+1) = -(1/2)*(2-2*2^(1/2))^n*(-1+2^(1/2))-(1/2)*(2+2*2^(1/2))^n(-1-2^(1/2)); G.f.: x*(1+2*x)/(1-4*x-4*x^2).
a(n) = sum{k=0..n, (-1)^k*C(n-1, k)*(Pell(2n-2k)-Pell(2n-2k-1))}, n>0, where Pell(n) = A000129(n). - Paul Barry, Jun 07 2005
a(n+1) = ((3+sqrt18)(2+sqrt8)^n+(3-sqrt18)(2-sqrt8)^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009, index corrected Jul 11 2012
a(n) = 2^(n-1) * A001333(n), n>0. - Ralf Stephan, Dec 02 2010
a(n) = A057087(n-1) + 2*A057087(n-2). - R. J. Mathar, Jul 11 2012

A125651 Numbers k such that A125650(k) is a perfect square.

Original entry on oeis.org

1, 3, 24, 147, 864, 5043, 29400, 171363, 998784, 5821347, 33929304, 197754483, 1152597600, 6717831123, 39154389144, 228208503747, 1330096633344, 7752371296323, 45184131144600, 263352415571283, 1534930362283104, 8946229758127347, 52142448186480984

Offset: 1

Alexander Adamchuk, Nov 29 2006

Corresponding numbers m such that m^2 = A125650(a(n)) are listed in A125652.
3 divides a(n) for n>1. For n>1 a(n) = 3*A001108(n-1), where A001108(k) = {0, 1, 8, 49, 288, 1681, ...}, A001108(k)-th triangular number is a square. - Alexander Adamchuk, Jan 19 2007
Disregarding the term 1, numbers k such that A071910(k) is a nonzero square; i.e., numbers k such that A000096(k) = k*(k+3)/2 is a nonzero square. - Rick L. Shepherd, Jul 13 2012

			a(2)=3 because A125650(3)=9=3^2; a(3)=24 because A125650(24)=81=9^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A125650, A125652.

Cf. A001108, A071910, A000096, A000217.

I:=[1, 3, 24, 147]; [n le 4 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, May 21 2012

Join[{1},LinearRecurrence[{7,-7,1},{3,24,147},35]] (* or *) CoefficientList[Series[(-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x) x)),{x,0,35}],x] (* Harvey P. Dale, May 15 2011 *)

For n>1, a(n+2) = 6*a(n+1) - a(n) + 6.
For n>1, a(n) = ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))*3/4 - 3/2.
For n>0, a(2*n) = 3*A002315(n-1)^2; a(2*n+1) = 6*A001542(n)^2.
a(n) = 3*A001108(n-1) for n>1. - Alexander Adamchuk, Jan 19 2007
From Harvey P. Dale, May 15 2011: (Start)
For n>1, a(2)=3, a(3)=24, a(4)=147, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).
G.f.: x*(-1+x*(4+(-10+x)*x))/((-1+x)*(1+(-6+x)*x)). (End)

Edited by Max Alekseyev, Jan 11 2007

A182439 Table a(k,i), read by antidiagonals, in which the n-th row comprises A214206(n) in position 0 followed by a second order recursive series G in which each product G(i)*G(i+1) lies in the same row of A001477 (interpreted as a square array - see below).

Original entry on oeis.org

0, 0, 4, 14, 1, 7, 110, 14, 2, 8, 672, 95, 14, 3, 10, 3948, 568, 84, 14, 4, 11, 23042, 3325, 492, 81, 14, 5, 12, 134330, 19394, 2870, 472, 74, 14, 6, 13, 782964, 113051, 16730, 2751, 424, 71, 14, 7, 14, 4563480, 658924, 97512, 16034, 2464, 404, 68, 14, 8, 15

Offset: 0

Kenneth J Ramsey, Apr 28 2012

This is a square array related to the square array of nonnegative integers, A001477. Each row k contains the positive argument of the largest triangular number equal to or less than 14*k in column 0 and a corresponding 2nd-order recursive sequence G(k) in the rest of the row. Each second-order recursive series term G(i) corresponds to a(k,i+1). If the product 14*k appears in row "r" of the square array A001477, then the product of adjacent terms G(i)*G(i+1), if greater than (r^2 + 3*r - 2)/2, is always in row "r" of square array A001477. If the product is less than (r^2 + 3*r -2)/2 then assuming the row can take negative indices, the product can still be said to lie in the same row r. For instance, 0, 1, 3, and 6 are each a triangular number and appear as the first 4 terms of row 0 of square array A001477. Note that in the next row and to the left of the 1, 3, and 6 are 2, 4 and 7 so going down a row and to the left in the square array increases the value by 1. Going down to the next row and to the left again would be 3, 5, and 8 so 3 which is 2 more than 1 would be in row 2 if that row were made to take the indices (2,-1).
A property of this table is that a(k+1,i)-a(k,i) directly depends on the value of a(k+1,0)-a(k,0) in the same manner regardless of the value of k. For example, a(k,2+n) - a(k,2+n) = A001652(n) for n=0,1,2,3,... whereever a(k+1,0) - a(k,0) = 1.
Also, a(k+1,2+n) - a(k,2+n) is divisible by A143608(n) for n>0 for all k.

