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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A004188 a(n) = n*(3*n^2 - 1)/2.

Original entry on oeis.org

0, 1, 11, 39, 94, 185, 321, 511, 764, 1089, 1495, 1991, 2586, 3289, 4109, 5055, 6136, 7361, 8739, 10279, 11990, 13881, 15961, 18239, 20724, 23425, 26351, 29511, 32914, 36569, 40485, 44671, 49136, 53889, 58939, 64295, 69966, 75961
Offset: 0

Views

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
(1), (4+7), (10+13+16), (19+22+25+28), ... - Jon Perry, Sep 10 2004

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
  • T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).

Crossrefs

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A236770 (partial sums).

Programs

Formula

Partial sums of n-1 3-spaced triangular numbers, e.g., a(4) = t(1) + t(4) + t(7) = 1 + 10 + 28 = 39. - Jon Perry, Jul 23 2003
a(n) = C(2*n+1,3) + C(n+1,3), n >= 0. - Zerinvary Lajos, Jan 21 2007
a(n) = A000447(n) + A000292(n). - Zerinvary Lajos, Jan 21 2007
G.f.: x*(1+7*x+x^2) / (x-1)^4. - R. J. Mathar, Oct 08 2011
From Miquel Cerda, Dec 25 2016: (Start)
a(n) = A000578(n) + A135503(n).
a(n) = A007588(n) - A135503(n). (End)
E.g.f.: (x/2)*(2 + 9*x + 3*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A006564 Icosahedral numbers: a(n) = n*(5*n^2 - 5*n + 2)/2.

Original entry on oeis.org

1, 12, 48, 124, 255, 456, 742, 1128, 1629, 2260, 3036, 3972, 5083, 6384, 7890, 9616, 11577, 13788, 16264, 19020, 22071, 25432, 29118, 33144, 37525, 42276, 47412, 52948, 58899, 65280, 72106, 79392, 87153, 95404, 104160, 113436, 123247, 133608
Offset: 1

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Author

Keywords

Comments

Schlaefli symbol for this polyhedron: {3,5}.
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010

References

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

Crossrefs

Cf. A000292 (tetrahedral numbers), A000578 (cubes), A005900 (octahedral numbers), A006566 (dodecahedral numbers).

Programs

  • Haskell
    a006564 n = n * (5 * n * (n - 1) + 2) `div` 2
    -- Reinhard Zumkeller, Jun 16 2013
    
  • Magma
    [(5*n^3-5*n^2+2*n)/2: n in [1..100]] // Vincenzo Librandi, Nov 21 2010
    
  • Maple
    A006564:=(1+8*z+6*z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[n (5n^2-5n+2)/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,12,48,124},40] (* Harvey P. Dale, May 26 2011 *)
  • PARI
    a(n)=5*n^2*(n-1)/2+n \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = C(n+2,3) + 8*C(n+1,3) + 6*C(n,3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) with a(0)=1, a(1)=12, a(2)=48, a(3)=124. - Harvey P. Dale, May 26 2011
G.f.: x*(6*x^2 + 8*x + 1)/(x-1)^4. - Harvey P. Dale, May 26 2011
a(n) = A006566(n) - A035006(n). - Peter M. Chema, May 04 2016
E.g.f.: x*(2 + 10*x + 5*x^2)*exp(x)/2. - Ilya Gutkovskiy, May 04 2016
Sum_{n>=1} 1/a(n) = A175578. - Amiram Eldar, Jan 03 2022

A004466 a(n) = n*(5*n^2 - 2)/3.

