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 1-5 of 5 results.

A063886 Number of n-step walks on a line starting from the origin but not returning to it.

Original entry on oeis.org

1, 2, 2, 4, 6, 12, 20, 40, 70, 140, 252, 504, 924, 1848, 3432, 6864, 12870, 25740, 48620, 97240, 184756, 369512, 705432, 1410864, 2704156, 5408312, 10400600, 20801200, 40116600, 80233200, 155117520, 310235040, 601080390, 1202160780, 2333606220, 4667212440
Offset: 0

Views

Author

Henry Bottomley, Aug 28 2001

Keywords

Comments

A Chebyshev transform of A007877(n+1). The g.f. is transformed to (1+x)/((1-x)(1+x^2)) under the mapping G(x)->(1/(1+x^2))G(1/(1+x^2)). - Paul Barry, Oct 12 2004
a(n-1) = 2*C(n-2, floor((n-2)/2)) is also the number of bit strings of length n in which the number of 00 substrings is equal to the number of 11 substrings. For example, when n = 4 we have 4 such bit strings: 0011, 0101, 1010, and 1100. - Angel Plaza, Apr 23 2009
Hankel transform is A120617. - Paul Barry, Aug 10 2009
The Hankel transform of a(n) is (-2)^C(n+1,2). The Hankel transform of (-1)^C(n+1,2)*a(n) is (-1)^C(n+1,2)*A164584(n). - Paul Barry, Aug 17 2009
For n > 1, a(n) is also the number of n-step walks starting from the origin and returning to it exactly once. - Geoffrey Critzer, Jan 24 2010
-a(n) is the Z-sequence for the Riordan array A130777. (See the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices). - Wolfdieter Lang, Jul 12 2011
Number of subsets of {1,...,n} in which the even elements appear as often at even positions as at odd positions. - Gus Wiseman, Mar 17 2018

Examples

			a(4) = 6 because there are six length four walks that do not return to the origin: {-1, -2, -3, -4}, {-1, -2, -3, -2}, {-1, -2, -1, -2}, {1, 2, 1, 2}, {1, 2, 3, 2}, {1, 2, 3, 4}. There are also six such walks that return exactly one time: {-1, -2, -1, 0}, {-1, 0, -1, -2}, {-1, 0, 1, 2}, {1, 0, -1, -2}, {1, 0, 1, 2}, {1, 2, 1, 0}. - _Geoffrey Critzer_, Jan 24 2010
The a(5) = 12 subsets in which the even elements appear as often at even positions as at odd positions: {}, {1}, {3}, {5}, {1,3}, {1,5}, {2,4}, {3,5}, {1,2,4}, {1,3,5}, {2,4,5}, {1,2,4,5}. - _Gus Wiseman_, Mar 17 2018
		

Crossrefs

Programs

  • Magma
    [1] cat [2*Binomial(n-1, Floor((n-1)/2)): n in [1..40]]; // G. C. Greubel, Jun 07 2023
    
  • Maple
    seq(seq(binomial(2*j,j)*i, i=1..2),j=0..16); # Zerinvary Lajos, Apr 28 2007
    # second Maple program:
    a:= proc(n) option remember; `if`(n<2, n+1,
           4*a(n-2) +2*(a(n-1) -4*a(n-2))/n)
        end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Feb 10 2014
    # third program:
    A063886 := series(BesselI(0, 2*x)*(1 + x*2 + x*Pi*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x), x = 0, 34): seq(n!*coeff(A063886, x, n), n = 0 .. 33); # Mélika Tebni, Jun 17 2024
  • Mathematica
    Table[Length[Select[Map[Accumulate, Strings[{-1, 1}, n]], Count[ #, 0] == 0 &]], {n, 0, 20}] (* Geoffrey Critzer, Jan 24 2010 *)
    CoefficientList[Series[Sqrt[(1+2x)/(1-2x)],{x,0,40}],x] (* Harvey P. Dale, Apr 28 2016 *)
  • PARI
    a(n)=(n==0)+2*binomial(n-1,(n-1)\2)
    
  • PARI
    a(n) = 2^n*prod(k=0,n-1,(k/n+1/n)^((-1)^k)); \\ Michel Marcus, Dec 03 2013
    
  • Python
    from math import ceil
    from sympy import binomial
    def a(n):
        if n==0: return 1
        return 2*binomial(n-1,(n-1)//2)
    print([a(n) for n in range(18)])
    # David Nacin, Feb 29 2012
    
  • SageMath
    [2*binomial(n-1, (n-1)//2) + int(n==0) for n in range(41)] # G. C. Greubel, Jun 07 2023

