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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A349155 Numbers k such that the k-th composition in standard order has sum equal to negative twice its reverse-alternating sum.

Original entry on oeis.org

0, 9, 130, 135, 141, 153, 177, 193, 225, 2052, 2059, 2062, 2069, 2074, 2079, 2089, 2098, 2103, 2109, 2129, 2146, 2151, 2157, 2169, 2209, 2242, 2247, 2253, 2265, 2289, 2369, 2434, 2439, 2445, 2457, 2481, 2529, 2561, 2689, 2818, 2823, 2829, 2841, 2865, 2913
Offset: 1

Views

Author

Gus Wiseman, Nov 22 2021

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

Examples

			The terms and corresponding compositions begin:
     0: ()
     9: (3,1)
   130: (6,2)
   135: (5,1,1,1)
   141: (4,1,2,1)
   153: (3,1,3,1)
   177: (2,1,4,1)
   193: (1,6,1)
   225: (1,1,5,1)
  2052: (9,3)
  2059: (8,2,1,1)
  2062: (8,1,1,2)
  2069: (7,2,2,1)
  2074: (7,1,2,2)
  2079: (7,1,1,1,1,1)
  2089: (6,2,3,1)
  2098: (6,1,3,2)
  2103: (6,1,2,1,1,1)
		

Crossrefs

These compositions are counted by A224274 up to 0's.
An unordered version is A348617, counted by A001523 up to 0's.
The positive version is A349153, unreversed A348614.
The unreversed version is A349154.
Positive unordered unreversed: A349159, counted by A000712 up to 0's.
A positive unordered version is A349160, counted by A006330 up to 0's.
A003242 counts Carlitz compositions.
A011782 counts compositions.
A025047 counts alternating or wiggly compositions, complement A345192.
A034871, A097805, and A345197 count compositions by alternating sum.
A103919 counts partitions by alternating sum, reverse A344612.
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
Statistics of standard compositions:
- The compositions themselves are the rows of A066099.
- Number of parts is given by A000120, distinct A334028.
- Sum and product of parts are given by A070939 and A124758.
- Maximum and minimum parts are given by A333766 and A333768.
- Heinz number is given by A333219.
Classes of standard compositions:
- Partitions and strict partitions are ranked by A114994 and A333256.
- Multisets and sets are ranked by A225620 and A333255.
- Strict and constant compositions are ranked by A233564 and A272919.
- Carlitz compositions are ranked by A333489, complement A348612.
- Alternating compositions are ranked by A345167, complement A345168.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[ Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
    Select[Range[0,1000],Total[stc[#]]==-2*sats[stc[#]]&]

A349154 Numbers k such that the k-th composition in standard order has sum equal to negative twice its alternating sum.

Original entry on oeis.org

0, 12, 160, 193, 195, 198, 204, 216, 240, 2304, 2561, 2563, 2566, 2572, 2584, 2608, 2656, 2752, 2944, 3074, 3077, 3079, 3082, 3085, 3087, 3092, 3097, 3099, 3102, 3112, 3121, 3123, 3126, 3132, 3152, 3169, 3171, 3174, 3180, 3192, 3232, 3265, 3267, 3270, 3276
Offset: 1

Views

Author

Gus Wiseman, Nov 21 2021

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

Examples

			The terms and corresponding compositions begin:
       0: ()
      12: (1,3)
     160: (2,6)
     193: (1,6,1)
     195: (1,5,1,1)
     198: (1,4,1,2)
     204: (1,3,1,3)
     216: (1,2,1,4)
     240: (1,1,1,5)
    2304: (3,9)
    2561: (2,9,1)
    2563: (2,8,1,1)
    2566: (2,7,1,2)
    2572: (2,6,1,3)
    2584: (2,5,1,4)
		

Crossrefs

These compositions are counted by A224274 up to 0's.
Except for 0, a subset of A345919.
The positive version is A348614, reverse A349153.
An unordered version is A348617, counted by A001523.
The reverse version is A349155.
A positive unordered version is A349159, counted by A000712 up to 0's.
A000346 = even-length compositions with alt sum != 0, complement A001700.
A003242 counts Carlitz compositions.
A011782 counts compositions.
A025047 counts alternating or wiggly compositions, complement A345192.
A034871, A097805, and A345197 count compositions by alternating sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
Statistics of standard compositions:
- The compositions themselves are the rows of A066099.
- Number of parts is given by A000120, distinct A334028.
- Sum and product of parts are given by A070939 and A124758.
- Maximum and minimum parts are given by A333766 and A333768.
Classes of standard compositions:
- Partitions and strict partitions are ranked by A114994 and A333256.
- Multisets and sets are ranked by A225620 and A333255.
- Strict and constant compositions are ranked by A233564 and A272919.
- Carlitz compositions are ranked by A333489, complement A348612.
- Necklaces are ranked by A065609, dual A333764, reversed A333943.
- Alternating compositions are ranked by A345167, complement A345168.

