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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A162440 The pg(n) sequence that is associated with the Eta triangle A160464.

Original entry on oeis.org

2, 16, 144, 4608, 115200, 4147200, 203212800, 26011238400, 2106910310400, 210691031040000, 25493614755840000, 3671080524840960000, 620412608698122240000, 121600871304831959040000
Offset: 2

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Author

Johannes W. Meijer, Jul 06 2009

Keywords

Comments

The EG1 matrix coefficients are defined by EG1[2m-1,1] = 2*eta(2m-1) and the recurrence relation EG1[2m-1,n] = EG1[2m-1,n-1] - EG1[2m-3,n-1]/(n-1)^2 with m = .., -2, -1, 0, 1, 2, ... and n = 1, 2, 3, ... . As usual, eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. For the EG2 matrix, the even counterpart of the EG1 matrix, see A008955.
The coefficients in the columns of the EG1 matrix, for m >= 1 and n >= 2, can be generated with GFE(z;n) = ((-1)^(n-1)*r(n)*CFN1(z,n)*GFE(z;n=1) + ETA(z,n))/pg(n) for n >= 2.
The CFN1(z,n) polynomials depend on the central factorial numbers A008955 and the ETA(z,n) are the Eta polynomials which led to the Eta triangle, see for both A160464.
The pg(n) sequence can be generated with the first Maple program and the EG1[2m-1,n] matrix coefficients can be generated with the second Maple program.
The EG1 matrix is related to the ES1 matrix, see A160464 and the formulas below.

Examples

			The first few generating functions GFE(z;n) are:
GFE(z;n=2) = ((-1)*2*(z^2 - 1)*GFE(z;n=1) + (-1))/2,
GFE(z;n=3) = ((+1)*4*(z^4 - 5*z^2 + 4)*GFE(z;n=1) + (-11 + 2*z^2))/16,
GFE(z;n=4) = ((-1)*4*(z^6-14*z^4+49*z^2-36)*GFE(z;n=1) + (-114+29*z^2-2*z^4))/144.
		

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Crossrefs

The ETA(z, n) polynomials and the ES1 matrix lead to the Eta triangle A160464.
The CFN1(z, n), the t1(n, m) and the EG2 matrix lead to A008955.
The EG1[ -1, n] equal (1/2)*A001803(n-1)/A046161(n-1).
The r(n) sequence equals A062383(n) (n>=1).
The e(n) sequence equals A029837(n) (n>=1).
Cf. A160473 (p(n) sequence).
Cf. A162443 (BG1 matrix), A162446 (ZG1 matrix) and A162448 (LG1 matrix).

Programs

  • Maple
    nmax := 16; seq((n-1)!^2*2^floor(ln(n-1)/ln(2)+1), n=2..nmax);
    # End program 1
    nmax1 := 5; coln := 4; mmax1 := nmax1: for n from 0 to nmax1 do t1(n, 0) := 1 end do: for n from 0 to nmax1 do t1(n, n) := (n!)^2 end do: for n from 1 to nmax1 do for m from 1 to n-1 do t1(n, m) := t1(n-1, m-1)*n^2 + t1(n-1, m) end do: end do: for m from 1 to mmax1 do EG1[1-2*m, 1] := evalf((2^(2*m)-1)* bernoulli(2*m)/(m)) od: EG1[1, 1] := evalf(2*ln(2)): for m from 2 to mmax1 do EG1[2*m-1, 1] := evalf(2*(1-2^(1-(2*m-1))) * Zeta(2*m-1)) od: for m from -mmax1+coln to mmax1 do EG1[2*m-1, coln]:= (-1)^(coln+1)*sum((-1)^k*t1(coln-1, k) * EG1[1-2*coln+2*m+2*k, 1], k=0..coln-1)/(coln-1)!^2 od;
    # End program 2 (Edited by Johannes W. Meijer, Sep 21 2012)

Formula

pg(n) = (n-1)!^2*2^floor(log(n-1)/log(2)+1) for n >= 2.
r(n) = 2^e(n) = 2^floor(log(n-1)/log(2)+1) for n >= 2.
EG1[ -1,n] = 2^(1-2*n)*(2*n-1)!/((n-1)!^2) for n >= 1.
GFE(z;n) = sum (EG1[2*m-1,n]*z^(2*m-2), m=1..infinity).
GFE(z;n) = (1-z^2/(n-1)^2)*GFE(z;n-1)-EG1[ -1,n-1]/(n-1)^2 for n = >2. with GFE(z;n=1) = 2*log(2)-Psi(z)-Psi(-z)+Psi(z/2)+Psi(-z/2) and Psi(z) is the digamma function.
EG1[2m-1,n] = (2*2^(1-2*n)*(2*n-1)!/((n-1)!^2)) * ES1[2m-1,n].

A160466 Row sums of the Eta triangle A160464.

Original entry on oeis.org

-1, -9, -87, -2925, -75870, -2811375, -141027075, -18407924325, -1516052821500, -153801543183750, -18845978136851250, -2744283682352086875, -468435979952504313750, -92643070481933918821875
Offset: 2

Views

Author

Johannes W. Meijer, May 24 2009

Keywords

Comments

It is conjectured that the row sums of the Eta triangle depend on five different sequences.
Two Maple algorithms are given. The first one gives the row sums according to the Eta triangle A160464 and the second one gives the row sums according to our conjecture.

Crossrefs

A160464 is the Eta triangle.
Row sum factors A119951, A000466, A043529, A045896 and A160467.

