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.

Showing 1-6 of 6 results.

A323233 Coefficients of polynomials p(n, x) generating the columns of A323224, triangle read by rows, T(n, k) for n >= 1 and k >= 0.

Original entry on oeis.org

1, 2, 2, 6, 15, 3, 24, 140, 48, 4, 120, 1750, 775, 110, 5, 720, 28644, 14550, 2670, 210, 6, 5040, 588588, 323008, 68775, 7105, 357, 7, 40320, 14592864, 8388800, 1962632, 239120, 16016, 560, 8, 362880, 423227376, 250742700, 62531532, 8502921, 680904, 32130, 828, 9
Offset: 1

Views

Author

Peter Luschny, Jan 27 2019

Keywords

Examples

			The triangle starts:
[ 1]       1;
[ 2]       2,         2;
[ 3]       6,        15,         3;
[ 4]      24,       140,        48,        4;
[ 5]     120,      1750,       775,      110,       5;
[ 6]     720,     28644,     14550,     2670,     210,      6;
[ 7]    5040,    588588,    323008,    68775,    7105,    357,     7;
[ 8]   40320,  14592864,   8388800,  1962632,  239120,  16016,   560,   8;
[ 9]  362880, 423227376, 250742700, 62531532, 8502921, 680904, 32130, 828, 9;
The first few polynomials are:
p[1](x) = 1;
p[2](x) = 2*x + 2!;
p[3](x) = 3*x*(x + 5) + 3!;
p[4](x) = 4*x*(x + 5)*(x + 7) + 4!;
p[5](x) = 5*x*(x + 5)*(x + 7)*(x + 10) + 5!;
p[6](x) = 6*x*(x + 7)*(x + 11)*(x^2 + 17*x + 62) + 6!;
p[7](x) = 7*x*(x + 6)*(x + 7)*(x + 11)*(x + 13)*(x + 14) + 7!;

Crossrefs

Cf. A323224, A034856, A323221, A323220, A014137, A014138.

Programs

  • Mathematica
    ogf[n_] := (2/(1 + Sqrt[1 - 4 x] ))^n  x/(1 - x);
ser[n_, len_] := CoefficientList[Series[ogf[n], {x, 0, (n + 1) len + 1}], x];
tab[k_, len_] := Table[{n, ser[n, k + 1][[k + 1]]}, {n, 0, len - 1}];
pol[n_] := n! InterpolatingPolynomial[tab[n, n + 1], x] // Expand;
row[n_] := CoefficientList[pol[n], x]; Table[row[n], {n, 1, 9}]

Formula

A323224(n, k) = p(k, n)/k!.
T(n, k) = [x^k] p(n, x).
p(n, 1)/n! and p(n, -1)/n! are versions of the partial sums of the Catalan numbers.

A114277 Sum of the lengths of the second ascents in all Dyck paths of semilength n+2.

Original entry on oeis.org

1, 5, 19, 67, 232, 804, 2806, 9878, 35072, 125512, 452388, 1641028, 5986993, 21954973, 80884423, 299233543, 1111219333, 4140813373, 15478839553, 58028869153, 218123355523, 821908275547, 3104046382351, 11747506651599
Offset: 0

Views

Author

Emeric Deutsch, Nov 20 2005

Keywords

Comments

Also number of Dyck paths of semilength n+4 having length of second ascent equal to three. Example: a(1)=5 because we have UD(UUU)DUDDD, UD(UUU)DDUDD, UD(UUU)DDDUD, UUD(UUU)DDDD and UUDD(UUU)DDD (second ascents shown between parentheses). Partial sums of A002057. Column 3 of A114276. a(n)=absolute value of A104496(n+3).
Also number of Dyck paths of semilength n+3 that do not start with a pyramid (a pyramid in a Dyck path is a factor of the form U^j D^j (j>0), starting at the x-axis; here U=(1,1) and D=(1,-1); this definition differs from the one in A091866). Equivalently, a(n)=A127156(n+3,0). Example: a(1)=5 because we have UUDUDDUD, UUDUDUDD, UUUDUDDD, UUDUUDDD and UUUDDUDD. - Emeric Deutsch, Feb 27 2007

Examples

			a(3)=5 because the total length of the second ascents in UD(U)DUD, UD(UU)DD, UUDD(U)D, UUD(U)DD and UUUDDD (shown between parentheses) is 5.

