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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A054142 Triangular array binomial(2*n-k, k), k=0..n, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 6, 1, 1, 7, 15, 10, 1, 1, 9, 28, 35, 15, 1, 1, 11, 45, 84, 70, 21, 1, 1, 13, 66, 165, 210, 126, 28, 1, 1, 15, 91, 286, 495, 462, 210, 36, 1, 1, 17, 120, 455, 1001, 1287, 924, 330, 45, 1, 1, 19, 153, 680, 1820, 3003, 3003, 1716, 495, 55, 1
Offset: 0

Views

Author

Keywords

Comments

Row sums are odd-indexed Fibonacci numbers.
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k double rises. Mirror image of A085478. - Emeric Deutsch, May 31 2004
Diagonal sums are A052535. - Paul Barry, Jan 21 2005
Matrix inverse is the triangle of Salie numbers A098435. - Paul Barry, Jan 21 2005
Coefficients of Morgan-Voyce polynomial b(n,x); e.g., b(3,x)=x^3+5x^2+6x+1. See A172431 for coefficients of Morgan-Voyce polynomial B(n,x). - Clark Kimberling, Feb 13 2010
T(n,k) is the number of stack polyominoes of perimeter 2n+4 with k+1 columns. - Emanuele Munarini, Apr 07 2011
Roots of signed n-th polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths A003558(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697... = A116415; we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...; the 3 roots to the polynomial with cycle length 3 matching A003558(3) = 3. The operation (-2, x^2) is the reversal of the well known chaotic operation (x^2 - 2) [Kappraff, Adamson, 2004] starting with seed 2*cos(2*Pi/N). Check: given 2*cos(2*Pi/7) = 1.24697..., we obtain the 3-cycle using (x^2 - 2): (1.24697...-> -0.445041...-> 1.801937...; where the terms in either set are intermediate terms in the other, irrespective of sign. - Gary W. Adamson, Sep 22 2011
A054142 is jointly generated with A172431 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section of A172431. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle given by (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 01 2012
The o.g.f. for row n of the array A(n, k) = binomial(2*n-k,k), k >= 0, n >= 0 is G(n,x) = Sum_{k=0..n} T(n, k)*x^k + (-x)^(2*n+1) * c(-x)^(2*n+1) / sqrt(1-4*(-x)), for n >= 0. Here c(x) is the o.g.f. of A000108 (Catalan). For powers of c(x) see the W. Lang link in A115139. For the alternating sign case replace x by -x. - Wolfdieter Lang, Sep 12 2016
Multiplying the n-th diagonal by A001147(n) generates A001497. - Tom Copeland, Oct 04 2016

Examples

			Triangle begins:
  1;
  1,  1;
  1,  3,  1;
  1,  5,  6,   1;
  1,  7, 15,  10,   1;
  1,  9, 28,  35,  15,   1;
  1, 11, 45,  84,  70,  21,   1;
  1, 13, 66, 165, 210, 126,  28,  1;
  1, 15, 91, 286, 495, 462, 210, 36, 1; ...
...
(0, 1, 0, 0, 0, 0, ...) DELTA (1, 0, 1, 0, 0, 0, ...) begins:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 3,  1;
  0, 1, 5,  6,  1;
  0, 1, 7, 15, 10,  1;
  0, 1, 9, 28, 35, 15, 1. _Philippe Deléham_, Apr 01 2012
		

Crossrefs

These are the even-indexed rows of A011973, the odd-indexed rows form A053123.

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(2*n-k,k) ))); # G. C. Greubel, Aug 01 2019
  • Magma
    [Binomial(2*n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
    
  • Maple
    T:=(n,k)->binomial(2*n-k,k): seq(seq(T(n,k), k=0..n), n=0..11);
  • Mathematica
    Flatten[Table[Binomial[2n - k, k], {n, 0, 11}, {k, 0, n}]] (* Emanuele Munarini, Apr 07 2011 *)
  • Maxima
    create_list(binomial(2*n-k,k),n,0,10,k,0,n); /* Emanuele Munarini, Apr 07 2011 */
    
  • PARI
    T(n,k)=if(n<0,0,polcoeff(charpoly(matrix(n,n,i,j,-min(i,j))),k))
    
  • Sage
    [[binomial(2*n-k,k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
    

Formula

G.f.: (1-t*z)/((1-t*z)^2-z). - Emeric Deutsch, May 31 2004
Column k has g.f.: (Sum_{j=0..k+1} binomial(k+1, 2j)*x^j)*x^k/(1-x)^(k+1). - Paul Barry, Jun 22 2005
Recurrence: T(n+2,k+2) = T(n+1,k+2) + 2*T(n+1,k+1) - T(n,k). - Emanuele Munarini, Apr 07 2011
T(n, k) = binomial(2*n-k, k) = A085478(n, n-k), for n >= 0, k = 0..n. - Wolfdieter Lang, Mar 25 2020

A005314 For n = 0, 1, 2, a(n) = n; thereafter, a(n) = 2*a(n-1) - a(n-2) + a(n-3).

Original entry on oeis.org

0, 1, 2, 3, 5, 9, 16, 28, 49, 86, 151, 265, 465, 816, 1432, 2513, 4410, 7739, 13581, 23833, 41824, 73396, 128801, 226030, 396655, 696081, 1221537, 2143648, 3761840, 6601569, 11584946, 20330163, 35676949, 62608681, 109870576, 192809420, 338356945, 593775046
Offset: 0

Views

Author

Keywords

Comments

Number of compositions of n into parts congruent to {1,2} mod 4. - Vladeta Jovovic, Mar 10 2005
a(n)/a(n-1) tends to A109134; an eigenvalue of the matrix M and a root to the characteristic polynomial. - Gary W. Adamson, May 25 2007
Starting with offset 1 = INVERT transform of (1, 1, 0, 0, 1, 1, 0, 0, ...). - Gary W. Adamson, May 04 2009
a(n-2) is the top left entry of the n-th power of the 3 X 3 matrix [0, 1, 0; 0, 1, 1; 1, 0, 1] or of the 3 X 3 matrix [0, 0, 1; 1, 1, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014
Counts closed walks of length (n+2) at a vertex of a unidirectional triangle containing a loop on remaining two vertices. - David Neil McGrath, Sep 15 2014
Also the number of binary words of length n that begin with 1 and avoid the subword 101. a(5) = 9: 10000, 10001, 10010, 10011, 11000, 11001, 11100, 11110, 11111. - Alois P. Heinz, Jul 21 2016
Also the number of binary words of length n-1 such that every two consecutive 0s are immediately followed by at least two consecutive 1s. a(4) = 5: 010, 011, 101, 110, 111. - Jerrold Grossman, May 03 2024

Examples

			G.f. = x + 2*x^2 + 3*x^3 + 5*x^4 + 9*x^5 + 16*x^6 + 28*x^7 + 49*x^8 + ...
From _Gus Wiseman_, Nov 25 2019: (Start)
a(n) is the number of subsets of {1..n} containing n such that if x and x + 2 are both in the subset, then so is x + 1. For example, the a(1) = 1 through a(5) = 9 subsets are:
  {1}  {2}    {3}      {4}        {5}
       {1,2}  {2,3}    {1,4}      {1,5}
              {1,2,3}  {3,4}      {2,5}
                       {2,3,4}    {4,5}
                       {1,2,3,4}  {1,2,5}
                                  {1,4,5}
                                  {3,4,5}
                                  {2,3,4,5}
                                  {1,2,3,4,5}
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equals row sums of triangle A099557.
Equals row sums of triangle A224838.
Cf. A011973 (starting with offset 1 = Falling diagonal sums of triangle with rows displayed as centered text).
First differences of A005251, shifted twice to the left.