Programs

  • Maple
    nmax:=15; c(2) := -1/3: for n from 3 to nmax do c(n):=(2*n-2)*c(n-1)/(2*n-1)-1/ ((n-1)*(2*n-1)) end do: for n from 2 to nmax do GCS(n-1) := ln(1/(2^(-(2*(n-1)-1-floor(ln(n-1)/ ln(2))))))/ln(2); p(n):=2^(-GCS(n-1))*(2*n-1)!; ETA(n, 1) := p(n)*c(n) end do: mmax:=nmax: for m from 2 to mmax do ETA(2, m) := 0 end do: for n from 3 to nmax do for m from 2 to mmax do q(n) := (1+(-1)^(n-3)*(floor(ln(n-1)/ln(2)) - floor(ln(n-2)/ln(2)))): ETA(n, m) := q(n)*(-ETA(n-1, m-1)+(n-1)^2*ETA(n-1, m)) end do end do: for n from 2 to nmax do s1(n):=0: for m from 1 to n-1 do s1(n) := s1(n) + ETA(n, m) end do end do: seq(s1(n), n=2..nmax);
    # End first program.
    nmax:=nmax; A160467 := proc(n): denom(4*(4^n-1)*bernoulli(2*n)/n) end: A043529 := proc(n): ceil(frac(log[2](n+1))+1) end proc: A000466 := proc(n): 4*n^2-1 end proc: A045896 := proc(n): denom((n)/((n+1)*(n+2))) end proc: A119951 := proc(n) : numer(sum(((2*k1)!/(k1!*(k1+1)!))/2^(2*(k1-1)), k1=1..n)) end proc: for n from 1 to nmax do SF(2*n+1):= A000466(n)/A043529(n-1); SF(2*n+2) := A045896(n-1)/A160467(n+1) end do: FF(2):=1: for n from 3 to nmax do FF(n) := SF(n) * FF(n-1) end do: for n from 2 to nmax do s2(n):= (-1)*A119951(n-1)*FF(n) end do: seq(s2(n), n=2..nmax);
    # End second program.

Formula

Rowsums(n) = (-1) * A119951(n-1) * FF(n) for n >= 2.
FF(n) = SF(n) * FF(n-1) for n >= 3 with FF(2) =1.
SF(2*n) = A045896(n-2) / A160467(n) for n >= 2.
SF(2*n+1) = A000466(n) / A043529(n-1) for n >= 1.

A160473 The p(n) sequence that is associated with the Eta triangle A160464.

Original entry on oeis.org

3, 30, 315, 11340, 311850, 12162150, 638512875, 86837751000, 7424627710500, 779585909602500, 98617617564716250, 14792642634707437500, 2596108782391155281250, 527010082825404522093750, 122529844256906551386796875, 64695757767646659132228750000
Offset: 2

Views

Author

Johannes W. Meijer, May 24 2009

Keywords

Crossrefs

A160464 is the Eta triangle.
Equals 3*(n-2)!*A000457(n-2)/A054243(n-1)
Equals 2^(-A049039(n-1))*(2*n-1)!
Cf. The pg(n) sequence A162440. - Johannes W. Meijer, Jul 06 2009

Programs

  • Magma
    [2^(-(2*(n-1)-1-Floor(Log(n-1)/Log(2))))*Factorial(2*n-1)  : n in [2..30]]; // Vincenzo Librandi, Jul 06 2015
  • PARI
    vector(20, n, n++; 2^(-(2*(n-1)-1-floor(log(n-1)/log(2))))*(2*n-1)!) \\ Michel Marcus, Jul 06 2015
    

Formula

a(n) = 2^(-(2*(n-1)-1-floor(log(n-1)/log(2))))*(2*n-1)! for n = 2, 3, 4, ... .

A160465 First left hand column of the Eta triangle A160464.

Original entry on oeis.org

-1, -11, -114, -3963, -104745, -3926745, -198491580, -26045435115, -2153099119815, -219022225836750, -26891482281048000, -3921682257253270125, -670160622793156369875, -132649536458654226136125
Offset: 2

Views

Author

Johannes W. Meijer, May 24 2009

Keywords

Crossrefs

A160464 is the Eta triangle.
The GCS(n) sequence equals the geometric Connell sequence A049039(n).
The p(n) sequence is given by A160473.

Programs

  • Maple
    nmax:=15; c(2) := -1/3: for n from 3 to nmax do c(n) := (2*n-2)*c(n-1)/(2*n-1)-1/ ((n-1)*(2*n-1)) end do: for n from 2 to nmax do GCS(n-1) := ln(1/(2^(-(2*(n-1)-1-floor(ln(n-1)/ ln(2))))))/ln(2); p(n) := 2^(-GCS(n-1))*(2*n-1)!; ETA(n, 1) := p(n)*c(n) end do: seq(ETA(n, 1), n=2..nmax);

A053644 Most significant bit of n, msb(n); largest power of 2 less than or equal to n; write n in binary and change all but the first digit to zero.

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64
Offset: 0

Views

Author

Henry Bottomley, Mar 22 2000

Keywords

Comments

Except for the initial term, 2^n appears 2^n times. - Lekraj Beedassy, May 26 2005
a(n) is the smallest k such that row k in triangle A265705 contains n. - Reinhard Zumkeller, Dec 17 2015
a(n) is the sum of totient function over powers of 2 <= n. - Anthony Browne, Jun 17 2016
Given positive n, reverse the bits of n and divide by 2^floor(log_2 n). Numerators are in A030101. Ignoring the initial 0, denominators are in this sequence. - Alonso del Arte, Feb 11 2020

Crossrefs

See A000035 for least significant bit(n).
MASKTRANS transform of A055975 (prepended with 0), MASKTRANSi transform of A048678.
Bisection of A065267, A065279, A065291, A072376.
First differences of A063915. Cf. A076877, A073121.
This is Guy Steele's sequence GS(5, 5) (see A135416).
Equals for n >= 1 the first right hand column of A160464. - Johannes W. Meijer, May 24 2009
Diagonal of A088370. - Alois P. Heinz, Oct 28 2011