Original entry on oeis.org

0, 1, 12, 43, 104, 205, 356, 567, 848, 1209, 1660, 2211, 2872, 3653, 4564, 5615, 6816, 8177, 9708, 11419, 13320, 15421, 17732, 20263, 23024, 26025, 29276, 32787, 36568, 40629, 44980, 49631, 54592
Offset: 0

Views

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
Also as a(n)=(1/6)*(10*n^3-4*n), n>0: structured pentagonal anti-diamond numbers (vertex structure 11) (Cf. A051673 = alternate vertex A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n+1)-10*a(n) = (n+1)*(5*(n+1)^2-2)/3 - (10n(n+1)(n+2)/6) = n. The unit digits are 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,... . - Eric Desbiaux, Aug 18 2008

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Crossrefs

Cf. A062786 (first differences), A264853 (partial sums).
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

G.f.: x*(1+8*x+x^2)/(1-x)^4. - Colin Barker, Jan 08 2012
E.g.f.: (x/3)*(3 + 15*x + 5*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A063521 a(n) = n*(7*n^2-4)/3.

Original entry on oeis.org

0, 1, 16, 59, 144, 285, 496, 791, 1184, 1689, 2320, 3091, 4016, 5109, 6384, 7855, 9536, 11441, 13584, 15979, 18640, 21581, 24816, 28359, 32224, 36425, 40976, 45891, 51184, 56869, 62960, 69471, 76416, 83809, 91664, 99995, 108816, 118141, 127984, 138359, 149280, 160761
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Comments

Also as a(n)=(1/6)*(14*n^3-8*n), n>0: structured heptagonal anti-diamond numbers (vertex structure 15) (Cf. A100186 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Crossrefs

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Programs

Formula

G.f.: x*(1+12*x+x^2)/(1-x)^4. - Colin Barker, Jan 10 2012
E.g.f.: (x/3)*(3 + 21*x + 7*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A077042 Square array read by falling antidiagonals of central polynomial coefficients: largest coefficient in expansion of (1 + x + x^2 + ... + x^(n-1))^k = ((1-x^n)/(1-x))^k, i.e., the coefficient of x^floor(k*(n-1)/2) and of x^ceiling(k*(n-1)/2); also number of compositions of floor(k*(n+1)/2) into exactly k positive integers each no more than n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 6, 7, 4, 1, 1, 0, 1, 10, 19, 12, 5, 1, 1, 0, 1, 20, 51, 44, 19, 6, 1, 1, 0, 1, 35, 141, 155, 85, 27, 7, 1, 1, 0, 1, 70, 393, 580, 381, 146, 37, 8, 1, 1, 0, 1, 126, 1107, 2128, 1751, 780, 231, 48, 9, 1, 1, 0, 1, 252, 3139
Offset: 0

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Author

Henry Bottomley, Oct 22 2002

Keywords

Comments

From Michel Marcus, Dec 01 2012: (Start)
A pair of numbers written in base n are said to be comparable if all digits of the first number are at least as big as the corresponding digit of the second number, or vice versa. Otherwise, this pair will be defined as uncomparable. A set of pairwise uncomparable integers will be called anti-hierarchic.
T(n,k) is the size of the maximal anti-hierarchic set of integers written with k digits in base n.
For example, for base n=2 and k=4 digits:
- 0 (0000) and 15 (1111) are comparable, while 6 (0110) and 9 (1001) are uncomparable,
- the maximal antihierarchic set is {3 (0011), 5 (0101), 6 (0110), 9 (1001), 10 (1010), 12 (1100)} with 6 elements that are all pairwise uncomparable. (End)

Examples

			Rows of square array start:
  1,    0,    0,    0,    0,    0,    0, ...
  1,    1,    1,    1,    1,    1,    1, ...
  1,    1,    2,    3,    6,   10,   20, ...
  1,    1,    3,    7,   19,   51,  141, ...
  1,    1,    4,   12,   44,  155,  580, ...
  1,    1,    5,   19,   85,  381, 1751, ...
  ...
Read by antidiagonals:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 2, 1, 1;
  0, 1, 3, 3, 1, 1;
  0, 1, 6, 7, 4, 1, 1;
  ...
		

Crossrefs

Programs

Formula

By the central limit theorem, T(n,k) is roughly n^(k-1)*sqrt(6/(Pi*k)).
T(n,k) = Sum{j=0,h/n} (-1)^j*binomial(k,j)*binomial(k-1+h-n*j,k-1) with h=floor(k*(n-1)/2), k>0. - Michel Marcus, Dec 01 2012

A004126 a(n) = n*(7*n^2 - 1)/6.