Links

Crossrefs

Cf. A002057, A114276, A104496, A127156, A279557.
Cf. A014137 (n=1), A014138 (n=2), A001453 (n=3), this sequence (n=4), A143955 (n=5), A323224 (array).

Programs

  • Maple
    a:=n->4*sum(binomial(2*j+3,j)/(j+4),j=0..n): seq(a(n),n=0..28);
  • Mathematica
    Table[4*Sum[Binomial[2j+3,j]/(j+4),{j,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 19 2012 *)
  • Python
    from functools import cache
@cache
def B(n, k):
    if n <= 0 or k <= 0: return 0
    if n == k: return 1
    return B(n - 1, k) + B(n, k - 1)
def A114277(n): return B(n + 5, n + 1)
print([A114277(n) for n in range(24)]) # Peter Luschny, May 16 2022

Formula

a(n) = 4*Sum_{j=0..n} binomial(2*j+3, j)/(j+4).
G.f.: C^4/(1-z), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function.
a(n) = c(n+3) - (c(0) + c(1) + ... + c(n+2)), where c(k)=binomial(2k,k)/(k+1) is a Catalan number (A000108). - Emeric Deutsch, Feb 27 2007
D-finite with recurrence: n*(n+4)*a(n) = (5*n^2 + 14*n + 6)*a(n-1) - 2*(n+1)*(2*n+3)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(2*n+7)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012
a(n) = exp((2*i*Pi)/3)-4*binomial(2*n+5,n+1)*hypergeom([1,3+n,n+7/2],[n+2,n+6],4)/ (n+5). - Peter Luschny, Feb 26 2017
a(n-1) = Sum_{i+j+k+lA000108 Catalan number. - Yuchun Ji, Jan 10 2019

Extensions

More terms from Emeric Deutsch, Feb 27 2007

A143955 Sum of the altitudes of the leftmost valleys of all Dyck paths of semilength n (if path has no valley, then this altitude is taken to be 0).

Original entry on oeis.org

0, 0, 0, 1, 6, 26, 101, 376, 1377, 5017, 18277, 66727, 244377, 898129, 3312554, 12260129, 45526754, 169588754, 633580634, 2373550184, 8914719134, 33562602134, 126640791884, 478848661898, 1814142235028, 6885560250148
Offset: 0

Views

Author

Emeric Deutsch, Oct 14 2008

Keywords

Comments

The positive terms form the partial sums of A000344.

Examples

			a(4)=6 because the Dyck paths of semilength 4 with leftmost valley at a positive altitude are UUDUDDUD, UUDUDUDD, UUDUUDDD, UUUDDUDD and UUUDUDDD, where U=(1,1) and D=(1,-1); these altitudes are 1, 1, 1, 1 and 2, respectively.

Links

Crossrefs

Cf. A000108, A000344, A097607, A323224.

Programs

  • Maple
    C:=((1-sqrt(1-4*z))*1/2)/z: G:=z^3*C^5/(1-z): Gser:=series(G,z=0,32): seq(coeff(Gser,z,n),n=0..27);
  • Mathematica
    CoefficientList[Series[x^3 ((1 - (1 - 4 x)^(1/2))/(2 x))^5/(1 - x), {x, 0, 40}], x] (* Vaclav Kotesovec, Mar 21 2014 *)
  • Maxima
    a(n):=5*sum(binomial(2*k,k-2)/(k+3),k,2,n-1); /* Vladimir Kruchinin, Mar 15 2016 */
  • Python
    from functools import cache
@cache
def B(n, k):
    if n <= 0 or k <= 0: return 0
    if n == k: return 1
    return B(n - 1, k) + B(n, k - 1)
def A143955(k):
    return B(k + 3, k - 2)
print([A143955(n) for n in range(26)]) # Peter Luschny, May 15 2022