Programs

  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],Total[stc[#]]==-2*ats[stc[#]]&]

A117178 Riordan array (c(x^2)/sqrt(1-4*x^2), x*c(x^2)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 3, 0, 1, 0, 4, 0, 1, 10, 0, 5, 0, 1, 0, 15, 0, 6, 0, 1, 35, 0, 21, 0, 7, 0, 1, 0, 56, 0, 28, 0, 8, 0, 1, 126, 0, 84, 0, 36, 0, 9, 0, 1, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1, 462, 0, 330, 0, 165, 0, 55, 0, 11, 0, 1, 0, 792, 0, 495, 0, 220, 0, 66, 0, 12, 0, 1
Offset: 0

Views

Author

Paul Barry, Mar 01 2006

Keywords

Comments

Row sums are A058622(n+1). Diagonal sums are A001791(n+1), with interpolated zeros. Inverse is A117179.
De-aerated and rows reversed, this matrix apparently becomes A014462. The nonzero antidiagonals are embedded in several entries and apparently contain partial sums of previous nonzero antidiagonals. - Tom Copeland, May 30 2017

Examples

			Triangle begins
    1;
    0,  1;
    3,  0,  1;
    0,  4,  0,  1;
   10,  0,  5,  0,  1;
    0, 15,  0,  6,  0,  1;
   35,  0, 21,  0,  7,  0,  1;
    0, 56,  0, 28,  0,  8,  0,  1;
  126,  0, 84,  0, 36,  0,  9,  0,  1;
		

Crossrefs

Programs

  • Magma
    [(1+(-1)^(n-k))*Binomial(n+1, Floor((n-k)/2))/2: k in [0..n], n in [0..15]]; // G. C. Greubel, Aug 08 2022
    
  • Mathematica
    T[n_, k_]:= Binomial[n+1, (n-k)/2]*(1+(-1)^(n-k))/2;
    Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 08 2022 *)
  • SageMath
    def A117178(n,k): return (1 + (-1)^(n-k))*binomial(n+1, (n-k)//2)/2
    flatten([[A117178(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Aug 08 2022

Formula

T(n,k) = C(n+1, (n-k)/2) * (1 + (-1)^(n-k))/2.
Column k has e.g.f. Bessel_I(k,2x) + Bessel_I(k+2, 2x).
From G. C. Greubel, Aug 08 2022: (Start)
Sum_{k=0..n} T(n, k) = A058622(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = ((1+(-1)^n)/2) * A001791((n+2)/2).
T(2*n, n) = ((1+(-1)^n)/2) * A052203(n/2).
T(2*n+1, n) = ((1-(-1)^n)/2) * A224274((n+1)/2).
T(2*n-1, n-1) = ((1+(-1)^n)/2) * A224274(n/2). (End)

A227726 a(n) = [x^n] (1 + x)/(1 - x)^(2*n+1).

Original entry on oeis.org

1, 4, 20, 112, 660, 4004, 24752, 155040, 980628, 6249100, 40060020, 258048960, 1668903600, 10829900592, 70480305440, 459823234112, 3006465218196, 19694758782300, 129235131438140, 849311959095600, 5589126007740660, 36825913869817380, 242910890642347200
Offset: 0

Views

Author

Paul D. Hanna, Jul 22 2013

Keywords

Examples

			G.f.: A(x) = 1 + 4*x + 20*x^2 + 112*x^3 + 660*x^4 + 4004*x^5 + ...
where a(n) equals the coefficient of x^n in (1+x)/(1-x)^(2*n+1)
and forms the main diagonal in the following table of coefficients:
(1+x)/(1-x)^1: [1, 2, 2, 2, 2, 2, 2, 2, 2, ...];
(1+x)/(1-x)^3: [1, 4, 9, 16, 25, 36, 49, 64, 81, ...];
(1+x)/(1-x)^5: [1, 6, 20, 50, 105, 196, 336, 540, ...];
(1+x)/(1-x)^7: [1, 8, 35, 112, 294, 672, 1386, 2640, ...];
(1+x)/(1-x)^9: [1, 10, 54, 210, 660, 1782, 4290, 9438, ...];
(1+x)/(1-x)^11:[1, 12, 77, 352, 1287, 4004, 11011, 27456, ...];
(1+x)/(1-x)^13:[1, 14, 104, 546, 2275, 8008, 24752, 68952, ...];
(1+x)/(1-x)^15:[1, 16, 135, 800, 3740, 14688, 50388, 155040, ...]; ...
Related series is G(x) = 1 + x*G(x)^3, which begins:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 +...+ A001764(n)*x^n +...
		

