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.

User: Kenneth J Ramsey

Kenneth J Ramsey's wiki page.

Kenneth J Ramsey has authored 32 sequences. Here are the ten most recent ones:

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.

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

Author

Kenneth J Ramsey, Apr 25 2012

Keywords

Comments

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).

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} -> 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

A212329 Expansion of x*(5+x)/(1-7*x+7*x^2-x^3).

Original entry on oeis.org

0, 5, 36, 217, 1272, 7421, 43260, 252145, 1469616, 8565557, 49923732, 290976841, 1695937320, 9884647085, 57611945196, 335787024097, 1957110199392, 11406874172261, 66484134834180, 387497934832825, 2258503474162776, 13163522910143837, 76722633986700252
Offset: 1

Author

Kenneth J Ramsey, May 14 2012

Keywords

Comments

Table of differences re Table A182441.
This is a sequence of differences between rows k and k+1 of table A182441. That is if A182441(k+1,0)-A182441(k,0) = 1, a(n) = A182441(k+1,n+1) - A182441(k,n+1) for n = 0 to 3. The remainder of the sequence is a continuation using the recursive formula D(n) = 6D(n-1)- D(n-2) + 6.
It appears that for n > 0, a(n) is divisible by A213005(n).
It appears that if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p), and that if p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p).

Crossrefs

Programs

  • Mathematica
    m = 12; n = 1; c = 0;
    list3 = Reap[While[c < 22, t = 6 n - m + 6; Sow[t];m = n; n = t;c++]][[2,1]]
    CoefficientList[ Series[x (5 + x)/(1 - 7x + 7x^2 - x^3), {x, 0, 20}], x] (* or *)
    LinearRecurrence[{7, -7, 1}, {0, 5, 36}, 21] (* Robert G. Wilson v, Jun 24 2014 *)
  • PARI
    concat(0, Vec(x^2*(5+x)/((1-x)*(1-6*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 05 2016

Formula

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
From Colin Barker, Mar 05 2016: (Start)
a(n) = (-6+(5-3*sqrt(2))*(3+2*sqrt(2))^n + (3-2*sqrt(2))^n*(5+3*sqrt(2)))/4.
G.f.: x*(5+x) / ((1-x)*(1-6*x+x^2)).
(End)

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

Author

Kenneth J Ramsey, Apr 17 2012

Keywords

Comments

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.

Programs

  • Magma
    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
    
  • Maple
    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
  • Mathematica
    LinearRecurrence[{7,-7,1},{-1,1,19},30] (* Harvey P. Dale, Feb 09 2014 *)
  • PARI
    Vec(-(1-8*x-5*x^2)/((1-x)*(1-6*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 05 2016
    
  • Sage
    [lucas_number2(2*n,2,-1) - lucas_number1(2*n,2,-1) - 3 for n in (0..40)] # G. C. Greubel, May 24 2021

Formula

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

Extensions

More terms from Harvey P. Dale, Feb 09 2014

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

Author

Kenneth J Ramsey, Apr 17 2012

Keywords

Comments

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).

Programs

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

Formula

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

A182119 Table of triangular arguments such that if A002262(14*k) = "r" then the product A182439(k,i + 1) *A182439(k,i + 2) equals "r" + A000217(a(k,i)) for i<4, while a(k,i) = 0 for i>3.

Original entry on oeis.org

0, 55, 4, 384, 51, 7, 2303, 328, 48, 8, 0, 1943, 287, 47, 10, 0, 0, 1680, 276, 45, 11, 0, 0, 0, 1611, 250, 44, 12, 0, 0, 0, 0, 1445, 239, 43, 13, 0, 0, 0, 0, 0, 1376, 228, 42, 14, 0, 0, 0, 0, 0, 1307, 213, 41, 15
Offset: 0

Author

Kenneth J Ramsey, Apr 12 2012

Keywords

Comments

The triangular product A000217(a(k,i)) for i < 4 + A002262(14*k) = the product of adjacent terms G(k,i+1)*G(k,i+2) where G is table A182439. 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 of 1 or 0 in column 0 form related series A182188 and A182190.
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).

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 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

A182118 Table of triangular arguments such that if A002262(14*k) = "r" then the product A182440(k,i + 1) *A182440(k,i + 2) equals "r" + a(k,i)*(a(k,i)+1)/2.