Original entry on oeis.org

0, 1, 9, 31, 74, 145, 251, 399, 596, 849, 1165, 1551, 2014, 2561, 3199, 3935, 4776, 5729, 6801, 7999, 9330, 10801, 12419, 14191, 16124, 18225, 20501, 22959, 25606, 28449, 31495, 34751, 38224, 41921, 45849, 50015, 54426, 59089, 64011
Offset: 0

Views

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Keywords

Comments

3-dimensional analog of centered polygonal numbers.
Sum of n triangular numbers starting from T(n), where T = A000217. E.g., a(4) = T(4) + T(5) + T(6) + T(7) = 10 + 15 + 21 + 28 = 74. - Amarnath Murthy, Jul 16 2004
Also as a(n) = (1/6)*(7*n^3-n), n>0: structured heptagonal diamond numbers (vertex structure 8). Cf. A100179 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Partial sums of A069099, centered heptagonal numbers (A000566). - Jonathan Vos Post, Mar 16 2006
Binomial transform of (0, 1, 7, 7, 0, 0, 0, ...) and third partial sum of (0, 1, 6, 7, 7, 7, ...). - Gary W. Adamson, Oct 05 2015

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

Crossrefs

Programs

Formula

a(n) = C(2*n+1,3)-C(n+1,3), n>=0. - Zerinvary Lajos, Jan 21 2007
a(n) = A000447(n) - A000292(n). - Zerinvary Lajos, Jan 21 2007
G.f.: x*(1+5*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
E.g.f.: (x/6)*(7*x^2 + 21*x + 6)*exp(x). - G. C. Greubel, Oct 05 2015
a(n) = Sum_{i = n..2*n-1} A000217(i). - Bruno Berselli, Sep 06 2017
a(n) = n^3 + Sum_{k=0..n-1} k*(k+1)/2. Alternately, a(n) = A000578(n) + A000292(n-1) for n>0. - Bruno Berselli, May 23 2018

A210391 Number A(n,k) of semistandard Young tableaux over all partitions of n with maximal element <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 9, 6, 1, 0, 1, 5, 16, 19, 9, 1, 0, 1, 6, 25, 44, 39, 12, 1, 0, 1, 7, 36, 85, 116, 69, 16, 1, 0, 1, 8, 49, 146, 275, 260, 119, 20, 1, 0, 1, 9, 64, 231, 561, 751, 560, 189, 25, 1, 0
Offset: 0

Views

Author

Alois P. Heinz, Mar 20 2012

Keywords

Examples

			Square array A(n,k) begins:
  1,  1,   1,   1,   1,    1,    1, ...
  0,  1,   2,   3,   4,    5,    6, ...
  0,  1,   4,   9,  16,   25,   36, ...
  0,  1,   6,  19,  44,   85,  146, ...
  0,  1,   9,  39, 116,  275,  561, ...
  0,  1,  12,  69, 260,  751, 1812, ...
  0,  1,  16, 119, 560, 1955, 5552, ...
		

Crossrefs

Columns k=0-8 give: A000007, A000012, A002620(n+2), A038163, A054498, A181477, A181478, A181479, A181480.
Main diagonal gives: A209673.