Formula

G.f.: sqrt((1+2*x)/(1-2*x)).
a(n+1) = 2*C(n, floor(n/2)) = 2*A001405(n); a(2n) = C(2n, n) = A000984(n) = 4*a(2n-2)-|A002420(n)| = 4*a(2n-2)-2*A000108(n-1) = 2*A001700(n-1); a(2n+1) = 2*a(2n) = A028329(n).
2*a(n) = A047073(n+1).
a(n) = Sum_{k=0..n} abs(A106180(n,k)). - Philippe Deléham, Oct 06 2006
a(n) = Sum_{k=0..n} (k+1)binomial(n, (n-k)/2) ( 1-cos((k+1)*Pi/2) (1+(-1)^(n-k))/(n+k+2) ). - Paul Barry, Oct 12 2004
G.f.: 1/(1-2*x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+ ... (continued fraction). - Paul Barry, Aug 10 2009
G.f.: 1 + 2*x/(G(0)-x+x^2) where G(k)= 1 - 2*x^2 - x^4/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 10 2012
D-finite with recurrence: n*a(n) = 2*a(n-1) + 4*(n-2)*a(n-2). - R. J. Mathar, Dec 03 2012
From Sergei N. Gladkovskii, Jul 26 2013: (Start)
G.f.: 1/G(0), where G(k) = 1 - 2*x/(1 + 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: G(0), where G(k) = 1 + 2*x/(1 - 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: W(0)/2*(1+2*x), where W(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/W(k+1) )), abs(x) < 1/2; (continued fraction). (End)
a(n) = 2^n*Product_{k=0..n-1} (k/n + 1/n)^((-1)^k). - Peter Luschny, Dec 02 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k+1)/((2*k+1)*(1+2*x) - (2*k+1)*(4*k+3)*x*(1+2*x)/((4*k+3)*x + (k+1)*(1+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 19 2014
From Peter Bala, Mar 29 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-1)^(n+k)*binomial(1/2, k)*binomial(- 1/2, n-k) = 2^n * A000246(n)/n!.
a(n) = (1/2^n) * binomial(2*n, n) * hypergeom([-1/2, -n], [1/2 - n], -1). (End)
E.g.f.: BesselI(0, 2*x)*(1 + x*(2 + Pi)*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x). - Stefano Spezia, May 11 2024
a(n) = A089849(n) + A138364(n). - Mélika Tebni, Jun 17 2024
From Amiram Eldar, Aug 15 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 + Pi/(9*sqrt(3)). (End)

A352833 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with k fixed points, k = 0, 1.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 2, 1, 3, 2, 4, 3, 6, 5, 8, 7, 12, 10, 16, 14, 23, 19, 30, 26, 42, 35, 54, 47, 73, 62, 94, 82, 124, 107, 158, 139, 206, 179, 260, 230, 334, 293, 420, 372, 532, 470, 664, 591, 835, 740, 1034, 924, 1288, 1148, 1588, 1422, 1962, 1756, 2404, 2161
Offset: 0

Views

Author

Gus Wiseman, Apr 08 2022

Keywords

Comments

A fixed point of a sequence y is an index y(i) = i. A fixed point of a partition is unique if it exists, so all columns k > 1 are zeros.
Conjecture:
(1) This is A064428 interleaved with A001522.
(2) Reversing rows gives A300788, the strict version of A300787.

Examples

			Triangle begins:
  0: {1,0}
  1: {0,1}
  2: {1,1}
  3: {2,1}
  4: {3,2}
  5: {4,3}
  6: {6,5}
  7: {8,7}
  8: {12,10}
  9: {16,14}
For example, row n = 7 counts the following partitions:
  (7)       (52)
  (61)      (421)
  (511)     (322)
  (43)      (3211)
  (4111)    (2221)
  (331)     (22111)
  (31111)   (1111111)
  (211111)
		

Crossrefs

Row sums are A000041.
The version for permutations is A008290, for nonfixed points A098825.
The columns appear to be A064428 and A001522.
The version counting strong nonexcedances is A114088.
The version for compositions is A238349, rank statistic A352512.
The version for reversed partitions is A238352.
Reversing rows appears to give A300788, the strict case of A300787.
A000700 counts self-conjugate partitions, ranked by A088902.
A115720 and A115994 count partitions by their Durfee square.
A330644 counts non-self-conjugate partitions, ranked by A352486.

Programs

  • Mathematica
    pq[y_]:=Length[Select[Range[Length[y]],#==y[[#]]&]];
    Table[Length[Select[IntegerPartitions[n],pq[#]==k&]],{n,0,15},{k,0,1}]

A026010 a(n) = number of (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n and s(0) = 2. Also a(n) = sum of numbers in row n+1 of array T defined in A026009.