Programs

  • Haskell
    a005314 n = a005314_list !! n
    a005314_list = 0 : 1 : 2 : zipWith (+) a005314_list
       (tail $ zipWith (-) (map (2 *) $ tail a005314_list) a005314_list)
    -- Reinhard Zumkeller, Oct 14 2011
    
  • Magma
    [0] cat [n le 3 select n else 2*Self(n-1) - Self(n-2) + Self(n-3):n in [1..35]]; // Marius A. Burtea, Oct 24 2019
    
  • Magma
    R:=PowerSeriesRing(Integers(), 36); [0] cat Coefficients(R!( x/(1-2*x+x^2-x^3))); // Marius A. Burtea, Oct 24 2019
    
  • Maple
    A005314 := proc(n)
        option remember ;
        if n <=2 then
            n;
        else
            2*procname(n-1)-procname(n-2)+procname(n-3) ;
        end if;
    end proc:
    seq(A005314(n),n=0..20) ; # R. J. Mathar, Feb 25 2024
  • Mathematica
    LinearRecurrence[{2, -1, 1}, {0, 1, 2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
    Table[Sum[Binomial[n - Floor[(k + 1)/2], n - Floor[(3 k - 1)/2]], {k, 0, n}], {n, 0, 100}] (* John Molokach, Jul 21 2013 *)
    Table[Sum[Binomial[n - Floor[(4 n + 15 - 6 k + (-1)^k)/12], n - Floor[(4 n + 15 - 6 k + (-1)^k)/12] - Floor[(2 n - 1)/3] + k - 1], {k, 1, Floor[(2 n + 2)/3]}], {n, 0, 100}] (* John Molokach, Jul 25 2013 *)
    a[ n_] := If[ n < 0, SeriesCoefficient[ x^2 / (1 - x + 2 x^2 - x^3), {x, 0, -n}], SeriesCoefficient[ x / (1 - 2 x + x^2 - x^3), {x, 0, n}]]; (* Michael Somos, Dec 13 2013 *)
    RecurrenceTable[{a[0]==0,a[1]==1,a[2]==2,a[n]==2a[n-1]-a[n-2]+a[n-3]},a,{n,40}] (* Harvey P. Dale, May 13 2018 *)
    Table[Length[Select[Subsets[Range[n]],MemberQ[#,n]&&!MatchQ[#,{_,x_,y_,_}/;x+2==y]&]],{n,0,10}] (* Gus Wiseman, Nov 25 2019 *)
  • PARI
    {a(n) = sum(k=0, (2*n-1)\3, binomial(n-1-k\2, k))}
    
  • PARI
    {a(n) = if( n<0, polcoeff( x^2 / (1 - x + 2*x^2 - x^3) + x * O(x^-n), -n), polcoeff( x / (1 - 2*x + x^2 - x^3) + x * O(x^n), n))}; /* Michael Somos, Sep 18 2012 */
    
  • SageMath
    def A005314(n): return sum( binomial(n-k, 2*k+1) for k in range(floor((n+2)/3)) )
    [A005314(n) for n in range(51)] # G. C. Greubel, Nov 10 2023

Formula

From Paul D. Hanna, Oct 22 2004: (Start)
G.f.: x/(1-2*x+x^2-x^3).
a(n) = Sum_{k=0..[(2n-1)/3]} binomial(n-1-[k/2], k), where [x]=floor(x). (End)
a(n) = Sum_{k=0..n} binomial(n-k, 2*k+1).
23*a_n = 3*P_{2n+2} + 7*P_{2n+1} - 2*P_{2n}, where P_n are the Perrin numbers, A001608. - Don Knuth, Dec 09 2008
G.f. (1-z)*(1+z^2)/(1-2*z+z^2-z^3) for the augmented version 1, 1, 2, 3, 5, 9, 16, 28, 49, 86, 151, ... was given in Simon Plouffe's thesis of 1992.
a(n) = a(n-1) + a(n-2) + a(n-4) = a(n-2) + A049853(n-1) = a(n-1) + A005251(n) = Sum_{i <= n} A005251(i).
a(n) = Sum_{k=0..floor((n-1)/3)} binomial(n-k, 2*k+1). - Richard L. Ollerton, May 12 2004
M^n*[1,0,0] = [a(n-2), a(n-1), a]; where M = the 3 X 3 matrix [0,1,0; 0,0,1; 1,-1,2]. Example M^5*[1,0,0] = [3,5,9]. - Gary W. Adamson, May 25 2007
a(n) = A000931(2*n + 4). - Michael Somos, Sep 18 2012
a(n) = A077954(-n - 2). - Michael Somos, Sep 18 2012
G.f.: 1/( 1 - Sum_{k>=0} x*(x-x^2+x^3)^k ) - 1. - Joerg Arndt, Sep 30 2012
a(n) = Sum_{k=0..n} binomial( n-floor((k+1)/2), n-floor((3k-1)/2) ). - John Molokach, Jul 21 2013
a(n) = Sum_{k=1..floor((2*n+2)/3)} binomial(n - floor((4*n+15-6*k+(-1)^k)/12), n - floor((4*n+15-6*k+(-1)^k)/12) - floor((2*n-1)/3) + k - 1). - John Molokach, Jul 24 2013
a(n) = round(A001608(2n+1)*r) where r is the real root of 23*x^3 - 23*x^2 + 8*x - 1 = 0, r = 0.4114955... - Richard Turk, Oct 24 2019
a(n+2) = n + 2 + Sum_{k=0..n} (n-k)*a(k). - Greg Dresden and Yichen P. Wang, Sep 16 2021
a(n) ~ (19 - r + 11*r^2) / (23 * r^(n-1)), where r = 0.569840290998... is the root of the equation r*(2 - r + r^2) = 1. - Vaclav Kotesovec, Apr 14 2024
a(n) = n*3F2(1/3-n/3,2/3-n/3,1-n/3;-n,3/2;27/4). - R. J. Mathar, Jun 27 2024
If p,q,r are the three solutions to x^3 = 2x^2 - x + 1, then a(n) = p^(n+1)/((p-q)*(p-r)) + q^(n+1)/((q-p)*(q-r)) + r^(n+1)/((r-p)*(r-q)). Compare to similar formula for A005251. - Greg Dresden and AnXing Yang, Aug 19 2025

Extensions

More terms and additional formulas from Henry Bottomley, Jul 21 2000
Plouffe's g.f. edited by R. J. Mathar, May 12 2008

A122848 Exponential Riordan array (1, x(1+x/2)).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 3, 1, 0, 0, 3, 6, 1, 0, 0, 0, 15, 10, 1, 0, 0, 0, 15, 45, 15, 1, 0, 0, 0, 0, 105, 105, 21, 1, 0, 0, 0, 0, 105, 420, 210, 28, 1, 0, 0, 0, 0, 0, 945, 1260, 378, 36, 1, 0, 0, 0, 0, 0, 945, 4725, 3150, 630, 45, 1, 0, 0, 0, 0, 0, 0, 10395, 17325, 6930, 990, 55, 1, 0, 0
Offset: 0

Views

Author

Paul Barry, Sep 14 2006

Keywords

Comments

Entries are Bessel polynomial coefficients. Row sums are A000085. Diagonal sums are A122849. Inverse is A122850. Product of A007318 and A122848 gives A100862.
T(n,k) is the number of self-inverse permutations of {1,2,...,n} having exactly k cycles. - Geoffrey Critzer, May 08 2012
Bessel numbers of the second kind. For relations to the Hermite polynomials and the Catalan (A033184 and A009766) and Fibonacci (A011973, A098925, and A092865) matrices, see Yang and Qiao. - Tom Copeland, Dec 18 2013.
Also the inverse Bell transform of the double factorial of odd numbers Product_{k= 0..n-1} (2*k+1) (A001147). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015

Examples

			Triangle begins:
    1
    0    1
    0    1    1
    0    0    3    1
    0    0    3    6    1
    0    0    0   15   10    1
    0    0    0   15   45   15    1
    0    0    0    0  105  105   21    1
    0    0    0    0  105  420  210   28    1
    0    0    0    0    0  945 1260  378   36    1
From _Gus Wiseman_, Jan 12 2021: (Start)
As noted above, a(n) is the number of set partitions of {1..n} into k singletons or pairs. This is also the number of set partitions of subsets of {1..n} into n - k pairs. In the first case, row n = 5 counts the following set partitions:
  {{1},{2,3},{4,5}}  {{1},{2},{3},{4,5}}  {{1},{2},{3},{4},{5}}
  {{1,2},{3},{4,5}}  {{1},{2},{3,4},{5}}
  {{1,2},{3,4},{5}}  {{1},{2,3},{4},{5}}
  {{1,2},{3,5},{4}}  {{1,2},{3},{4},{5}}
  {{1},{2,4},{3,5}}  {{1},{2},{3,5},{4}}
  {{1},{2,5},{3,4}}  {{1},{2,4},{3},{5}}
  {{1,3},{2},{4,5}}  {{1},{2,5},{3},{4}}
  {{1,3},{2,4},{5}}  {{1,3},{2},{4},{5}}
  {{1,3},{2,5},{4}}  {{1,4},{2},{3},{5}}
  {{1,4},{2},{3,5}}  {{1,5},{2},{3},{4}}
  {{1,4},{2,3},{5}}
  {{1,4},{2,5},{3}}
  {{1,5},{2},{3,4}}
  {{1,5},{2,3},{4}}
  {{1,5},{2,4},{3}}
In the second case, we have:
  {{1,2},{3,4}}  {{1,2}}  {}
  {{1,2},{3,5}}  {{1,3}}
  {{1,2},{4,5}}  {{1,4}}
  {{1,3},{2,4}}  {{1,5}}
  {{1,3},{2,5}}  {{2,3}}
  {{1,3},{4,5}}  {{2,4}}
  {{1,4},{2,3}}  {{2,5}}
  {{1,4},{2,5}}  {{3,4}}
  {{1,4},{3,5}}  {{3,5}}
  {{1,5},{2,3}}  {{4,5}}
  {{1,5},{2,4}}
  {{1,5},{3,4}}
  {{2,3},{4,5}}
  {{2,4},{3,5}}
  {{2,5},{3,4}}
(End)
		

