cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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)).

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

Views

Author

Kenneth J Ramsey, Apr 11 2012

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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

Formula

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).

A143608 A005319 and A002315 interleaved.

Original entry on oeis.org

0, 1, 4, 7, 24, 41, 140, 239, 816, 1393, 4756, 8119, 27720, 47321, 161564, 275807, 941664, 1607521, 5488420, 9369319, 31988856, 54608393, 186444716, 318281039, 1086679440, 1855077841, 6333631924, 10812186007, 36915112104, 63018038201, 215157040700
Offset: 0

Views

Author

Originally submitted by Clark Kimberling, Aug 27 2008. Merged with an essentially identical sequence submitted by Kenneth J Ramsey, Jun 01 2012, by N. J. A. Sloane, Aug 02 2012

Keywords

Comments

Also, numerators of the lower principal and intermediate convergents to 2^(1/2). The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496.
Sequence a(n) such that a(2*n) = sqrt(2*A001108(2*n)) and a(2*n+1) = sqrt(A001108(2*n+1)).
For n > 0, a(n) divides A******(k+1,n+1)-A******(k,n+1) where A****** is any one of A182431, A182439, A182440, A182441 and k is any nonnegative integer.
If p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p); if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p).
Numbers n such that sqrt(floor(n^2/2 + 1)) is an integer. The integer square roots are given by A079496. - Richard R. Forberg, Aug 01 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
a(2*n) = sqrt(2)*(1 o 1 o ... o 1) (2*n terms). Cf. A084068.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = sqrt(2)*U(2*n) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ). (End)

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Crossrefs

Programs

  • Magma
    I:=[0,1,4,7]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 27 2018
  • Maple
    A143608 := proc(n)
        option remember;
        if n <= 3 then
            op(n+1,[0,1,4,7]) ;
        else
            6*procname(n-2)-procname(n-4) ;
        end if;
    end proc: # R. J. Mathar, Jul 22 2012
  • Mathematica
    a = -4; b = -1; Reap[While[b<2000000000, t = 4*b-a; Sow[t]; a=b; b=t; t = 2*b-a; Sow[t]; a=b; b=t]][[2,1]]
    CoefficientList[Series[x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 24 2014 *)
    LinearRecurrence[{0, 6, 0, -1}, {0, 1, 4, 7}, 31] (* Jean-François Alcover, Sep 21 2017 *)
  • PARI
    a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,6,0]^n*[0;1;4;7])[1,1] \\ Charles R Greathouse IV, Jun 11 2015
    
  • PARI
    concat(0, Vec(x*(1+4*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Mar 27 2016
    

Formula

a(2*n) = (a(2*n - 1) + a(2*n + 1))/2.
a(2*n + 1) = (a(2*n) + a(2*n + 2))/4.
a(2*n) = 4*A001109(n).
a(2*n + 1) = 4*A001109(n) + A001541(n).
From Colin Barker, Jun 29 2012: (Start)
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1 + 4*x + x^2)/((1 + 2*x - x^2)*(1 - 2*x - x^2)) = x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4). (End)
2*a(n) = A078057(n) - A123335(n-1). - R. J. Mathar, Jul 04 2012
a(2n) = A005319(n); a(2n+1) = A002315(n). - R. J. Mathar, Jul 17 2009
a(n)*a(n+1) + 1 = A001653(n+1). - Charlie Marion, Dec 11 2012
a(n) = (((-2 - sqrt(2) + (-1)^n * (-2+sqrt(2))) * ((-1+sqrt(2))^n - (1+sqrt(2))^n)))/(4*sqrt(2)). - Colin Barker, Mar 27 2016
a(n) = A084068(n) - A079496(n). - César Aguilera, Feb 14 2023

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

Original entry on oeis.org

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

Views

Author

Kenneth J Ramsey, Apr 17 2012

Keywords

Comments

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

Crossrefs

Programs

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

Formula

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

A182189 a(n) = 6*a(n-1) - a(n-2) - 4 with n > 1, a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 13, 71, 409, 2379, 13861, 80783, 470833, 2744211, 15994429, 93222359, 543339721, 3166815963, 18457556053, 107578520351, 627013566049, 3654502875939, 21300003689581, 124145519261543, 723573111879673, 4217293152016491, 24580185800219269, 143263821649299119
Offset: 0

