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.

Previous Showing 31-38 of 38 results.

A061290 Square array read by antidiagonals of T(n,k) = T(n-1,k) + T(n-1, floor(k/2)) with T(0,0)=1.

Original entry on oeis.org

1, 0, 2, 0, 1, 4, 0, 0, 3, 8, 0, 0, 1, 7, 16, 0, 0, 1, 4, 15, 32, 0, 0, 0, 4, 11, 31, 64, 0, 0, 0, 1, 11, 26, 63, 128, 0, 0, 0, 1, 5, 26, 57, 127, 256, 0, 0, 0, 1, 5, 16, 57, 120, 255, 512, 0, 0, 0, 1, 5, 16, 42, 120, 247, 511, 1024, 0, 0, 0, 0, 5, 16, 42, 99, 247, 502, 1023, 2048, 0, 0
Offset: 0

Views

Author

Henry Bottomley, May 22 2001

Keywords

Comments

Row sums give 3^n.

Examples

			T(9,3) = T(8,3) + T(8,floor(3/2)) = T(8,3) + T(8,1) = 247 + 255 = 502. Rows start (1,0,0,0,0,...), (2,1,0,0,0,...), (4,3,1,1,0,...), (8,7,4,4,1,...), etc.
		

Crossrefs

Row sums are A000244. Columns are A000079, A000225, A000295 twice, A002662 four times, A002663 eight times, A002664 sixteen times, A035038 thirty two times, etc.

Formula

T(n, k) = C(n, 0) + C(n, 1) + ... + C(n, n-ceiling(log_2(k+1))) = 2^n - C(n, 0) - C(n, 1) - ... - C(n, floor(log_2(k))) = A008949(n, n-A029837(k+1)) = A000079(n) - A008949(n, A000523(k)).

A132795 Triangle of Gely numbers, read by rows.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 5, 0, 1, 0, 16, 6, 1, 1, 0, 42, 56, 21, 0, 1, 0, 99, 316, 267, 36, 1, 1, 0, 219, 1408, 2367, 960, 85, 0, 1, 0, 466, 5482, 16578, 14212, 3418, 162, 1, 1, 0, 968, 19624, 99330, 153824, 77440, 11352, 341, 0, 1, 0, 1981, 66496, 534898, 1364848, 1233970, 389104, 36829, 672, 1
Offset: 0

Views

Author

Olivier Gérard, Aug 31 2007

Keywords

Comments

First row is for n=0. First column is for k=0.
Sum of rows is n! = permutations of n symbols (A000142)
These numbers are related to the Eulerian numbers A(n,k).
Third Column (k=2) is A002662(n+1).
Second Diagonal (k=n-1) is A132796.
Binomial transform of this triangle gives set partitions without singletons (in a form very close to array A105794).

Examples

			Triangle starts:
1;
1, 0;
1, 0, 1;
1, 0, 5, 0;
1, 0, 16, 6, 1;
1, 0, 42, 56, 21, 0;
...
		

References

  • Charles O. Gely, Un tableau de conversion des polynomes cyclotomiques cousin des nombres Euleriens, Preprint Univ. Paris 7, 1999.
  • Olivier Gérard, Quelques facons originales de compter les permutations, submitted to Knuth07.
  • Olivier Gérard and Karol Penson, Set partitions, Multiset permutations and bi-permutations, in preparation.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 1990, p. 269.

Crossrefs

Programs

  • PARI
    T(n,k)= sum(j=0, k, (-1)^j*binomial(n+1, j)*sum(m=0, n, (k-j)^m)); \\ Michel Marcus, Jun 04 2014

Formula

T(n,k) = sum(j=0..k, (-1)^j*C(n+1,j)*sum(m=0..n, (k-j)^m) ).

A156732 Triangle T(n, k) = ((n-2*k)^2/(n-k+1))*binomial(n+1, k+1), read by rows.

