A182189 -id:A182189 - cp's OEIS frontend

A182190 a(n) = 6*a(n-1) - a(n-2) + 4 with n > 1, a(0)=0, a(1)=4.

0, 4, 28, 168, 984, 5740, 33460, 195024, 1136688, 6625108, 38613964, 225058680, 1311738120, 7645370044, 44560482148, 259717522848, 1513744654944, 8822750406820, 51422757785980, 299713796309064

Offset: 0

Kenneth J Ramsey, Apr 17 2012

Also, nonnegative m such that 2m(m+2)+1 is a square. - Bruno Berselli, Oct 22 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A182189, A182431.

I:=[0,4]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2)+4: n in [1..20]]; // Bruno Berselli, Jun 07 2012

m = 4;n = 0; c = 0;
list3 = Reap[While[c < 22, t = 6 n - m + 4; Sow[t];m = n; n = t;c++]][[2,1]]
Table[Fibonacci[2*n+1, 2] -1, {n,0,40}] (* G. C. Greubel, May 24 2021 *)

[lucas_number1(2*n+1,2,-1) -1 for n in (0..40)] # G. C. Greubel, May 24 2021

G.f.: 4*x/((1-x)*(1-6*x+x^2)). - Bruno Berselli, Jun 07 2012
a(n) = 4*A053142(n). - Bruno Berselli, Jun 07 2012
a(n) = A001653(n+1) - 1. - Kiran S. Kedlaya, Mar 14 2021
a(n) = A000129(2*n+1) - 1. - G. C. Greubel, May 24 2021
a(n) = A143608(n)*A143608(n+1). - R. J. Mathar, Jan 31 2024
a(n)-a(n-1) = A005319(n). - R. J. Mathar, Jan 31 2024

A182431 Table, read by antidiagonals, in which the n-th row comprises A214206(n) 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, 12, 1, 14, 8, 98, 4, 2, 14, 10, 602, 35, 0, 3, 14, 11, 3540, 218, 0, 4, 4, 14, 12, 20664, 1285, 2, 21, 4, 5, 14, 13, 120470, 7504, 14, 122, 14, 8, 6, 14, 14, 702182, 43751, 84, 711, 74, 35, 12, 7, 14, 15

Offset: 0

Kenneth J Ramsey, Apr 28 2012

This is a table 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) beginning at position a(k,1). Each term G(i) is the same as 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 table A001477. If the product is less than (r^2 +3*r -2)/2, then the product less r would be a triangular number, i.e., still lie in the same row assumed to contain all numbers n that equal a triangular number + r. For example, 3 is a triangular number and appears in row 0 of A001477, but if the rows could take negative indices, A001477(2,-1) would be a 3 so 3 can be said to also lie in row 2. See A182102 for a table of the arguments of the triangular numbers G(i)*G(i+1) - r.

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, if a(k+1,0) - a(k,0) = 1 then a(k+1,i+1) - a(k,i+1) equals A182435(i) for all i. Also, for i>0, A143608(i) divides a(k+1,i+1)-a(k,i+1) for all k.

			The Table begins:
0 14  0 12  98  602 3540 ...
4 14  1  4  35  218 1285 ...
7 14  2  0   0    2   14 ...
8 14  3  4  21  122  711 ...
10 14  4  4  14   74  424 ...
11 14  5  8  35  194 1121 ...
12 14  6 12  56  314 1818 ...
13 14  7 16  77  434 2515 ...
14 14  8 20  98  554 3212 ...
15 14  9 24 119  674 3909 ...
16 14 10 28 140  794 4606 ...
17 14 11 32 161  914 5303 ...
17 14 12 40 210 1202 6984 ...
...
Note that 14*0,0*12,12*98, 98*602 etc are each 0 more than a triangular number and are in row 0 of square array A001477; while 14*1, 1*4, 4*35, 35*218 etc are each 4 more than a triangular number and thus can be said to lie in row 4 of square array A001477.

Cf. A182189, A182190, A182435.

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 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[tab[[b,a+1-b]]]; b--]; a++]][[2,1]]; list5

a(k,0) equals the largest m such that m*(m+1)/2 is less than or equal to 14*k.
a(k,1) = 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*a(k,0).