Programs

  • Maple
    # First program:
    h:= (l, k)-> mul(mul((k+j-i)/(1+l[i] -j +add(`if`(l[t]>=j, 1, 0)
                     , t=i+1..nops(l))), j=1..l[i]), i=1..nops(l)):
    g:= proc(n, i, k, l)
          `if`(n=0, h(l, k), `if`(i<1, 0, g(n, i-1, k, l)+
          `if`(i>n, 0, g(n-i, i, k, [l[], i]))))
        end:
    A:= (n, k)-> `if`(n=0, 1, g(n, n, k, [])):
    seq(seq(A(n, d-n), n=0..d), d=0..12);
    # second program:
    gf:= k-> 1/((1-x)^k*(1-x^2)^(k*(k-1)/2)):
    A:= (n, k)-> coeff(series(gf(k), x, n+1), x, n):
    seq(seq(A(n, d-n), n=0..d), d=0..12);
  • Mathematica
    (* First program: *)
    h[l_, k_] := Product[Product[(k+j-i)/(1+l[[i]]-j + Sum[If[l[[t]] >= j, 1, 0], {t, i+1, Length[l]}]), {j, 1, l[[i]]}], {i, 1, Length[l]}]; g [n_, i_, k_, l_] := If[n == 0, h[l, k], If[i < 1, 0, g[n, i-1, k, l] + If[i > n, 0, g[n-i, i, k, Append[l, i]]]]]; a[n_, k_] := If[n == 0, 1, g[n, n, k, {}]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten
    (* second program: *)
    gf[k_] := 1/((1-x)^k*(1-x^2)^(k*(k-1)/2)); a[n_, k_] := Coefficient[Series[gf[k], {x, 0, n+1}], x, n]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Dec 09 2013, translated from Maple *)

Formula

G.f. of column k: 1/((1-x)^k*(1-x^2)^(k*(k-1)/2)).
A(n,k) = Sum_{i=0..k} C(k,i) * A138177(n,k-i). - Alois P. Heinz, Apr 06 2015

A001794 Negated coefficients of Chebyshev T polynomials: [x^n](-T(n+6, x)), n >= 0.

Original entry on oeis.org

1, 7, 32, 120, 400, 1232, 3584, 9984, 26880, 70400, 180224, 452608, 1118208, 2723840, 6553600, 15597568, 36765696, 85917696, 199229440, 458752000, 1049624576, 2387607552, 5402263552, 12163481600, 27262976000, 60850962432
Offset: 0

Views

Author

Keywords

Comments

A negated subdiagonal of A053120.
If X_1,X_2,...,X_n are 2-blocks of a (2n+1)-set X then a(n-2) is the number of (n+3)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 18 2007
The third corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n-2) is the number of strings of length n defined on {0, 1, 2, 3} that have exactly two 2's and no 3's or exactly three 3's and no 2's. For example, for n=4, a(2)=32 since the strings are the 6 permutations of 2200, the 12 permutations of 2201, the 6 permutations of 2211, the 4 permutations of 3330, and the 4 permutations of 3331. - Enrique Navarrete, Jun 03 2025

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A039991 (negative of column 6), A028297, A008310, A053120.
With alternating signs, the o.g.f. (with offset 1) is the inverse of the o.g.f. of A065097.
Cf. A001789 (partial sums), A081279 (binomial transform), A005900 (0 followed by inverse binomial transform).

Programs

  • GAP
    List([0..25],n->2^(n-2)*(n+1)*(n+2)*(n+6)/3); # Muniru A Asiru, Mar 20 2018
  • Magma
    [2^(n-1)/3*Binomial(n+2,2)*(n+6) : n in [0..25]]; // Brad Clardy, Mar 08 2012
    
  • Maple
    [seq(coeftayl((1-x)/(1-2*x)^4, x = 0, k), k=0..25)]; # Muniru A Asiru, Mar 20 2018
  • Mathematica
    a[n_] := 2^(n-2)*(n+1)*(n+2)*(n+6)/3; a /@ Range[0, 20] (* Giovanni Resta, Mar 25 2017 *)
    LinearRecurrence[{8,-24,32,-16},{1,7,32,120},30] (* Harvey P. Dale, Oct 08 2024 *)
  • PARI
    a(n) = sum(i=0, n+1, sum(k=0, i, k^2*binomial(n+1, i))); \\ Michel Marcus, Mar 25 2017
    
  • PARI
    a(n) = - polcoeff(polchebyshev(n+6), n); \\ Michel Marcus, Mar 20 2018
    