Original entry on oeis.org

1, 2, 4, 7, 14, 25, 50, 91, 182, 336, 672, 1254, 2508, 4719, 9438, 17875, 35750, 68068, 136136, 260338, 520676, 999362, 1998724, 3848222, 7696444, 14858000, 29716000, 57500460, 115000920, 222981435, 445962870, 866262915, 1732525830, 3370764540
Offset: 0

Views

Author

Keywords

Comments

Conjecture: a(n) is the number of integer compositions of n + 2 in which the even parts appear as often at even positions as at odd positions (confirmed up to n = 19). - Gus Wiseman, Mar 17 2018

Examples

			The a(3) = 7 compositions of 5 in which the even parts appear as often at even positions as at odd positions are (5), (311), (131), (113), (221), (122), (11111). Missing are (41), (14), (32), (23), (212), (2111), (1211), (1121), (1112). - _Gus Wiseman_, Mar 17 2018
		

Crossrefs

Programs

  • Magma
    [(&+[Binomial(Floor((n+k)/2), Floor(k/2)): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Nov 08 2018
  • Mathematica
    Array[Sum[Binomial[Floor[(# + k)/2], Floor[k/2]], {k, 0, #}] &, 34, 0] (* Michael De Vlieger, May 16 2018 *)
    Table[2^(-1 + n)*(((2 + 3*#)*Gamma[(1 + #)/2])/(Sqrt[Pi]*Gamma[2 + #/2]) &[n + Mod[n, 2]]), {n,0,40}] (* Peter Pein, Nov 08 2018 *)
    Table[(1/2)^((5 - (-1)^n)/2)*(6*n + 7 - 3*(-1)^n)*CatalanNumber[(2*n + 1 - (-1)^n)/4], {n, 0, 40}] (* G. C. Greubel, Nov 08 2018 *)
  • PARI
    vector(40, n, n--; sum(k=0,n, binomial(floor((n+k)/2), floor(k/2)))) \\ G. C. Greubel, Nov 08 2018
    

Formula

a(2*n) = ((3*n + 1)/(2*n + 1))*C(2*n + 1, n)= A051924(1+n), n>=0, a(2*n-1) = a(2*n)/2 = A097613(1+n), n >= 1. - Herbert Kociemba, May 08 2004
a(n) = Sum_{k=0..n} binomial(floor((n+k)/2), floor(k/2)). - Paul Barry, Jul 15 2004
Inverse binomial transform of A005774: (1, 3, 9, 26, 75, 216, ...). - Gary W. Adamson, Oct 22 2007
Conjecture: (n+3)*a(n) - 2*a(n-1) + (-5*n-3)*a(n-2) + 2*a(n-3) + 4*(n-3)*a(n-4) = 0. - R. J. Mathar, Jun 20 2013
a(n) = (1/2)^((5 - (-1)^n)/2)*(6*n + 7 - 3*(-1)^n)*Catalan((2*n + 1 - (-1)^n)/4), where Catalan is the Catalan number = A000108. - G. C. Greubel, Nov 08 2018

A300788 Number of strict integer partitions of n in which the even parts appear as often at even positions as at odd positions.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 19, 23, 26, 30, 35, 42, 47, 54, 62, 73, 82, 94, 107, 124, 139, 158, 179, 206, 230, 260, 293, 334, 372, 420, 470, 532, 591, 664, 740, 835, 924, 1034, 1148, 1288, 1422, 1588, 1756, 1962, 2161, 2404
Offset: 0

Views

Author

Gus Wiseman, Mar 12 2018

Keywords

Examples

			The a(9) = 3 strict partitions: (9), (621), (531). Missing are: (81), (72), (63), (54), (432).
		

Crossrefs

Programs

  • Mathematica
    cobal[y_]:=Sum[(-1)^x,{x,Join@@Position[y,_?EvenQ]}];
    Table[Length[Select[IntegerPartitions[n],cobal[#]===0&&UnsameQ@@#&]],{n,0,40}]

Extensions

a(41)-a(58) from Alois P. Heinz, Mar 13 2018

A300789 Heinz numbers of integer partitions whose Young diagram can be tiled by dominos.

Original entry on oeis.org

1, 3, 4, 7, 9, 10, 12, 13, 16, 19, 21, 22, 25, 27, 28, 29, 34, 36, 37, 39, 40, 43, 46, 48, 49, 52, 53, 55, 57, 61, 62, 63, 64, 70, 71, 75, 76, 79, 81, 82, 84, 85, 87, 88, 89, 90, 91, 94, 100, 101, 107, 108, 111, 112, 113, 115, 116, 117, 118, 121, 129, 130, 131
Offset: 1

Views

Author

Gus Wiseman, Mar 12 2018

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
This sequence is conjectured to be the Heinz numbers of integer partitions in which the odd parts appear as many times in even as in odd positions.

Examples

			Sequence of integer partitions whose Young diagram can be tiled by dominos begins: (), (2), (11), (4), (22), (31), (211), (6), (1111), (8), (42), (51), (33), (222), (411).
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local k; for k from 1+
          `if`(n=1, 0, a(n-1)) while (l-> add(`if`(l[i]::odd,
           (-1)^i, 0), i=1..nops(l))<>0)(sort(map(i->
           numtheory[pi](i[1])$i[2], ifactors(k)[2]))) do od; k
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, May 22 2018
  • Mathematica
    primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Select[Range[100],Total[(-1)^Flatten[Position[primeMS[#],_?OddQ]]]===0&] (* Conjectured *)
Showing 1-5 of 5 results.