Crossrefs

Programs

  • Mathematica
    Join[{1},Table[Binomial[3n,n]+Binomial[3n-1,n-1],{n,30}]] (* Harvey P. Dale, Jan 15 2015 *)
  • PARI
    {a(n)=binomial(3*n,n)+binomial(3*n-1,n-1)}
    
  • PARI
    {a(n)=2*binomial(3*n-1, n) - 0^n}
    
  • PARI
    {a(n)=polcoeff((1+x)/(1-x+x*O(x^n))^(2*n+1),n)}
    for(n=0,30,print1(a(n),", "))

Formula

a(n) = binomial(3*n, n) + binomial(3*n-1, n-1).
G.f.: (2*G(x) - 1) / (3 - 2*G(x)), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
2*n*(2*n - 1)*a(n) - 3*(3*n - 1)*(3*n - 2)*a(n-1)=0. - R. J. Mathar, Jul 28 2013
a(n) = Sum_{r = 0..n-1} C(n-1,r)*C(2*n,r) + Sum_{r = 0..n} C(n,r)*C(2*n,n + r) - J. M. Bergot, Mar 18 2014
From Peter Bala, Jul 12 2016: (Start)
For n >= 1, a(n) = 4*binomial(3*n - 1, n - 1) = 4*A025174(n).
a(n) = [x^n]( 1/C(-x)^4 )^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274. (End)

A260687 Triangular array with n-th row giving coefficients of polynomial Product_{k = 2..n} (k + n*t) for n >= 1.

Original entry on oeis.org

1, 2, 2, 6, 15, 9, 24, 104, 144, 64, 120, 770, 1775, 1750, 625, 720, 6264, 20880, 33480, 25920, 7776, 5040, 56196, 250096, 571095, 708295, 453789, 117649, 40320, 554112, 3127040, 9433088, 16486400, 16744448, 9175040, 2097152, 362880, 5973264, 41229324, 156498804
Offset: 1

Views

Author

Peter Bala, Nov 16 2015

Keywords

Comments

Related to A220883 and A251592.

Examples

			Triangle begins
...1
...2      2
...6     15       9
..24    104     144      64
.120    770    1775    1750     625
.720   6264   20880   33480   25920    7776
5040  56196  250096  571095  708295  453789  117649
...
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998

Crossrefs

A000142 (column 0), A000169 (main diagonal), A006675 (column 1). Cf. A001700, A025174, A056856, A163456, A220883, A224274, A251592.

Programs

  • Maple
    seq(seq(coeff(mul(n*t + k, k = 2 .. n), t, i), i = 0..n-1), n = 1..10);

Formula

E.g.f. (with constant term 1 included): A(x,t) = [ 1/x*Revert( x*(1 - x)^t ) ]^(1/t) = Sum_{n >= 0} 1/(n*t + 1)*binomial(n*t + n,n)*x^n = 1 + x + (2 + 2*t)*x^2/2! + (2 + 3*t)*(3 + 3*t)*x^3/3! + (2 + 4*t)*(3 + 4*t)*(4 + 4*t)*x^4/4! + ..., where Revert denotes the series reversion operator with respect to x.
In the notation of the Bala link, A(x,t) = I^t(1/(1 - x)) where I^t is a fractional inversion operator.
A(x,t) = B_(1+t)(x), where B_t(x) is the e.g.f. for A251592 and is the generalized binomial series of Lambert. See Graham et al., Section 5.4 and Section 7.5.
A(x,t)^m = Sum_{n >= 0} m/(n*t + m)*binomial(n*t + n + m - 1,n)*x^n = 1 + m*x + m*(2*t + m + 1)*x^2/2! + m*(3*t + m + 1)*(3*t + m + 2)*x^3/3! + m*(4*t + m + 1)*(4*t + m + 2)*(4*t + m + 3)*x^4/4! + ....
A(x,t)^t = 1 + t*x + t(1 + 3*t)*x^2/2! + t*(1 + 4*t)*(2 + 4*t)*x^3/3! + t*(1 + 5*t)*(2 + 5*t)*(3 + 5*t)*x^4/4! + ... is the e.g.f for A220883 with an extra constant term 1 and an extra factor of t included.
t*log( A(x,t) ) = t*x + t*(1 + 2*t)*x^2/2! + t*(1 + 3*t)*(2 + 3*t)*x^3/3! + t*(1 + 4*t)*(2 + 4*t)*(3 + 4*t)*x^4/4! + ... is the e.g.f for A056856.
For n = 1,2,3,..., the sequence [x^n] A(x,t)^n = [1, (2*t + 3), (3*t + 4)*(3*t + 5)/2!, (4*t + 5)*(4*t + 6)*(4*t + 7)/3!, ...]. This sequence has the following specializations:
t = 0: [1, 3, 10, 35, 126, ...] = A001700 (with different offset).
t = 1: [1, 5, 28, 165, 1001, ...] = A025174.
t = 2: [1, 7, 55, 455, 3876, ...] = A224274.
t = 3: [1, 9, 91, 969, 10626, ...] = A163456.

