A087204 -id:A087204 - cp's OEIS frontend

A277627 Square array read by antidiagonals downwards: T(n,k), n>=0, k>=0, in which column 0 is equal to A057427: 0, 1, 1, 1, ..., and for k > 0 column k lists two zeros followed by the partial sums of column k-1.

0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 3, 1, 0, 0, 0, 0, 0, 1, 4, 1, 0, 0, 0, 0, 0, 0, 3, 5, 1, 0, 0, 0, 0, 0, 0, 0, 6, 6, 1, 0, 0, 0, 0, 0, 0, 0, 1, 10, 7, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 15, 8, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 21, 9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 20, 28, 10, 1

Offset: 0

Paul Curtz, Oct 24 2016

In other words, for n > 0 the column k lists 2*k+1 zeros together with the partial sums of the positive terms of column k-1. - Omar E. Pol, Oct 25 2016
Comments from the author:
1) ZSPEC =
  0, 0, 0, 0, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, 0, 0, 0, ...
  1, 1, 0, 0, 0, 0, 0, 0, ...
  1, 2, 0, 0, 0, 0, 0, 0, ...
  1, 3, 1, 0, 0, 0, 0, 0, ...
  1, 4, 3, 0, 0, 0, 0, 0, ...
  1, 5, 6, 1, 0, 0, 0, 0, ...
etc.
The columns are the autosequences of the first kind of the title (column 1: 0, 0, followed by A001477(n); column 2: 0, 0, 0, 0, followed by A000217(n), etc) .
The positive terms are the Pascal triangle written by diagonals (A011973).
First column: A060576(n+1). Or A057427(n), n>-1, thanks to Omar E. Pol.
Row sums: A000045(n), autosequence of the first kind.
Alternated row sums and subtractions: 0, 1, 1, 0, -1, -1, 0 = A128834(n), autosequence of the first kind.
Antidiagonal sums: 0, 1, 1, 1, 2, 3, 4, 6, ... = A078012(n+2).
Application.
Numbers in triangle leading to the Genocchi numbers -A226158(n).
We multiply the columns of ZSPEC by d(n) = 1, -1, 2, -8, 56, -608, ... from A005439.
Hence, with only the first 0,
  0,
  1,
  1,
  1, -1,
  1, -2,
  1, -3,  2,
  1, -4,  6,
  1, -5, 12,   -8,
  1, -6, 20,  -32,
  1, -7, 30,  -80,  56,
  1, -8, 42, -160, 280,
  etc.
The row sums is -A226158(n).
2) Now consider the case of the autosequences of the second kind.
First step.
            2, 1, 1,  1,  1,  1, ...        = A054977(n)
      0, 0, 2, 3, 4,  5,  6,  7, ...        = A199969(n) with offset 0
0, 0, 0, 0, 2, 5, 9, 14, 20, 27, ...        see A000096
etc.
The positive terms are ASPEC in A191302. By triangle, they are either A029653(n) with A029653(0) = 2 instead of 1 or A029635(n).
Second step. YSPEC =
  2, 0,  0, 0, 0, 0, ...
  1, 0,  0, 0, 0, 0, ...
  1, 2,  0, 0, 0, 0, ...
  1, 3,  0, 0, 0, 0, ...
  1, 4,  2, 0, 0, 0, ...
  1, 5,  5, 0, 0, 0, ...
  1, 6,  9, 2, 0, 0, ...
  1, 7, 14, 7, 0, 0, ...
  etc.
Diagonals by triangle: A029635(n).
This is the companion to ZSPEC.
Row sums: A000032(n), autosequence of the second kind.
Alternated row sums and subtractions: period 6 repeat 2, 1, -1, -2, -1, 1 = A087204(n), autosequence of the second kind.
Application.
Numbers in triangle leading to A230324(n), a companion to -A226158(n).
We multiply the columns of YSPEC by d(n) 1, -1, 2, -8, 56, ... (see above).
Hence, without zeros:
  2,
  1,
  1, -2,
  1, -3,
  1, -4,  4,
  1, -5, 10,
  1, -6, 18,  -16,
  1, -7, 28,  -56,
  1, -8, 40, -128, 112,
  1, -9, 54, -240, 504,
  etc.
The row sum is A230324(n).

G. C. Greubel, Table of n, a(n) for the first 25 antidiagonals, flattened
OEIS Wiki, Autosequence

Cf. A011973 (without 0's), A007318 (Pascal's triangle).
Cf. A000045 (row sums), A078012 (antidiagonal sums).
Columns: A060576 or A057427 (k=0), A001477 (k=1), A000217 (k=2).
Cf. A000004, A000012, A000027, A000032, A005439, A029635, A029653, A054977, A087204, A128834, A191302, A199969, A226158, A230324.

kMax = 13; col[0] = Join[{0}, Array[1&, kMax]]; col[k_] := col[k] = Join[{0, 0}, col[k-1][[1 ;; -3]] // Accumulate]; T[n_, k_] := col[k][[n+1]]; Table[T[n-k, k], {n, 0, kMax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 15 2016 *)

Better definition from Omar E. Pol, Oct 25 2016

A115605 Expansion of -x^2(2 + x - 2x^2 - x^3 + 2x^4) / ( (x-1)(1+x)(1 + x + x^2)(x^2 - x + 1)(x^2 + 4x - 1)*(x^2 - x - 1) ).