Programs

  • Haskell
    a053644 n = if n <= 1 then n else 2 * a053644 (div n 2)
    -- Reinhard Zumkeller, Aug 28 2014
    a053644_list = 0 : concat (iterate (\zs -> map (* 2) (zs ++ zs)) [1])
    -- Reinhard Zumkeller, Dec 08 2012, Oct 21 2011, Oct 17 2010
    
  • Magma
    [0] cat [2^Ilog2(n): n in [1..90]]; // Vincenzo Librandi, Dec 11 2018
    
  • Maple
    a:= n-> 2^ilog2(n):
    seq(a(n), n=0..80);  # Alois P. Heinz, Dec 20 2016
  • Mathematica
    A053644[n_] := 2^(Length[ IntegerDigits[n, 2]] - 1); A053644[0] = 0; Table[A053644[n], {n, 0, 74}] (* Jean-François Alcover, Dec 01 2011 *)
    nv[n_] := Module[{c = 2^n}, Table[c, {c}]]; Join[{0}, Flatten[Array[nv, 7, 0]]] (* Harvey P. Dale, Jul 17 2012 *)
  • PARI
    a(n)=my(k=1);while(k<=n,k<<=1);k>>1 \\ Charles R Greathouse IV, May 27 2011
    
  • PARI
    a(n) = if(!n, 0, 2^exponent(n)) \\ Iain Fox, Dec 10 2018
    
  • Python
    def a(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1) # Indranil Ghosh, May 25 2017
    
  • Python
    def A053644(n): return 1<Chai Wah Wu, Jul 27 2022
  • Scala
    (0 to 127).map(Integer.highestOneBit()) // _Alonso del Arte, Feb 26 2020
    

Formula

a(n) = a(floor(n / 2)) * 2.
a(n) = 2^A000523(n).
From n >= 1 onward, A053644(n) = A062383(n)/2.
a(0) = 0, a(1) = 1 and a(n+1) = a(n)*floor(n/a(n)). - Benoit Cloitre, Aug 17 2002
G.f.: 1/(1 - x) * (x + Sum_{k >= 1} 2^(k - 1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = (A003817(n) + 1)/2 = A091940(n) + 1. - Reinhard Zumkeller, Feb 15 2004
a(n) = Sum_{k = 1..n} (floor(2^k/k) - floor((2^k - 1)/k))*A000010(k). - Anthony Browne, Jun 17 2016
a(2^m+k) = 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 07 2016

A062383 a(0) = 1: for n>0, a(n) = 2^floor(log_2(n)+1) or a(n) = 2*a(floor(n/2)).

Original entry on oeis.org

1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 128, 128, 128, 128, 128, 128
Offset: 0

Views

Author

Antti Karttunen, Jun 19 2001

Keywords

Comments

Informally, write down 1 followed by 2^k 2^(k-1) times, for k = 1,2,3,4,... These are the denominators of the binary van der Corput sequence (see A030101 for the numerators). - N. J. A. Sloane, Dec 01 2019
a(n) is the denominator of the form 2^k needed to make the ratio (2n-1)/2^k lie in the interval [1-2], i.e. such ratios are 1/1, 3/2, 5/4, 7/4, 9/8, 11/8, 13/8, 15/8, 17/16, 19/16, 21/16, ... where the numerators are A005408 (The odd numbers).
Let A_n be the upper triangular matrix in the group GL(n,2) that has zero entries below the diagonal and 1 elsewhere. For example for n=4 the matrix is / 1,1,1,1 / 0,1,1,1 / 0,0,1,1 / 0,0,0,1 /. The order of this matrix as an element of GL(n,2) is a(n-1). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 14 2001
A006257(n)/a(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
a(n) = maximum of row n+1 in A240769. - Reinhard Zumkeller, Apr 13 2014
This is the discriminator sequence for the odious numbers. - N. J. A. Sloane, May 10 2016
From Jianing Song, Jul 05 2025: (Start)
a(n) is the period of {binomial(N,n) mod 2: N in Z}. For the general result, see A349593.
Since the modulus (2) is a prime, the remainder of binomial(N,n) is given by Lucas's theorem. (End)

Crossrefs

Apart from the initial term, equals 2 * A053644. MASKTRANSi(A062383) seems to give a signed form of A038712. (See identities at A053644). floor_log_2 given in A054429.
Equals A003817(n)+1. Cf. A002884.
Bisection of A065285. Cf. A076877.
Equals for n>=1 the r(n) sequence of A160464. - Johannes W. Meijer, May 24 2009
Equals the r(n) sequence of A162440 for n>=1. - Johannes W. Meijer, Jul 06 2009
Discriminator of the odious numbers (A000069). - Jeffrey Shallit, May 08 2016
Column 2 of A349593. A064235 (if offset 0), A385552, A385553, and A385554 are respectively columns 3, 5, 6, and 10.

Programs

  • Haskell
    import Data.List (transpose)
    a062383 n = a062383_list !! n
    a062383_list = 1 : zs where
       zs = 2 : (map (* 2) $ concat $ transpose [zs, zs])
    -- Reinhard Zumkeller, Aug 27 2014, Mar 13 2014
    
  • Magma
    [2^Floor(Log(2,2*n+1)): n in [0..70]]; // Bruno Berselli, Mar 04 2016
    
  • Maple
    [seq(2^(floor_log_2(j)+1),j=0..127)]; or [seq(coerce1st_octave((2*j)+1),j=0..127)]; or [seq(a(j),j=0..127)];
    coerce1st_octave := proc(r) option remember; if(r < 1) then coerce1st_octave(2*r); else if(r >= 2) then coerce1st_octave(r/2); else (r); fi; fi; end;
    A062383 := proc(n)
        option remember;
        if n = 0 then
            1 ;
        else
            2*procname(floor(n/2));
        end if;
    end proc:
    A062383 := n -> 1 + Bits:-Iff(n, n):
    seq(A062383(n), n=0..69); # Peter Luschny, Sep 23 2019
  • Mathematica
    a[n_] := a[n] = 2 a[n/2 // Floor]; a[0] = 1; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 04 2016 *)
    Table[2^Floor[Log2[n] + 1], {n, 0, 20}] (* Eric W. Weisstein, Nov 17 2017 *)
    2^Floor[Log2[Range[0, 20]] + 1] (* Eric W. Weisstein, Nov 17 2017 *)
    2^BitLength[Range[0, 100]] (* Paolo Xausa, Jan 29 2025 *)
  • PARI
    { a=1; for (n=0, 1000, write("b062383.txt", n, " ", a*=ceil((n + 1)/a)) ) } \\ Harry J. Smith, Aug 06 2009
    