Formula

a(n) = Sum_{k>=0} k*A097607(n,k).
G.f.: z^3*C^5/(1-z), where C=(1-sqrt(1-4*z))/(2*z) is the generating function of the Catalan numbers (A000108).
Conjecture: (n+2)*a(n) -4*(2*n+1)*a(n-1) +2*(10*n-9)*a(n-2) +17*(2-n)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Jul 24 2012
a(n) ~ 5*4^n/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 21 2014
a(n) = 5*Sum_{k=2..n-1}(binomial(2*k,k-2)/(k+3)). - Vladimir Kruchinin, Mar 15 2016

A323222 A(n, k) = [x^k] (1 - 4*x)^(-n/2)*x/(1 - x), square array read by ascending antidiagonals with n >= 0 and k >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 5, 9, 1, 0, 1, 7, 21, 29, 1, 0, 1, 9, 37, 85, 99, 1, 0, 1, 11, 57, 177, 341, 351, 1, 0, 1, 13, 81, 313, 807, 1365, 1275, 1, 0, 1, 15, 109, 501, 1593, 3579, 5461, 4707, 1, 0, 1, 17, 141, 749, 2811, 7737, 15591, 21845, 17577, 1
Offset: 0

Views

Author

Peter Luschny, Jan 24 2019

Keywords

Comments

General asymptotic formula for g.f. (1 - 4*x)^(-j/2)*x/(1 - x) and fixed j>0 is a(n) ~ n^(j/2 - 1) * 4^n / (3*Gamma(j/2)). - Vaclav Kotesovec, Jan 29 2019

Examples

			[n\k] 0  1   2    3     4      5       6       7        8         9
-------------------------------------------------------------------
[0]   0, 1,  1,   1,    1,     1,      1,      1,       1,        1, ... A057427
[1]   0, 1,  3,   9,   29,    99,    351,   1275,    4707,    17577, ... A006134
[2]   0, 1,  5,  21,   85,   341,   1365,   5461,   21845,    87381, ... A002450
[3]   0, 1,  7,  37,  177,   807,   3579,  15591,   67071,   285861, ... A277178
[4]   0, 1,  9,  57,  313,  1593,   7737,  36409,  167481,   757305, ... A014916
[5]   0, 1, 11,  81,  501,  2811,  14823,  74883,  366603,  1752273, ... A323223
[6]   0, 1, 13, 109,  749,  4589,  26093, 140781,  730605,  3679725, ...
[7]   0, 1, 15, 141, 1065,  7071,  43107, 247311, 1355847,  7175661, ...
[8]   0, 1, 17, 177, 1457, 10417,  67761, 411825, 2377905, 13191345, ...
[9]   0, 1, 19, 217, 1933, 14803, 102319, 656587, 3982195, 23104441, ...
Triangle given by antidiagonals:
0;
0, 1;
0, 1,  1;
0, 1,  3,   1;
0, 1,  5,   9,   1;
0, 1,  7,  21,  29,    1;
0, 1,  9,  37,  85,   99,    1;
0, 1, 11,  57, 177,  341,  351,    1;
0, 1, 13,  81, 313,  807, 1365, 1275,    1;
0, 1, 15, 109, 501, 1593, 3579, 5461, 4707, 1;

Crossrefs

Sums of antidiagonals are A323217. Main diagonal is A323219.
Rows: A057427 (n=0), A006134 (n=1), A002450 (n=2), A277178 (n=3), A014916 (n=4), A323223 (n=5).
Columns: A005408 (k=2), A059993 (k=3), A323218 (k=4).
Similar array based on Catalan numbers is A323224.

Programs

  • Maple
    Row := proc(n, len) local ogf, ser; ogf := (1 - 4*x)^(-n/2)*x/(1 - x);
ser := series(ogf, x, (n+1)*len+1); seq(coeff(ser, x, j), j=0..len) end:
for n from 0 to 9 do Row(n, 9) od;
  • Mathematica
    BF[N_, K_] := Module[{}, r[n_, k_] := FrobeniusSolve[ConstantArray[1, n], k];
X[n_] := Flatten[Table[r[N, j], {j, 0, n - 1}], 1];
CentralBinomial[n_] := Binomial[2 n, n];
Sum[Product[CentralBinomial[m[[i]]], {i, 1, N}], {m , X[K]}]];
Trow[n_] := Table[BF[n, k], {k, 0, 9}]; Table[Trow[n], {n, 1, 9}]

Formula

For n>0 and k>0 let X(n, k) denote the set of all tuples of length n with elements from {0, ..., k-1} with sum < k. Let b(m) = binomial(2*m, m). Then A(n, k) = Sum_{(j1,...,jn) in X(n, k)} b(j1)*b(j2)*...*b(jn).