A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

Original entry on oeis.org

3, 143, 4899, 166463, 5654883, 192099599, 6525731523, 221682772223, 7530688524099, 255821727047183, 8690408031080163, 295218051329678399, 10028723337177985443, 340681375412721826703, 11573138040695364122499, 393146012008229658338303, 13355391270239113019379843

Offset: 1

Bruno Berselli, Nov 17 2016

Equivalently, both m+1 and 2*m+3 are squares for nonnegative m.
Corresponding triangular numbers T(m): 6, 10296, 12002550, 13855048416, 15988853699286, 18451128064030200, 21292585958400815526, ...
Square roots of T(m) + 3*T(m+1) are listed by A082405 (after 0).
Negative values of m for which T(m) + 3*T(m+1) is a square: -1, -2, -26, -842, -28562, -970226, -32959082, ...

			3 is in the sequence because T(3) + 3*T(4) = 6 + 3*10 = 6^2.
For n=5 is a(5) = 5654883, therefore floor(sqrt(5654883)) = 2377 = A182189(5) - 2 = 2379 - 2.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Subsequence of A000466.
Cf. A000217, A001109, A029546, A046176, A082405, A156164, A182189.
Cf. A278438: numbers m such that T(m) + 2*T(m+1) is a square.
Cf. A078522: numbers m such that 3*T(m) + T(m+1) is a square.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), A076218 (k=3), this sequence (k=-5).

Iv:=[3,143]; [n le 2 select Iv[n] else 34*Self(n-1)-Self(n-2)+40: n in [1..20]];

P:=proc(q) local n; for n from 3 to q do if type(sqrt(2*n^2+5*n+3),integer) then print(n); fi; od; end: P(10^9); # Paolo P. Lava, Nov 18 2016

Table[((1 + Sqrt[2])^(4 n) + (1 - Sqrt[2])^(4 n))/8 - 5/4, {n, 1, 20}]
RecurrenceTable[{a[1] == 3, a[2] == 143, a[n] == 34 a[n - 1] - a[n - 2] + 40}, a, {n, 1, 20}]
LinearRecurrence[{35, -35, 1}, {3, 143, 4899}, 50] (* G. C. Greubel, Nov 20 2016 *)

Vec(x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)) + O(x^50)) \\ G. C. Greubel, Nov 20 2016

def A278310():
    a, b = 3, 143
    yield a
    while True:
        yield b
        a, b = b, 34*b - a + 40
a = A278310(); print([next(a) for  in range(18)]) # _Peter Luschny, Nov 18 2016

O.g.f.: x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)).
E.g.f.: (exp((1-sqrt(2))^4*x) + exp((1+sqrt(2))^4*x) - 10*exp(x))/8 + 1.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
a(n) = 34*a(n-1) - a(n-2) + 40 for n>2.
a(n) = a(-n) = ((1 + sqrt(2))^(4*n) + (1 - sqrt(2))^(4*n))/8 - 5/4.
a(n) = 4*A001109(n)^2 - 1.
a(n) = -A029546(n) + 38*A029546(n-1) + 3*A029546(n-2) for n>1.
Lim_{n -> infinity} a(n)/a(n-1) = A156164.
Floor(sqrt(a(n))) = A182189(n) - 2.
a(n) - a(n-1) = 4*A046176(n) for n>1.

A182188 A sequence of row differences for table A182119.

Original entry on oeis.org

1, -1, -11, -69, -407, -2377, -13859, -80781, -470831, -2744209, -15994427, -93222357, -543339719, -3166815961, -18457556051, -107578520349, -627013566047, -3654502875937, -21300003689579

Offset: 0

Kenneth J Ramsey, Apr 17 2012

This is a list of row differences corresponding to a difference of 1 in table A182119, column 0. If A181119(k+1,0) - A182119(k,0) = 1, then a(n) = A182119(k+1,n) - A182119(k,n).