Original entry on oeis.org

-1, 0, -5, 63, 8, -8, 440, 151, 15, -9, 0, 996, 224, 20, -11, 0, 0, 1455, 267, 26, -12, 0, 0, 0, 1720, 325, 31, -13, 0, 0, 0, 0, 2082, 368, 36, -14, 0, 0, 0, 0, 0, 2347, 411, 41, -15, 0, 0, 0, 0, 0, 0, 2612, 454, 46
Offset: 0

Author

Kenneth J Ramsey, Apr 12 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 A182191 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.

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} -> 0,{1,12} -> 0}];
    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<12, b=a; While[b>4,Sow[0];b--];While[b>0, Sow[tt[[b, a+1-b]]]; b--]; a++]][[2, 1]];list2

A214229 a(n) equals gcd(r,2*n+1) where r is 1 + (A143608(i+1) mod (2*n+1)) where A143608(i) is the first zero mod 2n+1 other than 0.

Original entry on oeis.org

3, 5, 1, 9, 11, 13, 3, 17, 19, 3, 1, 25, 27, 29, 1, 33, 5, 37, 3, 1, 43, 9, 1, 1, 17, 53, 11, 57, 59, 61, 9, 65, 67, 3, 1, 73, 3, 11, 1, 81, 83, 17, 3, 89, 13, 3, 19, 97, 99, 101, 1, 3, 107, 109, 3, 113, 5, 9, 17, 121, 3, 125, 1, 129, 131, 19
Offset: 1

Author

Kenneth J Ramsey, Jul 07 2012

Keywords

Comments

It appears that a(n) * b(n) either equals 2*n+1 or 1 where b is the companion sequence A214228.

Examples

			a(7) = 3 which is a factor of 2*7 + 1.
		

Programs

  • Maple
    A214229 := proc(n)
        local i,r ;
        i := 1;
        while A143608(i) mod (2*n+1) <> 0 do
            i := i+1 ;
        end do;
        r := 1+(A143608(i+1) mod (2*n+1)) ;
        gcd(r,2*n+1) ;
    end proc: # R. J. Mathar, Jul 22 2012
  • Mathematica
    gcdN2[x_,y_] = GCD[y - x + 1,y];
    r0 = 3;
    table=Reap[While[r0 < 200,s1=1;s0=0;count=1;While[True,count++;temp=Mod[4*s1 - s0,r0];
    If[temp==0,Break[]];count++;s0 = s1; s1 = temp;
    temp=Mod[2*s1-s0,r0];If[temp == 0,Break[]];s0 = s1;s1 = temp;];
    Sow[gcdN2[s1,r0],d];
    r0+=2;]][[2]];
    table

A214228 a(n) = gcd(r,2*n+1) where r is 1 + (A143608(i-1) mod (2*n+1)) and A143608(i) is the first zero mod 2*n+1 other than i=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 5, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 13, 1, 1, 5, 1, 1, 3, 1, 5, 1, 1, 1, 7, 1, 1, 23, 1, 1, 25, 7, 1, 1, 1, 5, 29, 1, 7, 31, 5, 1, 1, 1, 1, 35, 1, 1, 37, 1, 23, 13, 7, 1, 41, 1, 1, 1, 1, 7, 5, 1, 1, 47, 13, 1, 49, 1, 1, 9, 31, 1, 53
Offset: 1

Author

Kenneth J Ramsey, Jul 07 2012

Keywords

Comments

It appears that a(n) * gcd(s,2*n+1) is either 2*n+1 or 1; where s is 1 + (A143608(i+1) mod (2*n+1)) and A143608(i) is as stated in the definition.

Examples

			a(7) = 5 which is a factor of 2*7+1.
		

Crossrefs

Cf. A143608.

Programs

  • Maple
    A214228 := proc(n)
        local i,r ;
        i := 1;
        while A143608(i) mod (2*n+1) <> 0 do
            i := i+1 ;
        end do;
        r := 1+(A143608(i-1) mod (2*n+1)) ;
        gcd(r,2*n+1) ;
    end proc: # R. J. Mathar, Jul 22 2012
  • Mathematica
    gcdN1[x_,y_] = GCD[x + 1,y]; r0 = 3; Reap[While[r0 < 200, s1=1; s0=0; count=1; While[True, count++; temp=Mod[4*s1 - s0, r0]; If[temp==0, Break[]]; count++; s0 = s1; s1 = temp; temp=Mod[2*s1-s0,r0]; If[temp == 0, Break[]]; s0 = s1; s1 = temp;]; Sow[gcdN1[s1, r0], c]; r0+=2;]][[2,1]]

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

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).

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

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.
		

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