Original entry on oeis.org

0, 1, 1, 4, 0, 4, 9, 2, 2, 9, 16, 10, 0, 10, 16, 25, 27, 5, 5, 27, 25, 36, 56, 28, 0, 28, 56, 36, 49, 100, 84, 14, 14, 84, 100, 49, 64, 162, 192, 84, 0, 84, 192, 162, 64, 81, 245, 375, 270, 42, 42, 270, 375, 245, 81
Offset: 0

Views

Author

Roger L. Bagula, Feb 14 2009

Keywords

Comments

The general formula for integrals of the form int (x^(2*k))/((arcsin(x))^2) dx involves the triangular sequence t(2*k, n). For example, the solution to the integral Integral x^6/((arcsin(x))^2) dx involves the following sequence: -5*Si(arcsin(x)) + 27*Si(3*arcsin(x)) - 25*Si(5*arcsin(x)), where Si represents the sine integral. The sequence of integers 5, 27, 25 corresponds to the sixth row of this triangular sequence. The general formula for the integral Integral x^(2*k)/((arcsin(x))^2) dx is: (1/(2^(2*k)))*( -((2^(2*k)*sqrt(1-x^2)*(x^(2*k)))/arcsin(x) )+ (-1)^(k+1)*(1+(2*k))*Si((1+2*k)*arcsin(x)) + Sum_{n=1..k} (-1)^n*((1-2*n)^2/(k-n+1))*binomial(2*k, k+n)*Si((2*n-1)*arcsin(x)) ). - John M. Campbell, Sep 22 2010

Examples

			Triangle begins as:
   0;
   1,   1;
   4,   0,   4;
   9,   2,   2,   9;
  16,  10,   0,  10, 16;
  25,  27,   5,   5, 27, 25;
  36,  56,  28,   0, 28, 56,  36;
  49, 100,  84,  14, 14, 84, 100,  49;
  64, 162, 192,  84,  0, 84, 192, 162,  64;
  81, 245, 375, 270, 42, 42, 270, 375, 245, 81;
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968.

Crossrefs

Programs

  • Magma
    A156732:= func< n,k | ((n-2*k)^2/(n+2))*Binomial(n+2, k+1) >;
    [A156732(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 28 2021
  • Mathematica
    T[n_, k_]:= ((n-2*k)^2/(n-k+1))*Binomial[n+1, k+1];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Feb 28 2021 *)
  • Sage
    def A156732(n, k): return ((n-2*k)^2/(n+2))*binomial(n+2, k+1)
    flatten([[A156732(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 28 2021
    

Formula

T(n, k) = ((n-2*k)^2/(n-k+1))*binomial(n+1, k+1).
Sum_{k=0..floor(n/2)} T(n-1, k-1) = 2^n.
From G. C. Greubel, Feb 28 2021: (Start)
T(n, k) = T(n, n-k).
T(n, k) = ((n-2*k)^2/(n+2))*binomial(n+2, k+1).
Sum_{k=0..n} T(n, k) = 2*(2^(n+1) -n-2) = 4*A002662(n) + 2*n^2. (End)

Extensions

Edited by G. C. Greubel, Feb 28 2021

A193590 Augmentation of the Euler triangle A008292. See Comments.

Original entry on oeis.org

1, 1, 1, 1, 5, 2, 1, 16, 33, 8, 1, 42, 275, 342, 58, 1, 99, 1669, 6441, 5600, 718, 1, 219, 8503, 82149, 217694, 143126, 14528, 1, 466, 39076, 843268, 5466197, 10792622, 5628738, 466220, 1, 968, 168786, 7621160, 107506633, 509354984, 788338180
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2011

Keywords

Comments

For an introduction to the unary operation "augmentation" as applied to triangular arrays or sequences of polynomials, see A193091.
Regarding A193590, (column 1)=A002662, with general term 2^n-1-n(n+1)/2.