Crossrefs

Row sums are A000085.
Column sums are A001515.
Same as A049403 but with a first column k = 0.
The same set partitions counted by number of pairs are A100861.
Reversing rows gives A111924 (without column k = 0).
A047884 counts standard Young tableaux by size and greatest row length.
A238123 counts standard Young tableaux by size and least row length.
A320663/A339888 count unlabeled multiset partitions into singletons/pairs.
A322661 counts labeled covering half-loop-graphs.
A339742 counts factorizations into distinct primes or squarefree semiprimes.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> `if`(n<2,1,0), 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    t[n_, k_] := k!*Binomial[n, k]/((2 k - n)!*2^(n - k)); Table[ t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten
    (* Second program: *)
    rows = 12;
    t = Join[{1, 1}, Table[0, rows]];
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 0, rows}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 23 2018,after Peter Luschny *)
    sbs[{}]:={{}};sbs[set:{i_,_}]:=Join@@Function[s,(Prepend[#1,s]&)/@sbs[Complement[set,s]]]/@Cases[Subsets[set],{i}|{i,_}];
    Table[Length[Select[sbs[Range[n]],Length[#]==k&]],{n,0,6},{k,0,n}] (* Gus Wiseman, Jan 12 2021 *)
  • PARI
    {T(n,k)=if(2*kn, 0, n!/(2*k-n)!/(n-k)!*2^(k-n))} /* Michael Somos, Oct 03 2006 */
    
  • Sage
    # uses[inverse_bell_transform from A265605]
    multifact_2_1 = lambda n: prod(2*k + 1 for k in (0..n-1))
    inverse_bell_matrix(multifact_2_1, 9) # Peter Luschny, Dec 31 2015

Formula

Number triangle T(n,k) = k!*C(n,k)/((2k-n)!*2^(n-k)).
T(n,k) = A001498(k,n-k). - Michael Somos, Oct 03 2006
E.g.f.: exp(y(x+x^2/2)). - Geoffrey Critzer, May 08 2012
Triangle equals the matrix product A008275*A039755. Equivalently, the n-th row polynomial R(n,x) is given by the Type B Dobinski formula R(n,x) = exp(-x/2)*Sum_{k>=0} P(n,2*k+1)*(x/2)^k/k!, where P(n,x) = x*(x-1)*...*(x-n+1) denotes the falling factorial polynomial. Cf. A113278. - Peter Bala, Jun 23 2014
From Daniel Checa, Aug 28 2022: (Start)
E.g.f. for the m-th column: (x^2/2+x)^m/m!.
T(n,k) = T(n-1,k-1) + (n-1)*T(n-2,k-1) for n>1 and k=1..n, T(0,0) = 1. (End)

A097610 Triangle read by rows: T(n,k) is number of Motzkin paths of length n and having k horizontal steps.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 2, 0, 6, 0, 1, 0, 10, 0, 10, 0, 1, 5, 0, 30, 0, 15, 0, 1, 0, 35, 0, 70, 0, 21, 0, 1, 14, 0, 140, 0, 140, 0, 28, 0, 1, 0, 126, 0, 420, 0, 252, 0, 36, 0, 1, 42, 0, 630, 0, 1050, 0, 420, 0, 45, 0, 1, 0, 462, 0, 2310, 0, 2310, 0, 660, 0, 55, 0, 1, 132, 0, 2772, 0
Offset: 0

Views

Author

Emeric Deutsch, Aug 30 2004

Keywords

Comments

Row sums are the Motzkin numbers (A001006). Column 0 gives the aerated Catalan numbers (A000108).
Let P_n(x) = Sum_{k=0..n} T(n,k)*x^k. P_0(x) = 1, P_1(x) = x, P_n(x) = x*P_(n-1)(x) + Sum_{j=0..n-2} P_j(x)*P_(n-2-j)(x); essentially the same as A124027. - Philippe Deléham, Oct 03 2007
G. J. Chaitin's numbers of s-expressions of size n are given by the coefficients of polynomials p(k, x) satisfying: p(k, x) = Sum_{j=2..k-1} p(j, x)*p(k-j, x). The coefficients of these polynomials also give (essentially) the triangle shown here. - Roger L. Bagula, Oct 31 2006
Exponential Riordan array [Bessel_I(1,2x)/x,x]. - Paul Barry, Mar 24 2010
Diagonal sums are the aerated large Schroeder numbers. - Paul Barry, Apr 21 2010
Non-vanishing antidiagonals are rows of A060693. - Tom Copeland, Feb 03 2016
These polynomials are related to the Gegenbauer polynomials which in turn are specializations of the Jacobi polynomials. The o.g.f. of the Gegenbauer polynomials is 1 / [1-2tx+x^2]^a. For the generating function Gb(x,h1,h2,a) = [x / (1 + h1 x + h2 x^2)]^a, the compositional inverse in x is Gbinv(x,h1,h2,a) = [(1-h1*y) - sqrt[(1-h1*y)^2-4h2*y^2]]/(2*h2*y) with y = x^(1/a). The polynomials of this entry are generated by Gbinv(x,t,1,1). The Legendre polynomials are related to the o.g.f. Gb(x,-2t,1,1/2). Cf. A121448. - Tom Copeland, Feb 07 2016
The bivariate o.g.f. in Copeland's Jan 29 2016 formulas can be related to conformal mappings of the complex plane and a solution of the dKP hierarchy. Cf. p. 24 of the Takebe et al. paper. - Tom Copeland, May 30 2018

Examples

			Triangle begins:
1;
0,  1;
1,  0,  1;
0,  3,  0,  1;
2,  0,  6,  0,  1;
0, 10,  0, 10,  0,  1;
5,  0, 30,  0, 15,  0,  1;
Row n has n+1 terms.
T(4,2)=6 because we have HHUD, HUDH, UDHH, HUHD, UHDH, UHHD, where U=(1,1), D=(1,-1) and H=(1,0).
		

References

  • G. J. Chaitin, Algorithmic Information Theory, Cambridge Univ. Press, 1987, page 169.

Crossrefs

Cf. A001006, A000108. A124027 is an essentially identical triangle.
Cf. A001263.

Programs

  • Maple
    G:=(1-t*z-sqrt(1-2*t*z+t^2*z^2-4*z^2))/2/z^2:
    Gser:=simplify(series(G,z=0,16)): P[0]:=1:
    for n from 1 to 13 do P[n]:=sort(coeff(Gser,z^n)) od:
    seq(seq(coeff(t*P[n],t^k),k=1..n+1),n=0..13);
    # Maple program for the triangular array:
    T:=proc(n,k) if n-k mod 2 = 0 and k<=n then n!/k!/((n-k)/2)!/((n-k)/2+1)! else 0 fi end: TT:=(n,k)->T(n-1,k-1): matrix(10,10,TT);
  • Mathematica
    T[n_,k_]:=If[n>=k&&EvenQ[n-k],n!/(k!((n-k)/2)!((n-k)/2+1)!),0];
    Flatten[Table[T[n,k],{n,0,20},{k,0,n}]] (* Peter J. C. Moses, Apr 06 2013 *)
    T[n_,k_] := If[OddQ[n - k], 0, Binomial[n, k] CatalanNumber[(n - k)/2]]; (* Peter Luschny, Jun 06 2018 *)