  • PARI
    a(n)=1<<(log(2*n+1)\log(2)) \\ Charles R Greathouse IV, Dec 08 2011
    
  • Python
    def A062383(n): return 1 << n.bit_length() # Chai Wah Wu, Jun 30 2022

Formula

a(1) = 1 and a(n+1) = a(n)*ceiling(n/a(n)). - Benoit Cloitre, Aug 17 2002
G.f.: 1/(1-x) * (1 + Sum_{k>=0} 2^k*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = A142151(2*n)/2 + 1. - Reinhard Zumkeller, Jul 15 2008
log(a(n))/log(2) = A029837(n+1). - Johannes W. Meijer, Jul 06 2009
a(n+1) = a(n) + A099894(n). - Reinhard Zumkeller, Aug 06 2009
a(n) = A264619(n) - A264618(n). - Reinhard Zumkeller, Dec 01 2015
a(n) is the smallest power of 2 > n. - Chai Wah Wu, Nov 04 2016
a(n) = 2^ceiling(log_2(n+1)). - M. F. Hasler, Sep 20 2017

A008955 Triangle of central factorial numbers |t(2n,2n-2k)| read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 4, 1, 14, 49, 36, 1, 30, 273, 820, 576, 1, 55, 1023, 7645, 21076, 14400, 1, 91, 3003, 44473, 296296, 773136, 518400, 1, 140, 7462, 191620, 2475473, 15291640, 38402064, 25401600, 1, 204, 16422, 669188, 14739153, 173721912, 1017067024, 2483133696, 1625702400
Offset: 0

Views

Author

Keywords

Comments

Discussion of Central Factorial Numbers by N. J. A. Sloane, Feb 01 2011: (Start)
Here is Riordan's definition of the central factorial numbers t(n,k) given in Combinatorial Identities, Section 6.5:
For n >= 0, expand the polynomial
x^[n] = x*Product{i=1..n-1} (x+n/2-i) = Sum_{k=0..n} t(n,k)*x^k.
The t(n,k) are not always integers. The cases n even and n odd are best handled separately.
For n=2m, we have:
x^[2m] = Product_{i=0..m-1} (x^2-i^2) = Sum_{k=1..m} t(2m,2k)*x^(2k).
E.g. x^[8] = x^2(x^2-1^2)(x^2-2^2)(x^2-3^2) = x^8-14x^6+49x^4-36x^2,
which corresponds to row 4 of the present triangle.
So the m-th row of the present triangle gives the absolute values of the coefficients in the expansion of Product_{i=0..m-1} (x^2-i^2).
Equivalently, and simpler, the n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first.
For n odd, n=2m+1, we have:
x^[2m+1] = x*Product_{i=0..m-1}(x^2-((2i+1)/2)^2) = Sum_{k=0..m} t(2m+1,2k+1)*x^(2k+1).
E.g. x^[5] = x(x^2-(1/2)^2)(x^2-(3/2)^2) = x^5-10x^3/4+9x/16,
which corresponds to row 2 of the triangle in A008956.
We now rescale to get integers by replacing x by x/2 and multiplying by 2^(2m+1) (getting 1, -10, 9 from the example).
The result is that row m of triangle A008956 gives the coefficients in the expansion of x*Product_{i=0..m} (x^2-(2i+1)^2).
Equivalently, and simpler, the n-th row of A008956 gives the coefficients in the expansion of Product_{i=0..n-1} (x+(2i+1)^2), highest powers first.
Note that the n-th row of A182867 gives the coefficients in the expansion of Product_{i=1..n} (x+(2i)^2), highest powers first.
(End)
Contribution from Johannes W. Meijer, Jun 18 2009: (Start)
We define Beta(n-z,n+z)/Beta(n,n) = Gamma(n-z)*Gamma(n+z)/Gamma(n)^2 = sum(EG2[2m,n]*z^(2m), m = 0..infinity) with Beta(z,w) the Beta function. The EG2[2m,n] coefficients are quite interesting, see A161739. Our definition leads to EG2[2m,1] = 2*eta(2m) and the recurrence relation EG2[2m,n] = EG2[2m,n-1] - EG2[2m-2,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... , with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. We found for the matrix coefficients EG2[2m,n] = sum((-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2,k=1..n) with the central factorial numbers t1(n,m) as defined above, see also the Maple program.
From the EG2 matrix we arrive at the ZG2 matrix, see A161739 for its odd counterpart, which is defined by ZG2[2m,1] = 2*zeta(2m) and the recurrence relation ZG2[2m,n] = ZG2[2m-2,n-1]/(n*(n-1))-(n-1)*ZG2[2m,n-1]/n for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... . We found for the ZG2[2m,n] = Sum_{k=1..n} (-1)^(k+1)*t1(n-1,k-1)* 2* zeta(2*m-2*n+2*k)/((n-1)!*(n)!), and we see that the central factorial numbers t1(n,m) once again play a crucial role.
(End)

Examples

			Triangle begins:
  1;
  1,   1;
  1,   5,   4;
  1,  14,  49,  36;
  1,  30, 273, 820, 576;
  ...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part 1, Springer-Verlag 1985.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