If p is a prime of the form 8*n +- 3, then a(p) == 3 (mod p). If p is a prime of the form 8*n +- 1, then a(p) == -1 (mod p).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001109, A001333, A001542, A005409, A182189, A182190, A182119.

m = 13;n = 3; c = 0;
list3 = Reap[While[c < 22, t = 6 n - m - 4; Sow[t];m = n; n = t;c++]][[2,1]]
Table[1 -Fibonacci[2*n, 2], {n,0,40}] (* G. C. Greubel, May 24 2021 *)

[1 - lucas_number1(2*n,2,-1) for n in (0..40)] # G. C. Greubel, May 24 2021

a(n) = 6*a(n-1) - a(n-2) - 4. [corrected by Klaus Purath, Mar 19 2021]
a(n) = -(A182189(n-1) + 2*A182190(n-1)).
a(n) = 2 - A182189(n).
From Klaus Purath, Mar 19 2021: (Start)
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
a(n) = (-1)*Sum_{i=1..2*n-1} A001333(i) for n > 0.
a(n) = 1 - A001542(n) for n > 0.
a(n) = 1 - 2*A001109(n) for n > 0.
a(n) = (-1)*A005409(2*n) for n > 0. (End)
G.f.: (1 - 8*x + 3*x^2)/((1-x)*(1-6*x+x^2)). - Chai Wah Wu, Apr 08 2021
a(n) = 1 - Pell(2*n), where Pell(n) = A000129(n). - G. C. Greubel, May 24 2021

A182102 Table of triangular arguments such that if A002262(14k) = "r" then the product A182431(k,i + 1) A182431(k,i + 2) equals "r" + A000217(a(k,i)).

Original entry on oeis.org

0, -1, 4, 48, -1, 7, 343, 16, 0, 8, 2064, 123, -1, 3, 10, 12095, 748, 0, 12, 5, 11, 70560, 4391, 7, 71, 10, 8, 12, 411319, 25624, 48, 416, 45, 23, 11, 13, 2397408, 149379, 287, 2427, 250, 116, 36, 14, 14, 13973183, 870676

Offset: 0

Kenneth J Ramsey, Apr 11 2012

It is noted that the difference between adjacent rows of the respective elements, depends on the difference between the elements of column 0 in the respective rows. It is apparent that the series of differences corresponding to a difference of d in column 0, i.e. A(k+1,0) - A(k,0) = d, is defined as follows: D(0) = d, D(1) = 4 - d, D(n) = 6*D(n-1) - D(n-2) -8*d + 4. The sequence of differences corresponding to a difference of 1 or 0 in column 0 form related series A182189 and A182190.

The Mathematica program below first calculates an array containing only the first four nonnegative triangular arguments P of each row then changes at most 2 of the arguments to the corresponding negative value, N = -P -1 in order to obtain the relation a(k,i) -7*a(k,i+1) + 7*a(k,i+2) - a(k,i+3) = 0, then chooses the appropriate argument to continue this relationship with the remainder of the row. In this way, the sequence is finally determined. Thus in this table a few 0's have been changed to -1.

			The table begins as follows:
0   -1  48  343 2064 12095  70560 ...
4   -1  16  123  748  4391  25624 ...
7    0  -1    0    7    48    287 ...
8    3  12   71  416  2427  14148 ...
10   5  10   45  250  1445   8410 ...
11   8  23  116  659  3824  22271 ...
12  11  36  187 1068  6203  36132 ...
13  14  49  258 1477  8582  49993 ...
14  17  62  329 1886 10961  63854 ...
15  20  75  400 2295 13340  77715 ...
16  23  88  471 2704 15719  91576 ...
17  26 101  542 3113 18098 105437 ...
17  30 129  710 4097 23838 138897 ...
...
For n > 1, a(k,n) = 6*a(k,n-1) - a(k,n-2) + G_k where G_k is dependent on k.

Cf. A182189, A182190, A182431.