Formula

G.f.: [1-tz-sqrt(1-2tz+t^2*z^2-4z^2)]/(2z^2).
T(n, k) = n!/[k!((n-k)/2)!((n-k)/2-1)! ] = A055151(n, (n-k)/2) if n-k is a nonnegative even number; otherwise T(n, k) = 0.
T(n, k) = C(n, k)*C((n-k)/2)*(1+(-1)^(n-k))/2 if k <= n, 0 otherwise. - Paul Barry, May 18 2005
T(n,k) = A121448(n,k)/2^k. - Philippe Deléham, Aug 17 2006
Sum_{k=0..n} T(n,k)*2^k = A000108(n+1). - Philippe Deléham, Aug 22 2006
Sum_{k=0..n} T(n,k)*3^k = A002212(n+1). - Philippe Deléham, Oct 03 2007
G.f.: 1/(1-x*y-x^2/(1-x*y-x^2/(1-x*y-x^2/.... (continued fraction). - Paul Barry, Dec 15 2008
Sum_{k=0..n} T(n,k)*4^k = A005572(n). - Philippe Deléham, Dec 03 2009
T(n,k) = A007318(n,k)*A126120(n-k). - Philippe Deléham, Dec 12 2009
From Tom Copeland, Jan 23 2016: (Start)
E.g.f.: M(x,t) = e^(xt) AC(t) = e^(xt) I_1(2t)/t = e(xt) * e.g.f.(A126120(t)) = e^(xt) Sum_{n>=0} t^(2n)/(n!(n+1)!) = exp[t P(.,x)].
The e.g.f. of this Appell sequence of polynomials P(n,x) is e^(xt) times the e.g.f. AC(t) of the aerated Catalan numbers A126120. AC(t) = I_1(2t)/t, where I_n(x) = T_n(d/dx) I_0(x) are the modified Bessel functions of the first kind and T_n, the Chebyshev polynomials of the first kind.
P(n,x) has the lowering and raising operators L = d/dx = D and R = d/dD log{M(x,D)} = x + d/dD log{AC(D)} = x + Sum_{n>=0} c(n) D^(2n+1)/(2n+1)! with c(n) = (-1)^n A180874(n+1), i.e., L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
(P(.,x) + y)^n = P(n,x+y) = Sum_{k=0..n} binomial(n,k) P(k,x) y^(n-k) = (b.+x+y)^n, where (b.)^k = b_k = A126120(k).
Exp(b.D) e^(xt) = exp[(x+b.)t] = exp[P(.,x)t] = e^(b.t) e^(xt) = e^(xt) AC(t).
See p. 12 of the Alexeev et al. link and A055151 for a refinement.
Shifted o.g.f: G(x,t) = [1-tx-sqrt[(1-tx)^2-4x^2]] / 2x = x + t x^2 + (1+t) x^3 + ... has the compositional inverse Ginv(x,t) = x / [1 + tx + x^2] = x - t x^2 +(-1+t^2) x^3 + (2t-t^3) x^4 + (1-3t^2+t^4) x^5 + ..., a shifted o.g.f. for the signed Chebyshev polynomials of the second kind of A049310 (cf. also the Fibonacci polynomials of A011973). Then the inversion formula of A134264, involving non-crossing partitions and free probability with their multitude of interpretations (cf. A125181 also), can be used with h_0 = 1, h_1 = t, and h_2 = 1 to interpret the coefficients of the Motzkin polynomials combinatorially.
(End)
From Tom Copeland, Jan 29 2016: (Start)
Provides coefficients of the inverse of f(x) = x / [1 + h1 x + h2 x^2], a bivariate generating function of A049310 (mod signs).
finv(x) = [(1-h1*x) - sqrt[(1-h1*x)^2-4h2*x^2]]/(2*h2*x) = x + h1 x^2 + (h2 + h1^2) x^3 + (3 h1 h2 + h1^3) x^4 + ... is a bivariate o.g.f. for this entry.
The infinitesimal generator for finv(x) is g(x) d/dx with g(x) = 1 /[df(x)/dx] = x^2 / [(f(x))^2 (1 - h2 x^2)] = (1 + h1 x + h2 x^2)^2 / (1 - h2 x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomials FI(n,h1,h2) of the compositional inverse of f(x), i.e., exp[x g(u)d/du] u |_(u=0) = finv(x) = 1 / [1 -x FI(.,h1,h2)]. Cf. A145271. E.g.,
FI(0,h1,h2) = 0
FI(1,h1,h2) = 1
FI(2,h1,h2) = 1 h1
FI(3,h1,h2) = 1 h2 + 1 h1^2
FI(4,h1,h2) = 3 h2 h1 + 1 h1^3
FI(5,h1,h2) = 2 h2^2 + 6 h2 h1^2 + 1 h1^4
FI(6,h1,h2) = 10 h2^2 h1 + 10 h2 h1^3 + 1 h1^5.
And with D = d/dh1, FI(n+1, h1,h2) = MT(n,h1,h2) = (b.y + h1)^n = Sum_{k=0..n} binomial(n,k) b(k) y^k h1^(n-k) = exp[(b.y D] (h1)^n = AC(y D) (h1)^n, where b(k) = A126120(k), y = sqrt(h2), and AC(t) is defined in my Jan 23 formulas above. Equivalently, AC(y D) e^(x h1) = exp[x MT(.,h1,h2)].
The MT polynomials comprise an Appell sequence in h1 with e.g.f. e^(h1*x) AC(xy) = exp[x MT(.,h1,h2)] with lowering operator L = d/dh1 = D, i.e., L MT(n,h1,h2) = dMT(n,h1,h2)/dh1 = n MT(n-1,h1,h2) and raising operator R = h1 + dlog{AC(y L)}/dL = h1 + Sum_{n>=0} c(n) h2^(n+1) D^(2n+1)/(2n+1)! = h1 + h2 d/dh1 - h2^2 (d/dh1)^3/3! + 5 h2^3 (d/dh1)^5/5! - ... with c(n) = (-1)^n A180874(n+1) (consistent with the raising operator in the Jan 23 formulas).
The compositional inverse finv(x) is also obtained from the non-crossing partitions of A134264 (or A125181) with h_0 = 1, h_1 = h1, h_2 = h2, and h_n = 0 for all other n.
See A238390 for the umbral compositional inverse in h1 of MT(n,h1,h2) and inverse matrix.
(End)
From Tom Copeland, Feb 13 2016: (Start)
z1(x,h1,h2) = finv(x), the bivariate o.g.f. above for this entry, is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,h1,h2),z2(x,h1,h2)) = (z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-h1*x)/(h2*x)] z + 1/h2, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,h1,h2) = (1 - h1*x)/(h2*x) - z1(x,h1,h2) = [1 - h1*x + sqrt[(1-h1*x)^2 - 4 h2*x^2]] / (2h2*x).
The two are the nontrivial zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2), (see Landweber et al., p. 14, Ellingsrud, and A121448), and the zeros for the Riccati equation z' = (z - z1)(z - z2), associated to soliton solutions of the KdV equation (see Copeland link).
(End)
Comparing the shifted o.g.f. S(x) = x / (1 - h_1 x + h_2 x^2) for the bivariate Chebyshev polynomials S_n(h_1,h_2) of A049310 with the shifted o.g.f. H(x) = x / ((1 - a x)(1 - b x)) for the complete homogeneous symmetric polynomials H_n(a,b) = (a^(n+1)-b^(n+1)) / (a - b) shows that S_n(h_1,h_2) = H_n(a,b) for h_1 = a + b and h_2 = ab and, conversely, a = (h_1 + sqrt(h_1^2 - 4 h_2)) / 2 and b = (h_1 - sqrt(h_1^2 - 4 h_2)) / 2. The compositional inverse about the origin of S(x) gives a bivariate o.g.f. for signed Motzkin polynomials M_n(h_1,h_2) of this entry, and that of H(x) gives one for signed Narayana polynomials N_n(a,b) of A001263, thereby relating the bivariate Motzkin and Narayana polynomials by the indeterminate transformations. E.g., M_2(h_1,h_2) = h_2 + h_1^2 = ab + (a + b)^2 = a^2 + 3 ab + b^2 = N_2(a,b). - Tom Copeland, Jan 27 2024

A115139 Array of coefficients of polynomials related to integer powers of the generating function of Catalan numbers A000108.

Original entry on oeis.org

1, 1, 1, -1, 1, -2, 1, -3, 1, 1, -4, 3, 1, -5, 6, -1, 1, -6, 10, -4, 1, -7, 15, -10, 1, 1, -8, 21, -20, 5, 1, -9, 28, -35, 15, -1, 1, -10, 36, -56, 35, -6, 1, -11, 45, -84, 70, -21, 1, 1, -12, 55, -120, 126, -56, 7, 1, -13, 66, -165, 210, -126, 28, -1, 1, -14, 78, -220, 330, -252, 84, -8, 1, -15, 91, -286, 495, -462, 210, -36, 1
Offset: 1