Crossrefs

Cf. A036969.
Columns include A000330, A000596, A000597. Right-hand columns include A001044, A001819, A001820, A001821. Row sums are in A101686.
Appears in A160464 (Eta triangle), A160474 (Zeta triangle), A160479 (ZL(n)), A161739 (RSEG2 triangle), A161742, A161743, A002195, A002196, A162440 (EG1 matrix), A162446 (ZG1 matrix) and A163927. - Johannes W. Meijer, Jun 18 2009, Jul 06 2009 and Aug 17 2009
Cf. A234324 (central terms).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return 1;
        elif k=n then return (Factorial(n))^2;
        else return n^2*T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..8], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Sep 14 2019
  • Haskell
    a008955 n k = a008955_tabl !! n !! k
    a008955_row n = a008955_tabl !! n
    a008955_tabl = [1] : f [1] 1 1 where
       f xs u t = ys : f ys v (t * v) where
         ys = zipWith (+) (xs ++ [t^2]) ([0] ++ map (* u^2) (init xs) ++ [0])
         v = u + 1
    -- Reinhard Zumkeller, Dec 24 2013
    
  • Magma
    T:= func< n,k | Factorial(2*(n+1))*(&+[(-1)^j*Binomial(n,k-j)*(&+[2^(m-2*k)*StirlingFirst(2*(n-k+1)+m, 2*(n-k+1))*Binomial(2*(n-k+1)+2*j-1, 2*(n-k+1)+m-1)/Factorial(2*(n-k+1)+m): m in [0..2*j]]): j in [0..k]]) >;
    [T(n,k): k in [0..n], n in [0..8]]; // G. C. Greubel, Sep 14 2019
    
  • Maple
    nmax:=7: for n from 0 to nmax do t1(n, 0):=1: t1(n, n):=(n!)^2 end do: for n from 1 to nmax do for k from 1 to n-1 do t1(n, k) := t1(n-1, k-1)*n^2 + t1(n-1, k) end do: end do: seq(seq(t1(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jun 18 2009, Revised Sep 16 2012
    t1 := proc(n,k)
            sum((-1)^j*stirling1(n+1,n+1-k+j)*stirling1(n+1,n+1-k-j),j=-k..k) ;
    end proc: # Mircea Merca, Apr 02 2012
    # third Maple program:
    T:= proc(n, k) option remember; `if`(k=0, 1,
          add(T(j-1, k-1)*j^2, j=1..n))
        end:
    seq(seq(T(n, k), k=0..n), n=0..8);  # Alois P. Heinz, Feb 19 2022
  • Mathematica
    t[n_, 0]=1; t[n_, n_]=(n!)^2; t[n_ , k_ ]:=t[n, k] = n^2*t[n-1, k-1] + t[n-1, k]; Flatten[Table[t[n, k], {n,0,8}, {k,0,n}] ][[1 ;; 42]]
    (* Jean-François Alcover, May 30 2011, after recurrence formula *)
  • Maxima
    T(n,m):=(2*(n+1))!*sum((-1)^k*binomial(n,m-k)*sum((2^(i-2*m)*stirling1(2*(n-m+1)+i,2*(n-m+1))*binomial(2*(n-m+1)+2*k-1,2*(n-m+1)+i-1))/(2*(n-m+1)+i)!,i,0,2*k),k,0,m); /* Vladimir Kruchinin, Oct 05 2013 */
    
  • PARI
    T(n,k)=if(k==0,1, if(k==n, (n!)^2, n^2*T(n-1, k-1) + T(n-1, k)));
    for(n=0,8, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 14 2019
    
  • Sage
    # This triangle is (0,0)-based.
    def A008955(n, k) :
        if k==0 : return 1
        if k==n : return factorial(n)^2
        return n^2*A008955(n-1, k-1) + A008955(n-1, k)
    for n in (0..7) : print([A008955(n, k) for k in (0..n)]) # Peter Luschny, Feb 04 2012
    

Formula

The n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first (see Comments section).
The triangle can be obtained from the recurrence t1(n,k) = n^2*t1(n-1,k-1) + t1(n-1,k) with t1(n,0) = 1 and t1(n,n) = (n!)^2.
t1(n,k) = Sum_{j=-k..k} (-1)^j*s(n+1,n+1-k+j)*s(n+1,n+1-k-j) = Sum_{j=0..2*(n+1-k)} (-1)^(n+1-k+j)*s(n+1,j)*s(n+1,2*(n+1-k)-j), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 02 2012
E.g.f.: cosh(2/sqrt(t)*asin(sqrt(t)*z/2)) = 1 + z^2/2! + (1 + t)*z^4/4! + (1 + 5*t + 4*t^2)*z^6/6! + ... (see Berndt, p.263 and p.306). - Peter Bala, Aug 29 2012
T(n,m) = (2*(n+1))!*Sum_{k=0..m} ((-1)^k*binomial(n,m-k)*Sum_{i=0..2*k} ((2^(i-2*m)*stirling1(2*(n-m+1)+i,2*(n-m+1))*binomial(2*(n-m+1)+2*k-1, 2*(n-m+1)+i-1))/(2*(n-m+1)+i)!)). - Vladimir Kruchinin, Oct 05 2013

Extensions

There's an error in the last column of Riordan's table (change 46076 to 21076).
More terms from Vladeta Jovovic, Apr 16 2000
Link added and cross-references edited by Johannes W. Meijer, Aug 17 2009
Discussion of Riordan's definition of central factorial numbers added by N. J. A. Sloane, Feb 01 2011

A094665 Another version of triangular array in A083061: triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 4, 15, 15, 0, 34, 147, 210, 105, 0, 496, 2370, 4095, 3150, 945, 0, 11056, 56958, 111705, 107415, 51975, 10395, 0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 0, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490, 18918900, 2027025
Offset: 0