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)]];
tt = SparseArray[{{12,1} -> 1,{1,12} -> 1}];
K1 = 0;
m = 14;While[K1<12,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;
o = overTri[m*K1]; tt[[1,K1+1]] =highTri[m*K1];
tt[[2,K1+1]] = highTri[K1*K2-o];tt[[3,K1+1]] = highTri[K2*K3-o];tt[[4,K1+1]] = highTri[K3*K4-o];
K1++];k = 1;
While[k<13,z = 1; xx = 99; While[z<5 && xx == 99,
If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,Break[]];
If[z == 1,t = -tt[[z,k]]-1;tt[[z,k]] = t,s = -tt[[z-1,k]]-1;tt[[z-1,k]]=s;t =-tt[[z,k]]-1];tt[[z,k]] = t;
w = 1;While[w<5 && xx == 99,If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,xx =0;Break[]];If[w==z,w++];
t=-tt[[w,k]] - 1;tt[[w,k]]=t;If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,xx =0;Break[],
t = -tt[[w,k]] - 1];tt[[w,k]] = t;w++];z++];cc = tt[[1,k]] -6 tt[[2,k]] + tt[[3,k]];p = 5;While[p < 14-k,
tt[[p,k]] = 6 tt[[p-1,k]] - tt[[p-2,k]] + cc;p++]; k++];
a=1;list2 = Reap[While[a<11, b=a; While[b>0, Sow[tt[[b, a+1-b]]]; b--]; a++]][[2, 1]];list2

a(k,0) = Floor[(Sqrt[1 + 112*k] - 1)/2]
a(k,i) = A003056(A182431(k,i+1)*A182431(k,i+2) - A002262(14*k)) or
          -1 -   A003056(A182431(k,i+1)*A182431(k,i+2) - A002262(14*k)) .
For i>2, a(k,i) = 7*a(k,i-1)-7*a(k,i-2)+a(k,i-3).

A182193 Sequence of row differences related to table A182355.

Original entry on oeis.org

-1, 1, 19, 125, 743, 4345, 25339, 147701, 860879, 5017585, 29244643, 170450285, 993457079, 5790292201, 33748296139, 196699484645, 1146448611743, 6681992185825, 38945504503219, 226991034833501, 1323000704497799, 7711013192153305, 44943078448422043

Offset: 0

Kenneth J Ramsey, Apr 17 2012

Sequence of row differences in table A182355. If A182355(k + 1, 0) - A182355(k, 0) = -1, a(n) = A182355(k + 1, n) - A182355(k, n).

If p is a prime of the form 8r = +/- 3, a(p) = 5 mod p; if p is a prime of the form 8r = +/- 1, a(p) = 1 mod p.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A002203, A182188, A182189, A182191, A182355.

I:=[-1,1]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2)+12: n in [1..30]]; // Vincenzo Librandi, Feb 10 2014

Pell:= proc(n) option remember;
    if n<2 then n
  else 2*Pell(n-1) + Pell(n-2)
    fi; end:
seq(Pell(2*n) + 2*Pell(2*n-1) - 3, n=0..40); # G. C. Greubel, May 24 2021

LinearRecurrence[{7,-7,1},{-1,1,19},30] (* Harvey P. Dale, Feb 09 2014 *)

Vec(-(1-8*x-5*x^2)/((1-x)*(1-6*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 05 2016

[lucas_number2(2*n,2,-1) - lucas_number1(2*n,2,-1) - 3 for n in (0..40)] # G. C. Greubel, May 24 2021

a(n) = 6*a(n-1) - a(n-2) + 12.
a(0)=-1, a(1)=1, a(2)=19, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Harvey P. Dale, Feb 09 2014
From Colin Barker, Mar 05 2016: (Start)
a(n) = -3 + (1/4)*( (4-sqrt(2))*(3+2*sqrt(2))^n + (4+sqrt(2))*(3-2*sqrt(2))^n ).
G.f.: -(1-8*x-5*x^2) / ((1-x)*(1-6*x+x^2)).
(End)
a(n) = A002203(2*n) - A000129(2*n) - 3. - G. C. Greubel, May 24 2021

More terms from Harvey P. Dale, Feb 09 2014

A182355 Table of triangular arguments such that if A002262(14k) = "r" then the product A182441(k,i + 1) A182441(k,i + 2) equals "r" + a(k,i)*(a(k,i) + 1)/2 for i<4, while a(k,i) = 0 for i>3.