Views

Author

Wolfdieter Lang, Jan 13 2006

Keywords

Comments

This is a signed version of A011973 (Fibonacci polynomials) with different offset.
The sequence of row lengths is [1,1,2,2,3,3,4,4,5,5,6,6,...] = A008619(n-1), n>=1.
The row sums give the period 6 sequence [1,1,0,-1,-1,0,...] = A010892(n-1), n>=1.
The o.g.f. for the column m sequence (with leading zeros) is ((-1)^m)*x^(2*m+1)/(1-x)^(m+1).
The unsigned row sums give the Fibonacci numbers A000045(n-1), n>=1.
The row polynomial are P(n,x):= Sum_{m=0..ceiling(n/2)-1} a(n,m)*x^m = (sqrt(x)^(n-1))*S(n-1,1/sqrt(x)), n>=1, with Chebyshev's S(n,x) polynomials A049310.
These polynomials appear in the formula 1/c(x)^n = P(n+1,x) - x*P(n,x)*c(x), n>=1, with the o.g.f. c(x):=(1-sqrt(1-4*x))/(2*x) of A000108 (Catalan numbers). See the W. Lang reference, eqs. (1) and (2), p. 408, with P(n,x):=p(-n,x).
These polynomials also appear in the formula c(x)^n = (-P(n-1,x) + P(n,x)*c(x))/x^(n-1), n>=1, with the above given o.g.f. c(x) of A000108 (Catalan numbers). See the W. Lang reference, eq. (1), with P(n,x):=p(-n,x).
With offset n>=0 this array a(n,m) coincides with the row reversed coefficient table of Chebyshev's S-polynomials without interspersed zeros. See A049310 for the S(n,x) coefficient table with increasing powers of x.
The polynomials with this sequence as coefficients form the set of so-called "Catalan polynomials", having arisen from computations in looking at the problem of 'fitting' iterated generating function schemes to the Catalan sequence. A neighboring pair forms the basis of a first-order linear recurrence that generates, through a succession of iterated generating functions (polynomials in Z[x]), a predetermined number of Catalan numbers before 'failing' - see the Clapperton et al. 2008 reference in Utilitas Mathematica, where some of the essential mathematical properties of the Catalan polynomials are also listed (based mainly on existing results for Dickson and Chebyshev polynomials, to which they are related). - Peter J Larcombe, Sep 16 2008
In the Clapperton et al. 2008 Congressus Numerantium paper, a new class of nonlinear identities satisfied by Catalan polynomials are presented. They arise from the algebraic implementation of particular cases of a general root finding formulation due to Householder, of which the classic O(2) Newton-Raphson and O(3) Halley algorithms are special cases. The role of Catalan polynomials in forming Padé approximants to the Catalan sequence o.g.f. is also discussed. - Peter J Larcombe, Nov 02 2008
These polynomials appear in the following statements: (i) P(k+1,x)/P(k+2,x) is the g.f. of all ordered trees (Dyck paths) of height at most k; (ii) x^k/(P(k+1,x)*P(k+2,x)) is the g.f. of all ordered trees (Dyck paths) of height k. See the de Bruijn et al., the Kreweras, the Sedgewick and Flajolet (p. 258), and the Flajolet and Sedgewick (p. 326) references. - Emeric Deutsch, Jun 16 2011
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. (Cf. A011973, A169803, A115139, A092865, A098925, and A102426.) - Tom Copeland, Nov 04 2014
M. Sinan Kul, Dec 09 2015, observed that (in a rewritten form) Chebyshev's S polynomials A049310 are given by S(n, x) = Sum_{m=0..floor(n/2)} a(n+1, m)*x^(n-2*m), n >= 0. This formula is well known and can be proved from the S recurrence by induction using the recurrence for the binomial coefficients. - Wolfdieter Lang, Feb 01 2016
These are the coefficients of generalized Fibonacci polynomials (see link bellow). - Rigoberto Florez, Aug 28 2022

Examples

			The irregular triangle a(n, m) begins:
  n\m  0   1   2    3   4    5   6   7  8
  1:   1
  2:   1
  3:   1  -1
  4:   1  -2
  5:   1  -3   1
  6:   1  -4   3
  7:   1  -5   6   -1
  8:   1  -6  10   -4
  9:   1  -7  15  -10   1
  10:  1  -8  21  -20   5
  11:  1  -9  28  -35  15   -1
  12:  1 -10  36  -56  35   -6
  13:  1 -11  45  -84  70  -21   1
  14:  1 -12  55 -120 126  -56   7
  15:  1 -13  66 -165 210 -126  28  -1
  16:  1 -14  78 -220 330 -252  84  -8
  17:  1 -15  91 -286 495 -462 210 -36  1
  ... Reformatted and extended. - _Wolfdieter Lang_, Jan 27 2016
1/c(x) = P(2,x) - x*P(1,x)*c(x) = 1 - x*c(x), with the o.g.f. of A000108 (Catalan).
1/c(x)^2 = P(3,x) - x*P(2,x)*c(x) = (1-x) - x*c(x).
c(x)^2 = (-P(1,x) + P(2,x)*c(x))/x^1 = (-1 + 1*c(x))/x.
c(x)^3 = (-P(2,x) + P(3,x)*c(x))/x^2 = (-1 + (1-x)*c(x))/x^2.
P(3,x) = 1-x = x*S(2,1/sqrt(x)) with Chebyshev's S(2,y) = U(2,y/2) = y^2 - 1.
		

References

  • J. A. Clapperton, P. J. Larcombe and E. J. Fennessey, On iterated generating functions for integer sequences and Catalan polynomials, Utilitas Mathematica, 77 (2008), 3-33.
  • J. A. Clapperton, P. J. Larcombe, E. J. Fennessey and P. Levrie, A class of auto-identities for Catalan polynomials and Padé approximation, Congressus Numerantium, 189 (2008), 77-95.
  • N. G. de Bruijn, D. E. Knuth and S. O. Rice, The average height of planted plane trees, in: Graph Theory and Computing (ed. T. C. Read), Academic Press, New York, 1972, pp. 15-22.
  • R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.

Crossrefs

Programs

  • Maple
    seq(seq((-1)^k*binomial(n-k,k),k=0..floor(n/2)),n=0..16); # Peter Luschny, May 10 2016
  • Mathematica
    p[x_, n_] := p[x, n] = p[x, n - 1] + x*p[x, n - 2];
    p[x_, -1] = p[x_, 0] = 1; p[x_, 1] = 1 + x;
    Flatten[ Table[ CoefficientList[p[-x, n - 1], x], {n, 0, 16}]]
    (* Jean-François Alcover, Jun 20 2011 *)
    Flatten[Map[CoefficientList[#,x]&, Table[Sum[Binomial[t - i, i] x^(i) (-1)^i, {i, 0, t}], {t, 1,15}]]] (* Rigoberto Florez, Aug 28 2022 *)
  • Python
    import math
    L1 = [math.comb(t - i, i)*(-1)**i for t in range(16) for i in range(t)]
    L1 = list(filter((0)._ne_, L1))
    print(L1) # Rigoberto Florez, Sep 03 2022

Formula

a(n, m) = ((-1)^(m))*binomial(n-1-m, m), n>=1, m=0..ceiling(n/2)-1.
a(n, m) = [x^m]P(n,x), n>=1, m=0..ceiling(n/2)-1, with P(n,x) given above in terms of Chebyshev's S-polynomials.
P(n,x) = (u^(2*n) - v^(2*n))/(u^2 - v^2), where u and v are defined by u^2 + v^2 =1 and u*v = sqrt(x). Example: P(3,x) = (u^6 - v^6)/(u^2 - v^2) = u^4 + u^2*v^2 + v^4 = 1 - x. - Emeric Deutsch, Jun 16 2011
G.f.: 1/(1- x + y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1- x*y)*x/((2*k+2- x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
T(n, k) = GegenbauerC(k, (n+1)/2-k, -1) assuming the triangle (0,0) based. - Peter Luschny, May 10 2016

A055151 Triangular array of Motzkin polynomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 6, 2, 1, 10, 10, 1, 15, 30, 5, 1, 21, 70, 35, 1, 28, 140, 140, 14, 1, 36, 252, 420, 126, 1, 45, 420, 1050, 630, 42, 1, 55, 660, 2310, 2310, 462, 1, 66, 990, 4620, 6930, 2772, 132, 1, 78, 1430, 8580, 18018, 12012, 1716, 1, 91, 2002, 15015, 42042
Offset: 0

Views

Author

Michael Somos, Jun 14 2000

Keywords

Comments

T(n,k) = number of Motzkin paths of length n with k up steps. T(n,k)=number of 0-1-2 trees with n edges and k+1 leaves, n>0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) E.g., T(4,1)=6 because we have UDHH, UHDH, UHHD, HHUD, HUHD, HUDH, where U=(1,1), D(1,-1), H(1,0). - Emeric Deutsch, Nov 30 2003
Coefficients in series reversion of x/(1+H*x+U*D*x^2) corresponding to Motzkin paths with H colors for H(1,0), U colors for U(1,1) and D colors for D(1,-1). - Paul Barry, May 16 2005
Eigenvector equals A119020, so that A119020(n) = Sum_{k=0..[n/2]} T(n,k)*A119020(k). - Paul D. Hanna, May 09 2006
Row reverse of A107131. - Peter Bala, May 07 2012
Also equals the number of 231-avoiding permutations of n+1 for which descents(w) = peaks(w) = k, where descents(w) is the number of positions i such that w[i]>w[i+1], and peaks(w) is the number of positions i such that w[i-1]w[i+1]. For example, T(4,1) = 6 because 13245, 12435, 14235, 12354, 12534, 15234 are the only 231-avoiding permutations of 5 elements with descents(w) = peaks(w) = 1. - Kyle Petersen, Aug 02 2013
Apparently, a refined irregular triangle related to this triangle (and A097610) is given in the Alexeev et al. link on p. 12. This entry's triangle is also related through Barry's comment to A125181 and A134264. The diagonals of this entry are the rows of A088617. - Tom Copeland, Jun 17 2015
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 24 2015

Examples

			The irregular triangle T(n,k) begins:
n\k 0  1   2    3   4  5 ...
0:  1
1:  1
2:  1  1
3:  1  3
4:  1  6   2
5:  1 10  10
6:  1 15  30    5
7:  1 21  70   35
8:  1 28 140  140  14
9:  1 36 252  420 126
10: 1 45 420 1050 630 42
... reformatted. - _Wolfdieter Lang_, Aug 24 2015
		

References

  • Miklos Bona, Handbook of Enumerative Combinatorics, CRC Press (2015), page 617, Corollary 10.8.2
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 4.3.