Views

Author

Philippe Deléham, Jun 07 2004, Jun 12 2007

Keywords

Comments

Define polynomials P(n,x) = x(2x+1)P(n-1,x+1) - 2x^2P(n-1,x), P(0,x) = 1. Sequence gives triangle read by rows, defined by P(n,x) = Sum_{k = 0..n} T(n,k)*x^k. - Philippe Deléham, Jun 20 2004
From Johannes W. Meijer, May 24 2009: (Start)
In A160464 we defined the coefficients of the ES1 matrix by ES1[2*m-1,n=1] = 2*eta(2*m-1) and the recurrence relation ES1[2*m-1,n] = ((2*n-2)/(2*n-1))*(ES1[2*m-1,n-1] - ES1[2*m-3,n-1]/(n-1)^2) for m the positive and negative integers and n >= 1. As usual eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. It is well-known that ES1[1-2*m,n=1] = (4^m-1)*(-bernoulli(2*m))/m for m >= 1. and together with the recurrence relation this leads to ES1[-1,n] = 0.5 for n >= 1.
We discovered that the n-th term of the row coefficients ES1[1-2*m,n] for m >= 1, can be generated with the rather simple polynomials RES1(1-2*m,n) = (-1)^(m+1)*ECGP(1-2*m, n)/2^m. This discovery was enabled by the recurrence relation for the RES1(1-2*m,n) which we derived from the recurrence relation for the ES1[2*m-1,n] coefficients and the fact that RES1(-1,n) = 0.5. The coefficients of the ECGP(1-2*m,n) polynomials led to this triangle and subsequently to triangle A083061. (End)
From David Callan, Jan 03 2011: (Start)
T(n,k) is the number of increasing 0-2 trees (A002105) on 2n edges in which the minimal path from the root has length k.
Proof. The number a(n,k) of such trees satisfies the recurrence a(0,0)=1, a(1,1)=1 and, counting by size of the subtree rooted at the smaller child of the root,
a(n,k) = Sum_{j=1..n-1} C(2n-1,j)*a(j,k-1)*a(n-1-j)
for 2<=k<=n, where a(n) = Sum_{k>=0} a(n,k) is the reduced tangent number A002105 (indexed from 0). The recurrence translates into the differential equation
F_x(x,y) = y*F(x,y)*G(x)
for the GF F(x,y) = Sum_{n,k>=0} a(n,k)x^(2n)/(2n)!*y^k, where G(x):=Sum_{n>=0} a(n)x^(2n+1)/(2n+1)! is known to be sqrt(2)*tan(x/sqrt(2)). The differential equation has solution F(x,y) = sec(x/sqrt(2))^(2y). (End)

Examples

			Triangle begins:
.1;
.0, 1;
.0, 1, 3;
.0, 4, 15, 15;
.0, 34, 147, 210, 105;
.0, 496, 2370, 4095, 3150, 945;
.0, 11056, 56958, 111705, 107415, 51975, 10395;
.0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135;
From _Johannes W. Meijer_, May 24 2009: (Start)
The first few ECGP(1-2*m,n) polynomials are: ECGP(-1,n) = 1; ECGP(-3,n) = n; ECGP(-5,n) = n + 3*n^2; ECGP(-7,n) = 4*n + 15*n^2+ 15*n^3 .
The first few RES1(1-2*m,n) are: RES1(-1,n) = (1/2)*(1); RES1(-3,n) = (-1/4)*(n); RES1(-5,n) = (1/8)*(n+3*n^2); RES1(-7,n) = (-1/16)*(4*n+15*n^2+15*n^3).
(End)
		

Crossrefs

From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

Programs

  • Maple
    nmax:=7; imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2 * x^2 * T1(i-1, x)): dx:=degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx) od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do T(n+1, k+1) := A083061(n, k) od: od: T(0, 0):=1: for n from 1 to nmax do T(n, 0):=0 od: seq(seq(T(n, k), k=0..n), n=0..nmax);
    # Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012
  • Mathematica
    nmax = 8;
    T[n_, k_] := SeriesCoefficient[Sec[x/Sqrt[2]]^(2y), {x, 0, 2n}, {y, 0, k}]* (2n)!;
    Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)

Formula

Sum_{k = 0..n} T(n, k) = A002105(n+1).
Sum_{k = 0..n} T(n, k)*2^(n-k) = A000364(n); Euler numbers.
Sum_{k = 0..n} T(n, k)*(-2)^(n-k) = 1.
RES1(1-2*m,n) = n^2*RES1(3-2*m,n)-n*(2*n+1)*RES1(3-2*m,n+1)/2 for m >= 2, with RES1(-1,n) = 0.5 for n >= 1. - Johannes W. Meijer, May 24 2009
G.f.: Sum_{n,k>=0} T(n,k)x^n/n!*y^k = sec(x/sqrt(2))^(2y).

Extensions

Term corrected by Johannes W. Meijer, Sep 23 2012

A160487 The Lambda triangle.

Original entry on oeis.org

1, -107, 10, 59845, -7497, 210, -6059823, 854396, -35574, 420, 5508149745, -827924889, 41094790, -765534, 4620, -8781562891079, 1373931797082, -75405128227, 1738417252, -17219202, 60060
Offset: 2