Examples

			First 5 rows of A193589:
1
1....1
1....5....2
1....16...33....8
1....42...275...342....58
		

Crossrefs

Programs

  • Mathematica
    p[n_, k_] :=
    Sum[((-1)^j)*((k + 1 - j)^(n + 1))*Binomial[n + 2, j], {j, 0, k + 1}]
    (* A008292, Euler triangle *)
    Table[p[n, k], {n, 0, 5}, {k, 0, n}]
    m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
    TableForm[m[4]]
    w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
    v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
    v[n_] := v[n - 1].m[n]
    TableForm[Table[v[n], {n, 0, 6}]]  (* A193590  *)
    Flatten[Table[v[n], {n, 0, 8}]]

A229611 Expansion of 1/((1-x)^3*(1-11x)).

Original entry on oeis.org

1, 14, 160, 1770, 19485, 214356, 2357944, 25937420, 285311665, 3138428370, 34522712136, 379749833574, 4177248169405, 45949729863560, 505447028499280, 5559917313492216, 61159090448414529, 672749994932559990, 7400249944258160080, 81402749386839761090
Offset: 0

Views

Author

Yahia Kahloune, Sep 26 2013

Keywords

Comments

This sequence was chosen to illustrate a method of matching generating functions and closed-form solutions: The general term associated with the generating function 1/((1-s*x)^3*(1-r*x)) with r>s>=1 is a(n) = [r^(n+3) - s^(n+1)*(s^2 + (r-s)*s*binomial(n+3,1) +(r-s)^2*binomial(n+3,2))] / (r-s)^3 .

Examples

			a(3) = (11^6 - (50*3^2+260*3 + 331))/1000 = 1770 .
		

Crossrefs

Programs

  • Magma
    [(11^(n+3) - (50*n^2 + 260*n + 331))/1000: n in [0..25]]; // Vincenzo Librandi, Sep 27 2013
  • Mathematica
    CoefficientList[Series[1/((1 - x)^3 (1 - 11 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Sep 27 2013 *)
    LinearRecurrence[{14,-36,34,-11},{1,14,160,1770},30] (* Harvey P. Dale, Apr 09 2016 *)

Formula

a(n) = (11^(n+3) - (1 + 10*C(n+3,1) + 100*C(n+3,2)))/1000 = (11^(n+3) - (50*n^2 + 260*n + 331))/1000.
a(n) = 14*a(n-1) -36*a(n-2) +34*a(n-3) -11*a(n-4). - Vincenzo Librandi, Sep 27 2013

A232774 Triangle T(n,k), read by rows, given by T(n,0)=1, T(n,1)=2^(n+1)-n-2, T(n,n)=(-1)^(n-1) for n > 0, T(n,k)=T(n-1,k)-T(n-1,k-1) for 1 < k < n.

Original entry on oeis.org

1, 1, 1, 1, 4, -1, 1, 11, -5, 1, 1, 26, -16, 6, -1, 1, 57, -42, 22, -7, 1, 1, 120, -99, 64, -29, 8, -1, 1, 247, -219, 163, -93, 37, -9, 1, 1, 502, -466, 382, -256, 130, -46, 10, -1, 1, 1013, -968, 848, -638, 386, -176, 56, -11, 1, 2036, -1981, 1816, -1486, 1024
Offset: 0

Views

Author

Philippe Deléham, Nov 30 2013

Keywords

Comments

Row sums are A000079(n) = 2^n.
Diagonal sums are A024493(n+1) = A130781(n).
Sum_{k=0..n} T(n,k)*x^k = -A003063(n+2), A159964(n), A000012(n), A000079(n), A001045(n+2), A056450(n), (-1)^(n+1)*A232015(n+1) for x = -2, -1, 0, 1, 2, 3, 4 respectively.