Views

Author

Kenneth J Ramsey, Apr 17 2012

Keywords

Comments

If p is a prime of the form 8*n +- 1 then a(p) == 3 (mod p); if p is a prime of the form 8*n +- 3 then a(p) == -1 (mod p).
The terms a(n) > 1 satisfy a(n)^5 + b(n)^5 = c(n)^5 + d(n)^5 where b(n) = a(n) - 2, c(n) = (a(n)-1) + i*ceiling((a(n)-1)*sqrt(2)), and d(n) is the conjugate of c(n), where i is the imaginary unit. Note that Re(c(n)) is A001542(n) and Im(d(n)) is A001541(n). - Pedro Caceres, Dec 30 2017

Crossrefs

Programs

  • Magma
    I:=[1,3]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2)-4: n in [1..41]]; // Bruno Berselli, Jun 07 2012
    
  • Mathematica
    m = -11;n = -1; c = 0;
    list3 = Reap[While[c < 22, t = 6 n - m - 4; Sow[t];m = n; n = t;c++]][[2,1]]
    CoefficientList[Series[(1-4*x-x^2)/((1-x)*(1-6*x+x^2)),{x,0,40}],x] (* Vincenzo Librandi, Jul 26 2012 *)
    1 + Fibonacci[2*Range[0, 40], 2] (* G. C. Greubel, May 24 2021 *)
  • PARI
    my(x='x+O('x^40)); Vec((1-4*x-x^2)/((1-x)*(1-6*x+x^2))) \\ Altug Alkan, Dec 30 2017
    
  • Sage
    [1 + lucas_number1(2*n,2,-1) for n in (0..40)] # G. C. Greubel, May 24 2021

Formula

G.f.: (1-4*x-x^2)/((1-x)*(1-6*x+x^2)). - Bruno Berselli, Jun 07 2012
a(n) = 1 + A000129(2*n). - G. C. Greubel, May 24 2021

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

Views

Author

Kenneth J Ramsey, Apr 28 2012

Keywords

Comments

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.

Examples

			     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.
		

Crossrefs

Programs

  • Maple
    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
  • Mathematica
    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 *)

Formula

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

A182441 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 - see below).

Original entry on oeis.org

0, 0, 4, 14, 1, 7, 114, 14, 2, 8, 700, 131, 14, 3, 10, 4116, 820, 144, 14, 4, 11, 24026, 4837, 912, 149, 14, 5, 12, 140070, 28250, 5390, 948, 158, 14, 6, 13, 816424, 164711, 31490, 5607, 1012, 163, 14, 7, 14, 4758504
Offset: 0

Views

Author

Kenneth J Ramsey, Apr 28 2012

Keywords

Comments

This is a table related to the square array of the nonnegative 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). That is each term G(i) is a(k,i+1). If A002262(14*n) is "r", 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 the square array A001477. If the product is less than (r^2 + 3*r -2)/2, then the product could still be said to lie in the same row r since the product is equal to the sum of a triangular number + r, which is a property of all numbers in row r of the square array A002262.
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, wherever a(k+1,0)-a(k,0) = 0, a(k+1,i+1)-a(k,i+1) = A212329. Also, a(k+1,n+2)-a(k,n+2) is divisible by A143608(n).

Examples

			For i>0 a(0,i) * a(0,i+1) = 0*14,14*114,114*700,700*4116,etc. which are all triangular numbers and lie in row 0 of square array A001477, while a(1,i)*a(1.i+1) = 1*14, 14*131, 131*820, 820*4837 etc. which are all 4 more than a triangular number and lie in row 4 of square array A001477.
		

Crossrefs

Programs

  • Mathematica
    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

Formula

a(k,0) equals the largest m such that m*(m+1)/2 <= 14*k (A003056(14*k)).
a(k,1) equals 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 is a constant equal to 28 + 2*k + 2 + 4*A003056(14*k).

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

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Author

Kenneth J Ramsey, Apr 28 2012

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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

Formula

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).
Showing 1-7 of 7 results.