			     0,     0,    14,   110,   672,  3948, 23042,134330,782964,
     4,     1,    14,    95,   568,  3325, 19394,113051,658924,
     7,     2,    14,    84,   492,  2870, 16730, 97512,568344,
     8,     3,    14,    81,   472,  2751, 16034, 93453,544684,
    10,     4,    14,    74,   424,  2464, 14354, 83654,487564,
    11,     5,    14,    71,   404,  2345, 13658, 79595,463904,
    12,     6,    14,    68,   384,  2226, 12962, 75536,440244.
Note that 0*14, 14*110, 110*672, etc. are all triangular numbers and thus appear in row 0 of square array A001477; while, 1*14, 14*95, 95*568, 568*3325, etc. are all 4 more than a triangular number and appear in row 4 of square array A001477.

Cf. A001108, A001652, A182431, A182440, A182441, A143608.

A182439 := proc(n,k)
        if k = 0 then
                A003056(14*n) ;
        elif k = 1 then
                n;
        elif k = 2 then
                14;
        else
                6*procname(n,k-1)-procname(n,k-2)+ 28+2*n-2-4*procname(n,0) ;
        end if;
end proc: # R. J. Mathar, Jul 09 2012

highTri = Compile[{{S1,_Integer}}, Module[{xS0=0, xS1=S1}, While[xS1-xS0*(xS0+1)/2 > xS0, xS0++]; xS0]];
overTri = Compile[{{S2,_Integer}}, Module[{xS0=0, xS2=S2}, While[xS2-xS0*(xS0+1)/2 > xS0, xS0++]; xS2 - (xS0*(1+xS0)/2)]];
K1 = 0; m = 14; tab=Reap[While[K1<16,J1=highTri[m*K1]; X = 2*(m+K1-(J1*2+1)); K2 = (6 m - K1 + X); K3 = 6 K2 - m + X;
K4 = 6 K3 - K2 + X; K5 = 6 K4 -K3 + X; K6 = 6*K5 - K4 + X; K7 = 6*K6-K5+X; K8 = 6*K7-K6+X; Sow[J1,c]; Sow[K1,d]; Sow[m,e];
Sow[K2,f]; Sow[K3,g]; Sow[K4,h];
  Sow[K5,i]; Sow[K6,j]; Sow[K7,k]; Sow[K8,l]; K1++]][[2]]; a=1; list5 = Reap[While[a<11, b=a; While[b>0,
Sow[tab[[b,a+1-b]]]; b--]; a++]][[2,1]]; list5
(* Second program: *)
A003056[n_] := Floor[(Sqrt[1 + 8n] - 1)/2];
T[n_, k_] := Switch[k, 0, A003056[14n], 1, n, 2, 14, _, 6T[n, k-1] - T[n, k-2] + 28 + 2n - 2 - 4T[n, 0]];
Table[T[n-k, k], {n, 0, 9}, {k, n, 0, -1}] (* Jean-François Alcover, May 09 2023, after R. J. Mathar *)

a(k,0) equals the largest m such that m*(m+1)/2 is equal to or less than 14*k, A003056(14*k).
a(k,1) = k; a(k,2) = 14.
For i > 2, a(k,i) = 6*a(k,i-1) -a (k,i-2) + G_k where G_k = 28 + 2*k - 2 - 4*a(k,0).
a(k,i) = 7*a(k,i-1)-7*a(k,i-2)+a(k,i-3). - R. J. Mathar, Jul 09 2012

A115598 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.

Original entry on oeis.org

1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521

Offset: 1

N. J. A. Sloane, Mar 14 2006

Old name was A001653(n) - A046090(n).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Essentially a duplicate of A001108.