Crossrefs

A107131 (row reversed), A080159 (with trailing zeros), A001006 = row sums, A000108(n) = T(2n, n), A001700(n) = T(2n+1, n), A119020 (eigenvector), A001263 (Narayana numbers), A089627 (gamma vectors of type B associahedra), A101280 (gamma vectors of type A permutohedra).
Cf. A014531.

Programs

  • Maple
    b:= proc(x, y) option remember;
          `if`(y>x or y<0, 0, `if`(x=0, 1, expand(
           b(x-1, y) +b(x-1, y+1) +b(x-1, y-1)*t)))
        end:
    T:= n-> (p-> seq(coeff(p, t, i), i=0..degree(p)))(b(n, 0)):
    seq(T(n), n=0..20);  # Alois P. Heinz, Feb 05 2014
  • Mathematica
    m=(1-x-(1-2x+x^2-4x^2y)^(1/2))/(2x^2 y); Map[Select[#,#>0&]&, CoefficientList[ Series[m,{x,0,15}],{x,y}]]//Grid (* Geoffrey Critzer, Feb 05 2014 *)
    p[n_] := Hypergeometric2F1[(1-n)/2, -n/2, 2, 4 x]; Table[CoefficientList[p[n], x], {n, 0, 13}] // Flatten (* Peter Luschny, Jan 23 2018 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, n! / ((n-2*k)! * k! * (k+1)!))}
    
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, polcoeff( polcoeff( 2 / (1 - x + sqrt((1 - x)^2 - 4*y*x^2 + x * O(x^n))), n), k))} /* Michael Somos, Feb 14 2006 */
    
  • PARI
    {T(n, k) = n++; if( k<0 || 2*k>n, 0, polcoeff( polcoeff( serreverse( x / (1 + x + y*x^2) + x * O(x^n)), n), k))} /* Michael Somos, Feb 14 2006 */

Formula

T(n,k) = n!/((n-2k)! k! (k+1)!) = A007318(n, 2k)*A000108(k). - Henry Bottomley, Jan 31 2003
E.g.f. row polynomials R(n,y): exp(x)*BesselI(1, 2*x*sqrt(y))/(x*sqrt(y)). - Vladeta Jovovic, Aug 20 2003
G.f. row polynomials R(n,y): 2 / (1 - x + sqrt((1 - x)^2 - 4 *y * x^2)).
From Peter Bala, Oct 28 2008: (Start)
The rows of this triangle are the gamma vectors of the n-dimensional (type A) associahedra (Postnikov et al., p. 38). Cf. A089627 and A101280.
The row polynomials R(n,x) = Sum_{k = 0..n} T(n,k)*x^k begin R(0,x) = 1, R(1,x) = 1, R(2,x) = 1 + x, R(3,x) = 1 + 3*x. They are related to the Narayana polynomials N(n,x) := Sum_{k = 1..n} (1/n)*C(n,k)*C(n,k-1)*x^k through x*(1 + x)^n*R(n, x/(1 + x)^2) = N(n+1,x). For example, for n = 3, x*(1 + x)^3*(1 + 3*x/(1 + x)^2) = x + 6*x^2 + 6*x^3 + x^4, the 4th Narayana polynomial.
Recursion relation: (n + 2)*R(n,x) = (2*n + 1)*R(n-1,x) - (n - 1)*(1 - 4*x)*R(n-2,x), R(0,x) = 1, R(1,x) = 1. (End)
G.f.: M(x,y) satisfies: M(x,y)= 1 + x M(x,y) + y*x^2*M(x,y)^2. - Geoffrey Critzer, Feb 05 2014
T(n,k) = A161642(n,k)*A258820(n,k) = (binomial(n,k)/A003989(n+1, k+1))* A258820(n,k). - Tom Copeland, Jun 18 2015
Let T(n,k;q) = n!*(1+k)/((n-2*k)!*(1+k)!^2)*hypergeom([k,2*k-n],[k+2],q) then T(n,k;0) = A055151(n,k), T(n,k;1) = A008315(n,k) and T(n,k;-1) = A091156(n,k). - Peter Luschny, Oct 16 2015
From Tom Copeland, Jan 21 2016: (Start)
Reversed rows of A107131 are rows of this entry, and the diagonals of A107131 are the columns of this entry. The diagonals of this entry are the rows of A088617. The antidiagonals (bottom to top) of A088617 are the rows of this entry.
O.g.f.: [1-x-sqrt[(1-x)^2-4tx^2]]/(2tx^2), from the relation to A107131.
Re-indexed and signed, this triangle gives the row polynomials of the compositional inverse of the shifted o.g.f. for the Fibonacci polynomials of A011973, x / [1-x-tx^2] = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... . (End)
Row polynomials are P(n,x) = (1 + b.y)^n = Sum{k=0 to n} binomial(n,k) b(k) y^k = y^n M(n,1/y), where b(k) = A126120(k), y = sqrt(x), and M(n,y) are the Motzkin polynomials of A097610. - Tom Copeland, Jan 29 2016
Coefficients of the polynomials p(n,x) = hypergeom([(1-n)/2, -n/2], [2], 4x). - Peter Luschny, Jan 23 2018
Sum_{k=1..floor(n/2)} k * T(n,k) = A014531(n-1) for n>1. - Alois P. Heinz, Mar 29 2020

A102426 Triangle read by rows giving coefficients of polynomials defined by F(0,x)=0, F(1,x)=1, F(n,x) = F(n-1,x) + x*F(n-2,x).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 4, 1, 1, 6, 5, 1, 4, 10, 6, 1, 1, 10, 15, 7, 1, 5, 20, 21, 8, 1, 1, 15, 35, 28, 9, 1, 6, 35, 56, 36, 10, 1, 1, 21, 70, 84, 45, 11, 1, 7, 56, 126, 120, 55, 12, 1, 1, 28, 126, 210, 165, 66, 13, 1, 8, 84, 252, 330, 220, 78, 14, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1
Offset: 0

Views

Author

Russell Walsmith, Jan 08 2005

Keywords

Comments

Essentially the same as A098925: a(0)=0 followed by A098925. - R. J. Mathar, Aug 30 2008
F(n) + 2x * F(n-1) gives Lucas polynomials (cf. A034807). - Maxim Krikun (krikun(AT)iecn.u-nancy.fr), Jun 24 2007
After the initial 0, these are the nonzero coefficients of the Fibonacci polynomials; see the Mathematica section. - Clark Kimberling, Oct 10 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
Aside from the initial zeros, these are the antidiagonals read from bottom to top of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper, which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse of A011973. - Tom Copeland, Jul 02 2018

Examples

			The first few polynomials are:
  0
  1
  1
  x + 1
  2*x + 1
  x^2 + 3*x + 1
  3*x^2 + 4*x + 1
------------------
From _Tom Copeland_, Jan 19 2016: (Start)
[n]:
0:  0
1:  1
2:  1
3:  1  1
4:  2  1
5:  1  3  1
6:  3  4  1
7:  1  6  5   1
8:  4 10  6   1
9:  1 10 15   7   1
10: 5 20 21   8   1
11: 1 15 35  28   9  1
12: 6 35 56  36  10  1
13: 1 21 70  84  45 11 1
(End)
		

References

  • Dominique Foata and Guo-Niu Han, Multivariable tangent and secant q-derivative polynomials, Manuscript, Mar 21 2012.

Crossrefs

Upward diagonals sums are A062200. Downward rows are A102427. Row sums are A000045. Row terms reversed = A011973. Also A102428, A102429.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.