Formula

a(n) = 2^(n-2)*(n+1)*(n+2)*(n+6)/3. [See a comment in A053120 on subdiagonals. - Wolfdieter Lang, Jan 03 2020]
G.f.: (1-x)/(1-2*x)^4. - Simon Plouffe in his 1992 dissertation
a(n) = Sum_{k=0..floor((n+6)/2)} C(n+6, 2*k)*C(k, 3). - Paul Barry, May 15 2003
With a leading zero, the binomial transform of A000330. - Paul Barry, Jul 19 2003
a(n) = Sum_{i=0..n+1} (Sum{k=0..i} (k^2*binomial(n+1, i))). - Jon Perry, Feb 26 2004 [corrected by Michel Marcus, Mar 25 2017]
Binomial transform of a(n) = (2*n^3 + 6*n^2 + 7*n + 3)/3 offset 0. a(3)=120. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) = (2^(n-1)/3)*binomial(n+2,2)*(n+6). - Brad Clardy, Mar 08 2012
E.g.f.: (1/3)*exp(2*x)*(3 + 15*x + 12*x^2 + 2*x^3). - Stefano Spezia, Jan 03 2020
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=0} 1/a(n) = 156*log(2)/5 - 511/25.
Sum_{n>=0} (-1)^n/a(n) = 241/25 - 108*log(3/2)/5. (End)
E.g.f.: exp(2*x)*(x^2/2 + x^3/6) (with two leading zeros). - Enrique Navarrete, Jun 03 2025

Extensions

Name clarified by Wolfdieter Lang, Nov 26 2019

A063522 a(n) = n*(5*n^2 - 3)/2.

Original entry on oeis.org

0, 1, 17, 63, 154, 305, 531, 847, 1268, 1809, 2485, 3311, 4302, 5473, 6839, 8415, 10216, 12257, 14553, 17119, 19970, 23121, 26587, 30383, 34524, 39025, 43901, 49167, 54838, 60929, 67455, 74431, 81872, 89793, 98209, 107135, 116586, 126577, 137123, 148239, 159940
Offset: 0

Views

Author

N. J. A. Sloane, Aug 02 2001

Keywords

Crossrefs

(1/12)*t*(n^3 - n) + n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Bisections: A160674, A160699.

Programs

  • Magma
    [n*(5*n^2 -3)/2: n in [0..30]]; // G. C. Greubel, May 02 2018
  • Mathematica
    lst={};Do[AppendTo[lst, LegendreP[3, n]], {n, 10^2}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
    CoefficientList[Series[x*(1 + 13*x + x^2)/(1-x)^4, {x, 0, 50}], x] (* G. C. Greubel, Sep 01 2017 *)
    LinearRecurrence[{4,-6,4,-1},{0,1,17,63},40] (* Harvey P. Dale, Sep 06 2023 *)
  • PARI
    a(n) = { n*(5*n^2 - 3)/2 } \\ Harry J. Smith, Aug 25 2009
    

Formula

G.f.: x*(1 + 13*x + x^2)/(1-x)^4. - Colin Barker, Jan 10 2012
E.g.f.: (x/2)*(2 + 15*x + 5*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A142978 Table of figurate numbers for the n-dimensional cross polytopes.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 19, 16, 5, 1, 10, 33, 44, 25, 6, 1, 12, 51, 96, 85, 36, 7, 1, 14, 73, 180, 225, 146, 49, 8, 1, 16, 99, 304, 501, 456, 231, 64, 9, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10
Offset: 1

Views

Author

Peter Bala, Jul 15 2008

Keywords

Comments

The n-th row entries for this array are the regular polytope numbers for the n-dimensional cross polytope as defined by [Kim]. The rows are the partial sums of the rows of the square array of Delannoy numbers A008288.
The odd numbered rows of this array form A142977. For a triangular version of this table see A104698. Cf. also A101603.
The n-th row of the array is the binomial transform of n-th row of triangle A081277, followed by zeros. Example: row 4 (1, 6, 19, 44, 85, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 17 2008
The main diagonal of the array T(n,k) is A047781 Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600. The link from A099193 to J. V. Post, Table of polytope numbers, Sorted, Through 1,000,000, includes all n-D Hyperoctahedron (n-Cross Polytope) Numbers through 10-Cross(20) = 1669752016. - Jonathan Vos Post, Jul 16 2008