Views

Author

Johannes W. Meijer, May 24 2009, Sep 18 2012

Keywords

Comments

The coefficients of the LS1 matrix are defined by LS1[2*m,n] = int(y^(2*m)/(sinh(y))^(2*n-1),y=0..infinity)/factorial(2*m) for m = 1, 2, 3, .. and n = 1, 2, 3, .. under the condition that n <= m.
This definition leads to LS1[2*m,n=1] = 2*lambda(2*m+1), for m = 1, 2, .. , and the recurrence relation LS1[2*m,n] = ((2*n-3)/(2*n-2))*(LS1[2*m-2,n-1]/(2*n-3)^2- LS1[2*m,n-1]). As usual lambda(m) = (1-2^(-m))*zeta(m) with zeta(m) the Riemann zeta function.
These two formulas enable us to determine the values of the LS1[2*m,n] coefficients, for all integers m and all positive integers n, but not for all n. If we choose, somewhat but not entirely arbitrarily, LS1[m=0,n=1] = gamma, with gamma the Euler-Mascheroni constant, we can determine them all.
The coefficients in the columns of the LS1 matrix, for m = 0, 1, 2, .. , and n = 2, 3, 4 .. , can be generated with the GL(z;n) polynomials for which we found the following general expression GL(z;n) = (h(n)*CFN2(z;n)*GL(z;n=1) + LAMBDA(z;n))/p(n).
The CFN2(z;n) polynomials depend on the central factorial numbers A008956.
The LAMBDA(z;n) are the Lambda polynomials which lead to the Lambda triangle.
The zero patterns of the Lambda polynomials resemble a UFO. These patterns resemble those of the Eta, Zeta and Beta polynomials, see A160464, A160474 and A160480.
The first Maple algorithm generates the coefficients of the Lambda triangle. The second Maple algorithm generates the LS1[2*m,n] coefficients for m= -1, -2, -3, .. .
Some of our results are conjectures based on numerical evidence.

Examples

			The first few rows of the triangle LAMBDA(n,m) with n=2,3,.. and m=1,2,.. are
  [1]
  [ -107, 10]
  [59845, -7497, 210]
  [ -6059823, 854396, -35574, 420]
The first few LAMBDA(z;n) polynomials are
  LAMBDA (z;n=2) = 1
  LAMBDA (z;n=3) = -107 +10*z^2
  LAMBDA (z;n=4) = 59845-7497*z^2+210*z^4
The first few CFN2(z;n) polynomials are
  CFN2(z;n=2) = (z^2-1)
  CFN2(z;n=3) = (z^4-10*z^2+9)
  CFN2(z;n=4) = (z^6- 35*z^4+259*z^2-225)
The first few generating functions GL(z;n) are:
  GL(z;n=2) = (6*(z^2-1)*GL(z,n=1) + (1)) /12
  GL(z;n=3) = (60*(z^4-10*z^2+9)*GL(z,n=1)+ (-107+10*z^2)) / 1440
  GL(z;n=4) = (1260*( z^6- 35*z^4+259*z^2-225)*GL(z,n=1) + (59845-7497*z^2+ 210*z^4))/907200
		

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Crossrefs

A160488 equals the first left hand column.
A160476 equals the first right hand column and 6*h(n).
A160489 equals the rows sums.
A160490 equals the p(n) sequence.
A160479 equals the ZL(n) sequence.
A001620 is the Euler-Mascheroni constant gamma.
The LS1[ -2, n] coefficients lead to A002197, A002198 and A058962.
The LS1[ -2*m, 1] coefficients equal (-1)^(m+1)*A036282/A036283.
The CFN2(z, n) and the cfn2(n, k) lead to A008956.
Cf. The Eta, Zeta and Beta triangles A160464, A160474 and A160480.
Cf. A162448 (LG1 matrix)

Programs

  • Maple
    nmax:=7; for n from 0 to nmax do cfn2(n, 0) := 1: cfn2(n, n) := (doublefactorial(2*n-1))^2 od: for n from 1 to nmax do for k from 1 to n-1 do cfn2(n, k) := (2*n-1)^2*cfn2(n-1, k-1) + cfn2(n-1, k) od: od: for n from 1 to nmax do Delta(n-1) := sum((1-2^(2*k1-1))* (-1)^(n+1)*(-bernoulli(2*k1)/(2*k1))*(-1)^(k1+n)*cfn2(n-1,n-k1, n), k1=1..n) / (2*4^(n-1)*(2*n-1)!); LAMBDA(-2, n) := sum(2*(1-2^(2*k1-1))*(-bernoulli(2*k1) / (2*k1))*(-1)^(k1+n)* cfn2(n-1,n-k1), k1=1..n)/ factorial(2*n-2) end do: Lcgz(2) := 1/12: f(2) := 1/12: for n from 3 to nmax do Lcgz(n) := LAMBDA(-2, n-1)/((2*n-2)*(2*n-3)): f(n) := Lcgz(n)-((2*n-3)/(2*n-2))*f(n-1) end do: for n from 1 to nmax do b(n) := denom(Lcgz(n+1)) end do: for n from 1 to nmax do b(n) := 2*n*denom(Delta(n-1))/2^(2*n) end do: p(2) := b(1): for n from 2 to nmax do p(n+1) := lcm(p(n)*(2*n)*(2*n-1), b(n)) end do: for n from 2 to nmax do LAMBDA(n, 1) := p(n)*f(n) end do: mmax:=nmax: for n from 2 to nmax do LAMBDA(n, n) := 0 end do: for n from 1 to nmax do b(n) := (2*n)*(2*n-1)*denom(Delta(n-1))/ (2^(2*n)*(2*n-1)) end do: c(1) := b(1): for n from 1 to nmax-1 do c(n+1) := lcm(c(n)*(2*n+2)* (2*n+1), b(n+1)) end do: for n from 1 to nmax do cm(n) := c(n)/(6*(2*n)!) end do: for n from 1 to nmax-1 do ZL(n+2) := cm(n+1)/cm(n) end do: for m from 2 to mmax do for n from m+1 to nmax do LAMBDA(n, m) := ZL(n)*(LAMBDA(n-1, m-1)-(2*n-3)^2*LAMBDA(n-1, m)) end do end do; seq(seq(LAMBDA(n,m), m=1..n-1), n=2..nmax);
    # End first program.
    nmax1:=10; m:=1; LS1row:=-2*m; for n from 0 to nmax1 do cfn2(n, 0) := 1: cfn2(n, n) := (doublefactorial(2*n-1))^2 od: for n from 1 to nmax1 do for k from 1 to n-1 do cfn2(n, k) := (2*n-1)^2*cfn2(n-1, k-1) + cfn2(n-1, k) od: od: mmax1:=nmax1: for m1 from 1 to mmax1 do LS1[-2*m1, 1] := 2*(1-2^(-(-2*m1+1)))*(-bernoulli(2*m1)/(2*m1)) od: for n from 2 to nmax1 do for m1 from 1 to mmax1-n+1 do LS1[ -2*m1, n] := sum((-1)^(k1+1)*cfn2(n-1,k1-1)* LS1[2*k1-2*n-2*m1, 1], k1=1..n)/(2*n-2)! od: od: seq(LS1[ -2*m, n], n=1..nmax1-m+1);
    # End second program.