LinearRecurrence[{7,-7,1},{1,8,49},30] (* Harvey P. Dale, Oct 27 2013 *)
CoefficientList[Series[-(x + 1)/((x - 1) (x^2 - 6 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 28 2014 *)

a(n) = (-2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4.
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).
G.f.: -x*(x+1) / ((x-1)*(x^2-6*x+1)).
a(n)^2 + (a(n)+1)^2 = A001542(n)^2 + 1^2. - Hermann Stamm-Wilbrandt, Jul 27 2014

Corrected and edited by Colin Barker, Jul 31 2013

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

A182435 a(n) = 6*a(n-1) - a(n-2) - 2 with n>1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597

Offset: 0

Kenneth J Ramsey, Apr 28 2012

It appears that for n>0, A143608(n) divides a(n).
The sequence a(n)/A143608(n) appears to generate A001541 interleaved with A001653. - R. J. Mathar, Jul 04 2012
It also appears that if p equals a prime of the form 8*r +/- 3 then a(p + 1) == 0 (mod p); and that if p is a prime in the form of 8*r +/- 1 then a(p - 1) == 0 (mod p), inherited from A143608.

Bruno Berselli, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001108, A143608, A001541 (first differences).

Essentially a duplicate of A046090.

[n le 2 select n-1 else 6*Self(n-1)-Self(n-2)-2: n in [1..24]]; // Bruno Berselli, May 15 2012

m = -20;
n = -3;
c = 0;
list3 = Reap[While[c < 20,t = 6 n - m - 2;Sow[t];m = n;n = t; c++]][[2,1]]
LinearRecurrence[{7,-7,1},{0,1,4},30] (* Harvey P. Dale, May 11 2018 *)

concat(0,Vec((1-3*x)/(1-x)/(1-6*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Jun 11 2013

a(n) = A046090(n-1), for n>=1.
G.f.: x*(1-3*x)/((1-x)*(1-6*x+x^2)). - Bruno Berselli, May 15 2012
a(n) = A001652(n-1)+1 with A001652(-1)=-1. - Bruno Berselli, May 16 2012
2*a(n)*(a(n)-1)+1 = A001653(n)^2 for n>0. - Bruno Berselli, Oct 23 2012

A182440 Table, read by antidiagonals, in which the n-th row comprises A214206(n) in position 0 followed by a second order recursive series G in which each product G(i)*G(i+1) lies in the same row of A001477 (interpreted as a square array).

Original entry on oeis.org

0, 14, 4, 0, 14, 7, 16, 1, 14, 8, 126, 40, 2, 14, 10, 770, 287, 60, 3, 14, 11, 4524, 1730, 420, 72, 4, 14, 12, 26404, 10141, 2522, 497, 88, 5, 14, 13, 153930, 59164, 14774, 2978, 602, 100, 6, 14, 14, 897206

Offset: 0

Kenneth J Ramsey, Apr 28 2012

This is a table related to A001477 interpreted as a square array of the onnegative integers (A001477). Each row k contains A003056(14*k) in column 0 and a corresponding 2nd order recursive sequence G(k) beginning at position a(k,1) such that G(i) = a(k,i+1). If the product 14*k appears in row "r" of the square array A001477, then the product of adjacent terms G(i)*G(i+1) if greater than (r^2 + 3*r - 2)/2, is always in row "r" of square array A001477.
A property of this table is that a(k+1,i)-a(k,i) directly depends on the value of a(k+1,0)-a(k,0) in the same manner regardless of the value of k. For instance, a(k+1,i+1)-a(k,i+1 = A210695(i) if a(k + 1,0) - a(k,0) = 1; while a(k+1,i+1)-a(k,i+1 = A001108(i) if a(k+1,0) - a(k,0) = 0.
A related property is that a(k+1,1+n) - a(k,1+n) is divisible by A143608(n) for all k.

			For i = 1,2,3,4 ..., a(1,i)*a(1,i+1) = 14*1,1*40,40*287,287*1730, ...; and, each product is 4 more than a triangular number and thus lies in row 4 of square array A001477.

Cf. A001108, A210695, A143608, A182431, A182439, A182441

highTri = Compile[{{S1,_Integer}},Module[{xS0=0,xS1=S1},
While[xS1-xS0*(xS0+1)/2>xS0,xS0++];
xS0]];
overTri = Compile[{{S2,_Integer}},Module[{xS0=0,xS2=S2},
While[xS2-xS0*(xS0+1)/2>xS0,xS0++];
xS2 - (xS0*(1+xS0)/2)]];
K1 = 0;
m = 14;table=Reap[While[K1<16,J1=highTri[m*K1];X = 2*(m+K1+(J1*2+1));K2 = (6 K1 - m + X);K3 = 6 K2 - K1 + X;
K4 =  6 K3 - K2 + X; K5 = 6 K4 -K3 + X; K6 = 6*K5 - K4 + X;K7 = 6*K6-K5+X; K8 = 6*K7-K6+X; Sow[J1,c];Sow[m,d];
Sow[K1,e];Sow[K2,f];Sow[K3,g];Sow[K4,h];
  Sow[K5,i]; Sow[K6,j];Sow[K7,k];Sow[K8,l];
K1++]][[2]];
a=1;
list5 = Reap[While[a<11,b=a;
While[b>0,Sow[table[[b,a+1-b]]];b--];a++]][[2,1]];
list5