Examples

			The square array A(n, k) begins:
  n\k| 1   2    3     4     5       6
  ---+-------------------------------
   1 | 1   2    3     4      5      6    A000027
   2 | 1   4    9    16     25     36    A000290
   3 | 1   6   19    44     85    146    A005900
   4 | 1   8   33    96    225    456    A014820
   5 | 1  10   51   180    501   1182    A069038
   6 | 1  12   73   304    985   2668    A069039
   7 | 1  14   99   476   1765   5418    A099193
		

Crossrefs

Cf. A008288 (Delannoy numbers), A005900 (row 3), A014820 (row 4), A069038 (row 5), A069039 (row 6), A099193 (row 7), A099195 (row 8), A099196 (row 9), A099197 (row 10), A101603, A104698 (triangle version), A142977, A142983.

Programs

  • Haskell
    a142978 n k = a142978_tabl !! (n-1) !! (k-1)
    a142978_row n = a142978_tabl !! (n-1)
    a142978_tabl = map reverse a104698_tabl
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    with(combinat): T:=(n,k) -> add(binomial(n-1,i)*binomial(k+i,n),i = 0..n-1); for n from 1 to 10 do seq(T(n,k),k = 1..10) end do; # Program restored by Peter Bala, Oct 02 2008
    A := (n, k) -> k*hypergeom([1 - n, 1 - k], [2], 2):
    seq(print(seq(simplify(A(n, k)), k = 1..9)), n=1..7); # Peter Luschny, Mar 23 2023
  • Mathematica
    t[n_, k_] := Sum[ Binomial[n-1, i]*Binomial[k+i, n], {i, 0, n-1}]; Table[t[n-k, k], {n, 1, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n-1} C(n-1,i)*C(k+i,n).
Reciprocity law: n*T(n,k) = k*T(k,n).
Recurrence relation: T(n,1) = 1, T(1,k) = k, T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k), n,k > 1.
O.g.f. row n: x*(1 + x)^(n-1)/(1 - x)^(n+1).
O.g.f. for array: Sum_{n >= 1, k >= 1} T(n, k)*x^k*y^n = x*y/((1 - x)*(1 - x - y - x*y)).
The n-th row entries are the values [p_n(k)], k >= 1, of the polynomial function p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(x,k). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_n(x) is the unique polynomial solution of the difference equation x*( f(x+1) - f(x-1) ) = 2*n*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_n(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (-1)^(n+1) * (2*n)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n)). For example, n = 3 gives log(2) = 4/6 + (1/6)*(1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ...). See A142983 for further details.
From Peter Bala, Oct 02 2008: (Start)
The odd-indexed columns of this array form the array A142992 of crystal ball sequences for lattices of type C_n.
Conjectural congruences for main diagonal entries: Put A(n) = T(n,n). Calculation suggests the following congruences: for prime p > 3 and m, r >= 1, A(m*p^r) == A(m*p^(r-1)) (mod p^(3*r));
Sum_{k = 0..p-1} A(k)^2 == 0 (mod p) if p is a prime of the form 8*n+1 or 8*n+7;
Sum_{k = 0..p-1} A(k)^2 == -1 (mod p) if p is a prime of the form 8*n+3 or 8*n+5.
(End)
From Peter Bala, Sep 27 2021: (Start)
T(n,k) = (1/2)*Sum_{i = 0..k} binomial(k,i)*binomial(n+k-1-i,k-1).
T(n,k) = (1/2)*[x^n] ((1+x)/(1-x))^k = (1/2)*(k/n)*[x^k] ((1+x)/(1-x))^n.
n*T(n,k) = 2*k*T(n-1,k) + (n - 2)*T(n-2,k). (End)
A(n,k) = k*hypergeom([1 - n, 1 - k], [2], 2). - Peter Luschny, Mar 23 2023
T(n,k) = 2*(Sum_{j=1..k-1} T(n-1,j)) + T(n-1,k) for n > 1. - Robert FERREOL, Jun 25 2024
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