Formula

We discovered a remarkable relation between the Lambda triangle coefficients Lambda(n,m) = ZL(n)*(Lambda(n-1,m-1)-(2*n-3)^2*Lambda(n-1,m)) for n = 3, 4, .. and m = 2, 3, .. . See A160488 for LAMBDA(n,m=1) and furthermore LAMBDA(n,n) = 0 for n = 2, 3, .. .
We observe that the ZL(n) = A160479(n) sequence also rules the Zeta triangle A160474.
The generating functions GL(z;n) of the coefficients in the matrix columns are defined by
GL(z;n) = sum(LS1[2*m-2,n]*z^(2*m-2), m=1..infinity), with n = 1, 2, 3, .. .
This definition, and our choice of LS1[m=0,n=1] = gamma, leads to GL(z;n=1) = -2*Psi(1-z)+Psi(1-(z/2))-(Pi/2)*tan(Pi*z/2) with Psi(z) the digamma-function. Furthermore we discovered that GL(z;n) =GL(z;n-1)*(z^2/((2*n-2)*(2*n-3)) -(2*n-3)/((2*n-2)))+LS1[ -2,n-1]/((2*n-2)*(2*n-3)) for n = 2, 3 , .. . with LS1[ -2,n] = (-1)^(n-1)*4*A058962(n-1)*A002197(n-1)/A002198(n-1) for n = 1, 2, .. , with A058962(n-1) = 2^(2*n-2)*(2*n-1).
We found the following general expression for the GL(z;n) polynomials, for n = 2, 3, ..
GL(z;n) = (h(n)*CFN2(z;n)*GL(z;n=1) + LAMBDA(z;n))/p(n) with
h(n) = 6*A160476(n) and p(n) = A160490(n).

A083061 Triangle of coefficients of a companion polynomial to the Gandhi polynomial.

Original entry on oeis.org

1, 1, 3, 4, 15, 15, 34, 147, 210, 105, 496, 2370, 4095, 3150, 945, 11056, 56958, 111705, 107415, 51975, 10395, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490
Offset: 0

Views

Author

Hans J. H. Tuenter, Apr 19 2003

Keywords

Comments

This polynomial arises in the setting of a symmetric Bernoulli random walk and occurs in an expression for the even moments of the absolute distance from the origin after an even number of timesteps. The Gandhi polynomial, sequence A036970, occurs in an expression for the odd moments.
When formatted as a square array, first row is A002105, first column is A001147, second column is A001880.
Another version of the triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, 9, ...] = 1; 0, 1; 0, 1, 3; 0, 4, 15, 15; 0, 34, 147, 210, 105; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 07 2004
In A160464 we defined the coefficients of the ES1 matrix. Our discovery that the n-th term of the row coefficients ES1[1-2*m,n] for m>=1, can be generated with rather simple polynomials led to triangle A094665 and subsequently to this one. - Johannes W. Meijer, May 24 2009
Related to polynomials defined in A160485 by a shift of +-1/2 and scaling by a power of 2. - Richard P. Brent, Jul 15 2014

Examples

			Triangle starts (with an additional first column 1,0,0,...):
[1]
[0,      1]
[0,      1,       3]
[0,      4,      15,      15]
[0,     34,     147,     210,     105]
[0,    496,    2370,    4095,    3150,     945]
[0,  11056,   56958,  111705,  107415,   51975,  10395]
[0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135]
		

Crossrefs

From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
A002105 equals the row sums (n>=2) and the first left hand column (n>=1).
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

Programs

  • Maple
    imax := 6;
    T1(0, x) := 1:
    T1(0, x+1) := 1:
    for i from 1 to imax do
        T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2*x^2*T1(i-1, x)):
        dx := degree(T1(i, x)):
        for k from 0 to dx do
            c(k) := coeff(T1(i, x), x, k)
        od:
        T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1 = 0..dx):
    od:
    for i from 0 to imax do
        for j from 0 to i do
            a(i, j) := coeff(T1(i, x), x, j)
        od:
    od:
    seq(seq(a(i, j), j = 0..i), i = 0..imax);
    # Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012
  • Mathematica
    b[0, 0] = 1;
    b[n_, k_] := b[n, k] = Sum[2^j*(Binomial[k + j, 1 + j] + Binomial[k + j + 1, 1 + j])*b[n - 1, k - 1 + j], {j, Max[0, 1 - k], n - k}];
    a[0, 0] = 1;
    a[n_, k_] := b[n, k]/2^(n - k);
    Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 19 2018, after Philippe Deléham *)
  • Sage
    # uses[fr2_row from A088874]
    A083061_row = lambda n: [(-1)^(n-k)*m*2^(-n+k) for k,m in enumerate(fr2_row(n))]
    for n in (0..7): print(A083061_row(n)) # Peter Luschny, Sep 19 2017

Formula

Let T(i, x)=(2x+1)(x+1)T(i-1, x+1)-2x^2T(i-1, x), T(0, x)=1; so that T(1, x)=1+3x; T(2, x)=4+15x+15x^2; T(3, x)=34+147x+210x^2+105x^3, etc. Then the (i, j)-th entry in the table is the coefficient of x^j in T(i, x).
a(n, k)*2^(n-k) = A085734(n, k). - Philippe Deléham, Feb 27 2005
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