a(k,0) equals the positive argument of the largest triangular number equal to or less than 14*k (= A214206(k) which = A003056(14*k)).
a(k,1) equals 14; a(k,2) = k.
For i > 2, a(k,i) = 6*a(k,i-1) -a (k,i-2) + G_k where G_k is a constant equal to 28 + 2*k + 2 + 4*A214206(k).

A221073 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

2, 4, 1, 8, 1, 32, 1, 56, 1, 196, 1, 336, 1, 1152, 1, 1968, 1, 6724, 1, 11480, 1, 39200, 1, 66920, 1, 228484, 1, 390048, 1, 1331712, 1, 2273376, 1, 7761796, 1, 13250216, 1, 45239072, 1, 77227928, 1, 263672644, 1, 450117360, 1, 1536796800, 1, 2623476240, 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 = 2. For other cases see A221074 (m = 3), A221075 (m = 4) and A221076 (m = 5).

If we denote the present sequence by [2; 4, 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(2)*{(sqrt(2)-1)^(2*k+1)}^(4*n+3)]/[1 - sqrt(2)*{(sqrt(2)-1)^(2*k+1)}^(4*n+1)]. An example is given below.

			Product {n >= 0} {1 - sqrt(2)*(sqrt(2) - 1)^(4*n+3)}/{1 - sqrt(2)*(sqrt(2) - 1)^(4*n+1)} = 2.20409 39255 78752 05766 ...
= 2 + 1/(4 + 1/(1 + 1/(8 + 1/(1 + 1/(32 + 1/(1 + 1/(56 + ...))))))).
We have (sqrt(2) - 1)^3 = 5*sqrt(2) - 7 so product {n >= 0} {1 - sqrt(2)*(5*sqrt(2) - 7)^(4*n+3)}/{1 - sqrt(2)*(5*sqrt(2) - 7)^(4*n+1)} = 1.11117 34981 94843 98511 ... = 1 + 1/(8 + 1/(1 + 1/(196 + 1/(1 + 1/(1968 + 1/(1 + 1/(39200 + ...))))))).

G. C. Greubel, Table of n, a(n) for n = 0..1000
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,6,0,-6,0,-1,0,1).

Cf. A001108, A053141, A174500, A221074 (m = 3), A221075 (m = 4), A221076 (m = 5).

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2)/((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1)))); // G. C. Greubel, Jul 15 2018

NProduct[( Sqrt[2]*(Sqrt[2] - 1)^(4*n + 3) - 1)/( Sqrt[2]*(Sqrt[2] - 1)^(4*n + 1) - 1), {n, 0, Infinity}, WorkingPrecision -> 200] // ContinuedFraction[#, 37] & (* Jean-François Alcover, Mar 06 2013 *)
Join[{2},LinearRecurrence[{0,1,0,6,0,-6,0,-1,0,1},{4,1,8,1,32,1,56,1,196,1},60]] (* Harvey P. Dale, Feb 16 2014 *)

x='x+O('x^30); Vec((x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2)/((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1))) \\ G. C. Greubel, Jul 15 2018

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (sqrt(2) + 1)^(2*n) + (sqrt(2) - 1)^(2*n) - 2;
a(4*n - 1) = 1/sqrt(2)*{(sqrt(2) + 1)^(2*n + 1) + (sqrt(2) - 1)^(2*n + 1)} - 2.
a(4*n - 3) = 4*A001108(n); a(4*n - 1) = 4*A053141(n).
O.g.f.: 2 + x^2/(1 - x^2) + 4*x*(1 + x^2)^2/(1 - 7*x^4 + 7*x^8 - x^12) = 2 + 4*x + x^2 + 8*x^3 + x^4 + 32*x^5 + ....
O.g.f.: (x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2) / ((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1)). - Colin Barker, Jan 10 2014

More terms from Harvey P. Dale, Feb 16 2014

cp's OEIS Frontend

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A182439 Table a(k,i), read by antidiagonals, in which the n-th row comprises A214206(n) in position 0 followed by a second order recursive series G in which each product G(i)*G(i+1) lies in the same row of A001477 (interpreted as a square array - see below).

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