Examples

			Triangle begins:
  1;
  1,    1;
  1,    4,   -1;
  1,   11,   -5,   1;
  1,   26,  -16,   6,   -1;
  1,   57,  -42,  22,   -7,   1;
  1,  120,  -99,  64,  -29,   8,   -1;
  1,  247, -219, 163,  -93,  37,   -9,  1;
  1,  502, -466, 382, -256, 130,  -46, 10,  -1;
  1, 1013, -968, 848, -638, 386, -176, 56, -11, 1;
		

Crossrefs

Formula

G.f.: Sum_{n>=0, k=0..n} T(n,k)*y^k*x^n=(1+2*(y-1)*x)/((1-2*x)*(1+(y-1)*x)).
|T(2*n,n)| = 4^n = A000302(n).
T(n,k) = (-1)^(k-1) * (Sum_{i=0..n-k} (2^(i+1)-1) * binomial(n-i-1,k-1)) for 0 < k <= n and T(n,0) = 1 for n >= 0. - Werner Schulte, Mar 22 2019

A326292 Number of crossing integer partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 43, 57, 80, 105, 142, 186, 248, 320, 421, 539, 698, 889, 1140, 1438, 1827, 2291, 2882, 3593, 4489, 5559, 6902, 8503, 10484, 12853, 15763
Offset: 0

Views

Author

Gus Wiseman, Oct 03 2019

Keywords

Comments

A multiset partition is crossing if it has two blocks of the form {...x...y...}, {...z...t...} where x < z < y < t or z < x < t < y. An integer partition is crossing if, by replacing each part with its multiset of prime indices, we obtain a crossing multiset partition.

Examples

			The a(31) = 1 through a(36) = 7 partitions:
  21,10  21,10,1  21,10,2    21,10,3      21,10,4        21,10,5
                  21,10,1,1  21,10,2,1    21,10,2,2      21,10,3,2
                             21,10,1,1,1  21,10,3,1      21,10,4,1
                                          21,10,2,1,1    21,10,2,2,1
                                          21,10,1,1,1,1  21,10,3,1,1
                                                         21,10,2,1,1,1
                                                         21,10,1,1,1,1,1
		

Crossrefs

The Heinz numbers of these partitions are given by A324170.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    croXQ[stn_]:=MatchQ[stn,{_,{_,x_,_,y_,_},_,{_,z_,_,t_,_},_}/;x
    				

Extensions

More terms from Jinyuan Wang, Jun 28 2020

A357255 Triangular array: row n gives the recurrence coefficients for the sequence (c(k) = number of subsets of {1,2,...,n} that have at least k-1 elements) for k >= 1.

Original entry on oeis.org

2, 3, -2, 4, -5, 2, 5, -9, 7, -2, 6, -14, 16, -9, 2, 7, -20, 30, -25, 11, -2, 8, -27, 50, -55, 36, -13, 2, 9, -35, 77, -105, 91, -49, 15, -2, 10, -44, 112, -182, 196, -140, 64, -17, 2, 11, -54, 156, -294, 378, -336, 204, -81, 19, -2
Offset: 1

Views

Author

Clark Kimberling, Sep 24 2022

Keywords

Comments

n-th row sum = 1 for n >= 2.

Examples

			First 7 rows:
  2
  3      -2
  4      -5       2
  5      -9       7     -2
  6     -14      16     -9     2
  7     -20      30    -25    11     -2
  8     -27      50    -55    36    -13     2
Row 4 gives recurrence coefficients for the sequence
(r(k)) = (A002662(k)) = (0,0,0,1,5,16,42,99,219,...); i.e.,
r(k) = 5*r(k-1) - 9*r(k-2) + 7*r(k-3) - 2*r(k-4),
with initial values (r(0), r(1), r(2), r(3)) = (0,0,0,1).
(Here r(k) = number of subsets of {1,2,...,4} having at least 3 elements.)
		

Crossrefs

Cf. sequences generated by recurrences, by row, beginning with row 1: A000079, A000225, A000295, A002662, A002663, A002664, A035038, A035039.

Programs

  • Mathematica
    Table[Binomial[n, k]*(-1)^(k - 1)*(n + k)/n, {n, 1, 12}, {k, 1, n}]

Formula

T(n,k) = (-1)^(k-1) * (C(n,k) + C(n-1,k-1)), for n >= 1, k >= 1.
T(n,k) = (-1)^(k-1) * C(n,k)*(n+k)/n, for n >= 1, k >= 1.
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