Original entry on oeis.org

0, 0, 2, 7, 31, 128, 549, 2315, 9826, 41594, 176242, 746496, 3162334, 13395658, 56745250, 240376201, 1018250793, 4313378176, 18271765435, 77400436781, 327873517634, 1388894499108, 5883451527348, 24922700587008

Offset: 0

Roger L. Bagula, Mar 13 2006

Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1,0,1,-3,-6,3,1).

Cf. A000045, A079962.

A000035 := proc(n)
        n mod 2 ;
end proc:
A061347 := proc(n)
        op((n mod 3)+1,[-2,1,1]) ;
end proc:
A001076 := proc(n)
        option remember;
        if n <=1 then
                n;
        else
                4*procname(n-1)+procname(n-2) ;
        end if;
end proc:
A039834 := proc(n)
        (-1)^(n+1)*combinat[fibonacci](n) ;
end proc:
A087204 := proc(n)
        op((n mod 6)+1,[2,1,-1,-2,-1,1]) ;
end proc:
A115605 := proc(n)
        -A000035(n+1)/6 +A061347(n+2)/12 + A001076(n+1)/10 +3*A039834(n+1)/20 -A087204(n)/12 ;
end proc: # R. J. Mathar, Dec 16 2011

LinearRecurrence[{3,6,-3,-1,0,1,-3,-6,3,1},{0,0,2,7,31,128,549,2315,9826,41594},30] (* Harvey P. Dale, Dec 16 2011 *)

concat([0,0],Vec((2+x-2*x^2-x^3+2*x^4)/((1-x)*(1+x)*(1+x+x^2)*(x^2-x+1)*(x^2+4*x-1)*(x^2-x-1))+O(x^99))) \\ Charles R Greathouse IV, Sep 27 2012

Lim_{n->infinity} a(n+1)/a(n) = phi^3 = A098317.

a(n) = -A000035(n+1)/6 +A061347(n+2)/12 +A001076(n+1)/10 +3*A039834(n+1)/20 -A087204(n)/12. - R. J. Mathar, Dec 16 2011

A141212 a(n) = 1, if n == {1,3,4} mod 6; otherwise 0.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1

Offset: 1

Gary W. Adamson, Jun 14 2008

			a(7) = 1 since 7 == 1 mod 6.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A029739, A049347.

Table[If[MemberQ[{1,3,4},Mod[n,6]],1,0],{n,120}] (* or *) PadRight[{},120,{1,0,1,1,0,0}] (* Harvey P. Dale, May 03 2013 *)

a(n) = 1 if n is in A029739, otherwise 0.
Begin with the sequence S: (1,0,1,0,1,0,...) and create a hole every 3n-th place: 1,0_1,0_1,0_1,0_,... Then insert terms of the sequence S in the holes.
From R. J. Mathar, Jun 17 2008: (Start)
O.g.f.: x(1+x^2+x^3)/((1-x)(1+x)(x^2+x+1)(x^2-x+1)).
a(n) = 1/2+A049347(n-1)/2-(-1)^n/6-A087204(n)/6 = a(n-6). (End)
a(n) = ((Fibonacci(n+4) mod 4) mod 3) mod 2. - Gary Detlefs, Dec 29 2010

cp's OEIS Frontend

A277627 Square array read by antidiagonals downwards: T(n,k), n>=0, k>=0, in which column 0 is equal to A057427: 0, 1, 1, 1, ..., and for k > 0 column k lists two zeros followed by the partial sums of column k-1.

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A115605 Expansion of -x^2(2 + x - 2x^2 - x^3 + 2x^4) / ( (x-1)(1+x)(1 + x + x^2)(x^2 - x + 1)(x^2 + 4x - 1)*(x^2 - x - 1) ).

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A141212 a(n) = 1, if n == {1,3,4} mod 6; otherwise 0.

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cp's OEIS Frontend

A277627 Square array read by antidiagonals downwards: T(n,k), n>=0, k>=0, in which column 0 is equal to A057427: 0, 1, 1, 1, ..., and for k > 0 column k lists two zeros followed by the partial sums of column k-1.

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Author

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Programs

Mathematica

Extensions

A115605 Expansion of -x^2*(2 + x - 2*x^2 - x^3 + 2*x^4) / ( (x-1)*(1+x)*(1 + x + x^2)*(x^2 - x + 1)*(x^2 + 4*x - 1)*(x^2 - x - 1) ).

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Maple

Mathematica

PARI

Formula

A141212 a(n) = 1, if n == {1,3,4} mod 6; otherwise 0.

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Formula

A115605 Expansion of -x^2(2 + x - 2x^2 - x^3 + 2x^4) / ( (x-1)(1+x)(1 + x + x^2)(x^2 - x + 1)(x^2 + 4x - 1)*(x^2 - x - 1) ).