Original entry on oeis.org

-1, 56, -5, 399, 60, -8, 2400, 463, 63, -9, 0, 2816, 512, 64, -11, 0, 0, 3135, 531, 66, -12, 0, 0, 0, 3260, 565, 67, -13, 0, 0, 0, 0, 3482, 584, 68, -14, 0, 0, 0, 0, 0, 3607, 603, 69, -15, 0, 0, 0, 0, 0, 0, 3732, 622

Offset: 0

Kenneth J Ramsey, Apr 25 2012

The triangular product a(k,i)*(a(k,i)+1)/2 + A002262(14*k) for i<4 = the product of adjacent terms G(k,i+1)*G(k,i+2) where G is table A182441. The remainder of each row is padded with zeros. However, if for i > 3, a(k,i) were set to equal 7*a(k,i-1) - 7*a(k,i-2) + a(k,i-3) then the relation above would not be limited to i < 4.
Also, it is noted that the difference between adjacent rows of the respective elements, depends on the difference between the elements of column 0 in the respective rows. In the Mathematica program below, m is set to 14; however, regardless of it value of m, it is apparent that the series of differences corresponding to a difference of d in column 0, i.e. A(k+1,0) - A(k,0) = d, is defined as follows: D(0) = d, D(1) = - d, D(n) = 6*D(n-1) - D(n-2) -8*d + 4. The sequence of differences corresponding to a difference d of -1 is series A182193.
The Mathematica program below basically first computes only the nonnegative triangular arguments P. Then it changes at most two of the arguments P in each row k to the corresponding negative value, N = -P -1, in order to obtain the relation a(k,3) = a(k,0) - 7*a(k,1) + 7*a(k,2).

Cf. A182102, A182118, A182119, A182190, A182193, A182188, A182189.

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)]];
tt = SparseArray[{{12,1} -> 0,{1,12} -> 0}];
K1 = 0;
m = 14;While[K1<12,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;
o = overTri[m*K1]; tt[[1,K1+1]] =highTri[m*K1];
tt[[2,K1+1]] = highTri[m*K2-o];tt[[3,K1+1]] = highTri[K2*K3-o];tt[[4,K1+1]] = highTri[K3*K4-o];
K1++];k = 1;
While[k<13,z = 1; xx = 99; While[z<5 && xx == 99,
If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,Break[]];
If[z == 1,t = -tt[[z,k]]-1;tt[[z,k]] = t,s = -tt[[z-1,k]]-1;tt[[z-1,k]]=s;t =-tt[[z,k]]-1];tt[[z,k]] = t;
w = 1;While[w<5 && xx == 99,If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,xx =0;Break[]];If[w==z,w++];
t=-tt[[w,k]] - 1;tt[[w,k]]=t;If[tt[[1,k]]+ 7 tt[[3,k]] - 7 tt[[2,k]] - tt[[4,k]] == 0,xx =0;Break[],
t = -tt[[w,k]] - 1];tt[[w,k]] = t;w++];z++];cc = tt[[1,k]] -6 tt[[2,k]] + tt[[3,k]];p = 5;While[p < 14-k,
tt[[p,k]] = 6 tt[[p-1,k]] - tt[[p-2,k]] + cc;p++]; k++];
a=1;list2 = Reap[While[a<12, b=a; While[b>4,Sow[0];b--];While[b>0, Sow[tt[[b, a+1-b]]]; b--]; a++]][[2, 1]];list2

cp's OEIS Frontend

A182190 a(n) = 6*a(n-1) - a(n-2) + 4 with n > 1, a(0)=0, a(1)=4.

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A182431 Table, read by antidiagonals, in which the n-th row comprises A214206(n) 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).

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A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

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A182188 A sequence of row differences for table A182119.

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A182102 Table of triangular arguments such that if A002262(14*k) = "r" then the product A182431(k,i + 1) * A182431(k,i + 2) equals "r" + A000217(a(k,i)).

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A182193 Sequence of row differences related to table A182355.

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A182355 Table of triangular arguments such that if A002262(14*k) = "r" then the product A182441(k,i + 1) *A182441(k,i + 2) equals "r" + a(k,i)*(a(k,i) + 1)/2 for i<4, while a(k,i) = 0 for i>3.

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A182102 Table of triangular arguments such that if A002262(14k) = "r" then the product A182431(k,i + 1) A182431(k,i + 2) equals "r" + A000217(a(k,i)).

A182355 Table of triangular arguments such that if A002262(14k) = "r" then the product A182441(k,i + 1) A182441(k,i + 2) equals "r" + a(k,i)*(a(k,i) + 1)/2 for i<4, while a(k,i) = 0 for i>3.