A264773 Triangle T(n,k) = binomial(4*n - 3*k, 3*n - 2*k), 0 <= k <= n.

Original entry on oeis.org

1, 4, 1, 28, 5, 1, 220, 36, 6, 1, 1820, 286, 45, 7, 1, 15504, 2380, 364, 55, 8, 1, 134596, 20349, 3060, 455, 66, 9, 1, 1184040, 177100, 26334, 3876, 560, 78, 10, 1, 10518300, 1560780, 230230, 33649, 4845, 680, 91, 11, 1, 94143280, 13884156, 2035800, 296010, 42504, 5985, 816, 105, 12, 1
Offset: 0

Views

Author

Peter Bala, Nov 30 2015

Keywords

Comments

Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. for A002293 and f(x) = g(x)/(4 - 3*g(x)) = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + ... is the o.g.f. for A005810.
More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 3 and b = 2. See A092392, A264772, A264774 and A113139 for further examples.

Examples

			Triangle begins
  n\k |       0      1     2    3   4   5   6   7
------+-----------------------------------------------
   0  |       1
   1  |       4      1
   2  |      28      5     1
   3  |     220     36     6    1
   4  |    1820    286    45    7   1
   5  |   15504   2380   364   55   8   1
   6  |  134596  20349  3060  455  66   9   1
   7  | 1184040 177100 26334 3876 560  78  10   1
...
		

Crossrefs

A005810 (column 0), A052203 (column 1), A257633 (column 2), A224274 (column 3), A004331 (column 4). Cf. A002293, A007318, A092392 (C(2n-k,n)), A119301 (C(3n-k,n-k)), A264772, A264774.

Programs

  • Magma
    /* As triangle: */ [[Binomial(4*n-3*k, 3*n-2*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015
  • Maple
    A264773:= proc(n,k) binomial(4*n - 3*k, 3*n - 2*k); end proc:
    seq(seq(A264773(n,k), k = 0..n), n = 0..10);
  • Mathematica
    A264773[n_,k_] := Binomial[4*n - 3*k, n - k];
    Table[A264773[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Feb 06 2024 *)

Formula

T(n,k) = binomial(4*n - 3*k, n - k).
O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(4*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(3*n + 1)*binomial(4*n,n)*x^n.

A343093 Number of rooted toroidal maps with n edges and no isthmuses.

Original entry on oeis.org

1, 14, 159, 1680, 17147, 171612, 1696491, 16631840, 162090756, 1572801142, 15210259585, 146710561296, 1412132981778, 13569013500024, 130199055578307, 1247825314752768, 11947157409479180, 114288613130155608, 1092495810452593564, 10436544808441964352
Offset: 2

Views

Author

Andrew Howroyd, Apr 04 2021

Keywords

Crossrefs

Row sums of A343092.

Formula

Conjectural g.f.: A(x) = ( Sum_{n >= 1} (1/4)*binomial(4*n, n)*x^n )^2. - Peter Bala, Jul 24 2025

A359108 a(n) = A128899(2*n, n) = 2*binomial(4*n - 1, 3*n) for n >= 1 and a(0) = 1.

Original entry on oeis.org

1, 2, 14, 110, 910, 7752, 67298, 592020, 5259150, 47071640, 423830264, 3834669566, 34834267234, 317506779800, 2902365981900, 26597044596360, 244263468539790, 2247575790712824, 20716044882791720, 191230475831922200, 1767658071106087160, 16359617358545329440
Offset: 0

Views

Author

Peter Luschny, Dec 27 2022

Keywords

Crossrefs

Programs

  • Maple
    a := n -> ifelse(n = 0, 1, 2*binomial(4*n - 1, 3*n)):
    # Alternative:
    a := proc(n) option remember; ifelse(n < 2, n + 1, (8*(2*n - 1) * (4*n - 3) * (4*n - 1) * a(n - 1)) / (3 * n * (3*n - 2) * (3*n - 1))) end:
    seq(a(n), n = 0..19);
  • Mathematica
    A359108[n_] := If[n == 0, 1, 2*Binomial[4*n - 1, 3*n]];
    Array[A359108, 25, 0] (* Paolo Xausa, Sep 18 2024 *)

Formula

a(n) = (8*(2*n - 1) * (4*n - 3) * (4*n - 1) * a(n - 1)) / (3*n * (3*n - 2) * (3*n - 1)) for n >= 2.
a(n) = (1/2)*A005810(n) = 2*A224274(n) for n >= 1. - Peter Bala, Feb 08 2023
a(n) = [x^n] C(x)^(2*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. - Peter Bala, Apr 27 2023
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