A323221 a(n) = n*(n + 5)*(n + 7)/6 + 1.

Original entry on oeis.org

1, 9, 22, 41, 67, 101, 144, 197, 261, 337, 426, 529, 647, 781, 932, 1101, 1289, 1497, 1726, 1977, 2251, 2549, 2872, 3221, 3597, 4001, 4434, 4897, 5391, 5917, 6476, 7069, 7697, 8361, 9062, 9801, 10579, 11397, 12256, 13157, 14101, 15089, 16122, 17201, 18327, 19501
Offset: 0

Views

Author

Peter Luschny, Jan 25 2019

Keywords

Comments

a(n) is related to the total angular defect of certain polytopes. See Hilton and Pedersen, Cor. 1; compare A275874.

Examples

			For n = 2 the sum formula gives:
I(2) = {{0,0}, {0,1}, {1,0}, {0,2}, {1,1}, {2,0}, {0,3}, {1,2}, {2,1}, {3,0}};
a(2) = 1 + 1 + 1 + 2 + 1 + 2 + 5 + 2 + 2 + 5 = 22.

Links

Crossrefs

Çf. A323224 (column 4), A323233 (row 4), A034856 (first difference), A275874.

Programs

  • Maple
    a := n -> n*(35 + 12*n + n^2)/6 + 1:
seq(a(n), n = 0..45);
  • Mathematica
    a[n_] := n (35 + 12 n + n^2)/6 + 1;
Table[a[n], {n, 0, 45}]

Formula

Let I(n) denote the set of all tuples of length n with elements from {0, 1, 2, 3} with sum <= 3 and C(m) denote the m-th Catalan number. Then for n > 0
a(n) = Sum_{(j1,...,jn) in I(n)} C(j1)*C(j2)*...*C(jn).
a(n) = [x^n] (3*x^3 - 8*x^2 + 5*x + 1)/(x - 1)^4.
a(n) = n! [x^n] exp(x)*(x^3 + 15*x^2 + 48*x + 6)/6.
a(n) = a(n - 1)*(n*(n + 5)*(n + 7) + 6)/(n*(n + 2)*(n + 7) - 18) for n > 0.
a(n) = A323224(n, 4).
a(n) = A275874(n+4) + 1.

A323220 a(n) = n*(n + 5)*(n + 7)*(n + 10)/24 + 1.

Original entry on oeis.org

1, 23, 64, 131, 232, 376, 573, 834, 1171, 1597, 2126, 2773, 3554, 4486, 5587, 6876, 8373, 10099, 12076, 14327, 16876, 19748, 22969, 26566, 30567, 35001, 39898, 45289, 51206, 57682, 64751, 72448, 80809, 89871, 99672, 110251, 121648, 133904, 147061, 161162, 176251
Offset: 0

Views

Author

Peter Luschny, Jan 25 2019

Keywords

Crossrefs

Cf. A323224 (column 5), A323233 (row 5), A323221 (first diff.), A034856 (second diff.).

Programs

  • Maple
    a := n -> (n^4 + 22*n^3 + 155*n^2 + 350*n + 24)/24:
seq(a(n), n=0..40);

Formula

a(n) = [x^n] (8*x^4 - 31*x^3 + 41*x^2 - 18*x - 1)/(x - 1)^5.
a(n) = n! [x^n] exp(x)*(x^4 + 28*x^3 + 228*x^2 + 528*x + 24)/24.
a(n) = (1/3)*((2*n + 17)*a(n-3) - (3*n + 25)*a(n-2) + (n + 15)*a(n-1)) for n >= 3.
a(n) = A323224(n, 5).
Showing 1-6 of 6 results.

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