Programs

  • Magma
    [0] cat [Binomial(Floor(n/2)+k, Floor((n-1)/2-k) ): k in [0..Floor((n-1)/2)], n in [0..17]]; // G. C. Greubel, Oct 13 2019
    
  • Mathematica
    Join[{0}, Table[ Select[ CoefficientList[ Fibonacci[n, x], x], 0 < # &], {n, 0, 17}]//Flatten] (* Clark Kimberling, Oct 10 2013 and slightly modified by Robert G. Wilson v, May 03 2017 *)
  • PARI
    F(n) = if (n==0, 0, if (n==1, 1, F(n-1) + x*F(n-2)));
    tabf(nn) = for (n=0, nn, print(Vec(F(n)))); \\ Michel Marcus, Feb 10 2020

Formula

Alternatively, as n is even or odd: T(n-2, k) + T(n-1, k-1) = T(n, k), T(n-2, k) + T(n-1, k) = T(n, k)
T(n, k) = binomial(floor(n/2)+k, floor((n-1)/2-k) ). - Paul Barry, Jun 22 2005
Beginning with the second polynomial in the example and offset=0, P(n,t)= Sum_{j=0..n}, binomial(n-j,j)*x^j with the convention that 1/k! is zero for k=-1,-2,..., i.e., 1/k! = lim_{c->0} 1/(k+c)!. - Tom Copeland, Oct 11 2014
From Tom Copeland, Jan 19 2016: (Start)
O.g.f.: (x + x^2 - x^3) / (1 - (2+t)*x^2 + x^4) = (x^2 (even part) + x*(1-x^2) (odd)) / (1 - (2+t)*x^2 + x^4).
Recursion relations:
A) p(n,t) = p(n-1,t) + p(n-2,t) for n=2,4,6,8,...
B) p(n,t) = t*p(n-1,t) + p(n-2,t) for n=3,5,7,...
C) a(n,k) = a(n-2,k) + a(n-1,k) for n=4,6,8,...
D) a(n,k) = a(n-2,k) + a(n-1,k-1) for n=3,5,7,...
Relation A generalized to MV(n,t;r) = P(2n+1,t) + r R(2n,t) for n=1,2,3,... (cf. A078812 and A085478) is the generating relation on p. 229 of Andre-Jeannine for the generalized Morgan-Voyce polynomials, e.g., MV(2,t;r) = p(5,t) + r*p(4,t) = (1 + 3t + t^2) + r*(2 + t) = (1 + 2r) + (3 + r)*t + t^2, so P(n,t) = MV(n-4,t;1) for n=4,6,8,... .
The even and odd polynomials are also presented in Trzaska and Ferri.
Dropping the initial 0 and re-indexing with initial m=0 gives the row polynomials Fb(m,t) = p(n+1,t) below with o.g.f. G(t,x)/x, starting with Fb(0,t) = 1, Fb(1,t) = 1, Fb(2,t) = 1 + t, and Fb(3,t) = 2 + t.
The o.g.f. x/G(x,t) = (1 - (2+t)*x^2 + x^4) / (1 + x - x^2) then generates a sequence of polynomials IFb(t) such that the convolution Sum_{k=0..n} IFb(n-k,t) Fb(k,t) vanishes for n>1 and is one for n=0. These linear polynomials have the basic Fibonacci numbers A000045 as an overall factor:
IFb(0,t) = 1
IFb(1,t) = -1
IFb(2,t) = -t
IFb(3,t) = -1 (1-t)
IFb(4,t) = 2 (1-t)
IFb(5,t) = -3 (1-t)
IFb(6,t) = 5 (1-t)
IFb(7,t) = -8 (1-t)
IFb(8,t) = 13 (1-t)
... .
(End)

Extensions

Name corrected by John K. Sikora, Feb 10 2020

A128099 Triangle read by rows: T(n,k) is the number of ways to tile a 3 X n rectangle with k pieces of 2 X 2 tiles and 3n-4k pieces of 1 X 1 tiles (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 1, 2, 1, 4, 1, 6, 4, 1, 8, 12, 1, 10, 24, 8, 1, 12, 40, 32, 1, 14, 60, 80, 16, 1, 16, 84, 160, 80, 1, 18, 112, 280, 240, 32, 1, 20, 144, 448, 560, 192, 1, 22, 180, 672, 1120, 672, 64, 1, 24, 220, 960, 2016, 1792, 448, 1, 26, 264, 1320, 3360, 4032, 1792, 128, 1, 28
Offset: 0

Views

Author

Emeric Deutsch, Feb 18 2007

Keywords

Comments

Row sums are the Jacobsthal numbers (A001045).
Apparently, T(n,k)/2^n equals the probability P that n will occur as a partial sum in a randomly-generated infinite sequence of 1s and 2s with n compositions (ordered partitions) into (n-2k) 1s and k 2s. Example: T(6,2)=24; P = 3/8 (24/2^6) that 6 will occur as a partial sum in the sequence with 2 (6-2*2) 1s and 2 2s. - Bob Selcoe, Jul 06 2013
From Johannes W. Meijer, Aug 28 2013: (Start)
The antidiagonal sums are A077949 and the backwards antidiagonal sums are A052947.
Moving the terms in each column of this triangle, see the example, upwards to row 0 gives the Pell-Jacobsthal triangle A013609 as a square array. (End)
The numbers in rows of the triangle are along "first layer" skew diagonals pointing top-right in center-justified triangle given in A013609 ((1+2*x)^n) and along (first layer) skew diagonals pointing top-left in center-justified triangle given in A038207 ((2+x)^n), see links. - Zagros Lalo, Jul 31 2018
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.000..., when n approaches infinity. - Zagros Lalo, Jul 31 2018
It appears that the rows of this array are the coefficients of the Jacobsthal polynomials (see MathWorld link). - Michel Marcus, Jun 15 2019

Examples

			Triangle starts:
  1;
  1;
  1,  2;
  1,  4;
  1,  6,  4;
  1,  8, 12;
  1, 10, 24,  8;
  1, 12, 40, 32;
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 80-83, 357-358

Crossrefs

Cf. (Triangle sums) A001045, A095977, A077949, A052947, A113726, A052942, A077909.
Cf. (Similar triangles) A008315, A011973, A102541.

Programs

  • Maple
    T := proc(n,k) if k<=n/2 then 2^k*binomial(n-k,k) else 0 fi end: for n from 0 to 16 do seq(T(n,k),k=0..floor(n/2)) od; # yields sequence in triangular form
    T := proc(n, k) option remember: if k<0 or k > floor(n/2) then return(0) fi: if k = 0 then return(1) fi: 2*procname(n-2, k-1) + procname(n-1, k): end: seq(seq(T(n, k), k=0..floor(n/2)), n=0..13); # Johannes W. Meijer, Aug 28 2013
  • Mathematica
    Table[2^k*Binomial[n - k, k] , {n,0,25}, {k,0,Floor[n/2]}] // Flatten  (* G. C. Greubel, Dec 28 2016 *)
    t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, t[n - 1, k] + 2 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}] // Flatten (* Zagros Lalo, Jul 31 2018 *)

Formula

T(n, k) = 2^k*binomial(n-k,k) = 2^k*A011973(n,k).
G.f.: 1/(1-z-2*t*z^2).
Sum_{k=0..floor(n/2)} k*T(n,k) = A095977(n-1).
From Johannes W. Meijer, Aug 28 2013: (Start)
T(n, k) = 2*T(n-2, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, k) = 0 for k < 0 and k > floor(n/2).
T(n, k) = A013609(n-k, k), n >= 0 and 0 <= k <= floor(n/2). (End)

A119262 Number of B-trees of order infinity with n leaves, where a(n) = Sum_{k=1..floor(n/2)} a(k)*C(n-k-1,n-2*k) for n >= 2, with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 5, 8, 14, 25, 46, 85, 158, 294, 548, 1022, 1908, 3567, 6683, 12556, 23669, 44781, 85046, 162122, 310157, 595322, 1146057, 2212004, 4278908, 8292738, 16097018, 31286456, 60873574, 118543329, 231009934, 450434739, 878687665
Offset: 0

Views

Author

Paul D. Hanna, May 11 2006

Keywords

Comments

A B-tree of order m is an ordered tree such that every node has at most m children, the root has at least 2 children, every node except the root has 0 or at least m/2 children, all end-nodes are at the same level. This sequence is the limit of the B-trees as m --> infinity.
Starting with offset 2, the eigensequence of triangle A011973. - Gary W. Adamson, Jul 08 2012
Number of balanced series-reduced rooted plane trees with n leaves. A rooted tree is series-reduced if every non-leaf node has at least two branches, and balanced if all leaves are the same distance from the root. - Gus Wiseman, Oct 07 2018

Examples

			A(x) = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 5*x^6 + 8*x^7 + 14*x^8 + ...
Series form:
A(x) = x + x^2/(1-x) + x^4/((1-x)*((1-x)-x^2)) + x^8/((1-x)*((1-x)-x^2)*((1-x)*((1-x)-x^2)-x^4)) + ... + x^(2^n)/D(n,x) + x^(2^(n+1))/[D(n,x)*(D(n,x)-x^(2^n))] + ...
Terms also satisfy the series:
x = x*(1-x) + x^2*(1-x^2)/(1+x) + x^3*(1-x^3)/(1+x)^2 + 2*x^4*(1-x^4)/(1+x)^3 + 3*x^5*(1-x^5)/(1+x)^4 + 5*x^6*(1-x^6)/(1+x)^5 + 8*x^7*(1-x^7)/(1+x)^6 + 14*x^8*(1-x^8)/(1+x)^7 + 25*x^9*(1-x^9)/(1+x)^8 + ... + a(n)*x^n*(1-x^n)/(1+x)^(n-1) + ...
From _Gus Wiseman_, Oct 07 2018: (Start)
The a(1) = 1 through a(7) = 8 balanced series-reduced rooted plane trees:
  o  (oo)  (ooo)  (oooo)      (ooooo)      (oooooo)        (ooooooo)
                  ((oo)(oo))  ((oo)(ooo))  ((oo)(oooo))    ((oo)(ooooo))
                              ((ooo)(oo))  ((ooo)(ooo))    ((ooo)(oooo))
                                           ((oooo)(oo))    ((oooo)(ooo))
                                           ((oo)(oo)(oo))  ((ooooo)(oo))
                                                           ((oo)(oo)(ooo))
                                                           ((oo)(ooo)(oo))
                                                           ((ooo)(oo)(oo))
(End)
		

Crossrefs

Cf. A092684 (similar recurrence); B-trees: A014535 (order 3), A037026 (order 4), A058521 (order 5).
Cf. A011973.

Programs

  • Mathematica
    nn=38;f[x_]:=Sum[a[n]x^n,{n,0,nn}];a[0]=0;sol=SolveAlways[0==Series[f[x]-x-f[x^2/(1-x)],{x,0,nn}],x];Table[a[n],{n,0,nn}]/.sol  (* Geoffrey Critzer, Mar 28 2013 *)
  • PARI
    a(n)=if(n==0,0,if(n==1,1,sum(k=1,n\2,a(k)*binomial(n-k-1,n-2*k))))
    for(n=1, 20, print1(a(n), ", "))
    
  • PARI
    /* From: A(x) = x + A(x^2/(1-x)) */
    {a(n)=local(A=x);for(i=1,n,A=x+subst(A,x,x^2/(1-x+x*O(x^n))));polcoeff(A,n)}
    for(n=1, 20, print1(a(n), ", "))
    
  • PARI
    /* From: x = Sum_{n>=1} a(n)*x^n*(1-x^n)/(1+x)^(n-1) */
    a(n)=if(n==1, 1, -polcoeff(sum(k=1, n-1, a(k)*x^k*(1-x^k)/(1+x+x*O(x^n))^(k-1)), n))
    for(n=1, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Jul 31 2013

Formula

G.f. A(x) satisfies: A(x) = x + A(x^2/(1-x)).
G.f.: Sum_{n>=0} x^(2^n)/D(n,x) where D(0,x)=1, D(n+1,x) = D(n,x)*[D(n,x) - x^(2^n)].
G.f.: x = Sum_{n>=1} a(n) * x^n * (1-x^n) / (1+x)^(n-1). - Paul D. Hanna, Jul 31 2013
Conjecture: Let M_n be an n X n matrix whose elements are m_ij = 0 for i < j - 1, m_ij = -1 for i = j - 1, and m_ij = binomial(i - j, n - i) otherwise. Then a(n + 1) = Det(M_n). - Benedict W. J. Irwin, Apr 19 2017

A269254 To find a(n), define a sequence by s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n + 1; then a(n) is the smallest index k such that s(k) is prime, or -1 if no such k exists.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, -1, 2, 2, 1, 2, 1, 2, -1, 2, 1, 3, 1, 2, 2, 2, 1, -1, 2, 6, 2, 3, 1, 3, 1, 2, 9, 9, -1, 2, 1, 6, 2, 2, 1, 2, 1, 5, 2, 2, 1, -1, 2, 5, 2, 9, 1, 2, 2, 2, 2, 6, 1, 2, 1, 14, -1, 5, 2, 2, 1, 5, 2, 3, 1, 6, 1, 8, 3, 6, 2, 3, 1, -1, 3, 18, 1, 2, 3, 2, 2, 3, 1, 2, 9, 3, 5, 2, 2, 96, 1, 3, -1, 5, 1, 2, 1, 2, 15, 14, 1, 44, 1, 3, -1
Offset: 1

Views

Author

Arkadiusz Wesolowski, Jul 09 2016

Keywords

Comments

The s(k) sequences can be viewed in A294099, where they appear as rows. - Peter Munn, Aug 31 2020
For n >= 3, a(n) is that positive integer k yielding the smallest prime of the form (x^y - 1/x^y)/(x - 1/x), where x = (sqrt(n+2) +- sqrt(n-2))/2 and y = 2*k + 1, or -1 if no such k exists.
Every positive term belongs to A005097.
When n=7, the sequence {s(k)} is A033890, which is Fibonacci(4i+2), and since x|y <=> F_x|F_y, and 2i+1|4i+2, A033890 is never prime, and so a(7)=-1. For the other -1 terms below 100, see the theorem below and the Klee link - N. J. A. Sloane, Oct 20 2017 and Oct 22 2017
Theorem (Brad Klee): For all n > 2, a(n^2 - 2) = -1. See Klee link for a proof. - L. Edson Jeffery, Oct 22 2017
Theorem (Based on work of Hans Havermann, L. Edson Jeffery, Brad Klee, Don Reble, Bob Selcoe, and N. J. A. Sloane) a(110) = -1. [For proof see link. - N. J. A. Sloane, Oct 23 2017]
From Bob Selcoe, Oct 24 2017, edited by N. J. A. Sloane, Oct 27 2017: (Start)
Suppose n = m^2 - 2, where m >= 3, and let j = m-2, with j >= 1.
For this value of n, the sequence s(k) satisfies s(k) = (c(k) + d(k))*(c(k) - d(k)), where c(0) = 1, d(0) = 0; and for k >= 1: c(k) = (j+2)*c(k-1) - d(k-1), and d(k) = c(k-1). So (as Brad Klee already proved) a(n) = -1 .
We have s(0) = 1 and s(1) = n+1 = j^2 + 4j + 3. In general, the coefficients of s(k) when expanded in powers of j are given by the (4k+2)-th row of A011973 (the triangle of coefficients of Fibonacci polynomials) in reverse order. For example, s(2) = j^4 + 8j^3 + 21j^2 + 20j + 5, s(3) = j^6 + 12j^5 + 55j^4 + 120j^3 + 126j^2 + 56j + 7, etc.
Perhaps the above comments could be generalized to apply to a(110) or to other n for which a(n) = -1?
(End)
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Examples

			Let b(k) be the recursive sequence defined by the initial conditions b(0) = 1, b(1) = 16, and the recursive equation b(k) = 15*b(k-1) - b(k-2). a(15) = 2 because b(2) = 239 is the smallest prime in b(k).
Let c(k) be the recursive sequence defined by the initial conditions c(0) = 1, c(1) = 18, and the recursive equation c(k) = 17*c(k-1) - c(k-2). a(17) = 3 because c(3) = 5167 is the smallest prime in c(k).
		

Crossrefs

Programs

  • Magma
    lst:=[]; for n in [1..85] do if n gt 2 and IsSquare(n+2) then Append(~lst, -1); else a:=n+1; c:=1; t:=1; if IsPrime(a) then Append(~lst, t); else repeat b:=n*a-c; c:=a; a:=b; t+:=1; until IsPrime(a); Append(~lst, t); end if; end if; end for; lst;
    
  • Mathematica
    kmax = 100;
    a[1] = a[2] = 1;
    a[n_ /; IntegerQ[Sqrt[n+2]]] = -1;
    a[n_] := Module[{s}, s[0] = 1; s[1] = n+1; s[k_] := s[k] = n s[k-1] - s[k-2]; For[k=1, k <= kmax, k++, If[PrimeQ[s[k]], Return[k]]]; Print["For n = ", n, ", k = ", k, " exceeds the limit kmax = ", kmax]; -1];
    Array[a, 110] (* Jean-François Alcover, Aug 05 2018 *)
  • PARI
    allocatemem(2^30);
    default(primelimit,(2^31)+(2^30));
    s(n,k) = if(0==k,1,if(1==k,(1+n),((n*s(n,k-1)) - s(n,k-2))));
    A269254(n) = { my(k=1); if((n>2)&&issquare(2+n),-1,while(!isprime(s(n,k)),k++);(k)); }; \\ Antti Karttunen, Oct 20 2017

Formula

If n is prime then a(n-1) = 1.

Extensions

a(86)-a(94) from Antti Karttunen, Oct 20 2017
a(95)-a(109) appended by L. Edson Jeffery, Oct 22 2017
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