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.

A287318 Square array A(n,k) = (2*n)! [x^n] BesselI(0, 2*sqrt(x))^k read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 6, 0, 1, 6, 36, 20, 0, 1, 8, 90, 400, 70, 0, 1, 10, 168, 1860, 4900, 252, 0, 1, 12, 270, 5120, 44730, 63504, 924, 0, 1, 14, 396, 10900, 190120, 1172556, 853776, 3432, 0, 1, 16, 546, 19920, 551950, 7939008, 32496156, 11778624, 12870, 0
Offset: 0

Views

Author

Peter Luschny, May 23 2017

Keywords

Examples

			Arrays start:
  k\n| 0   1    2      3        4          5           6
  ---|---------------------------------------------------------
  k=0| 1,  0,   0,     0,       0,         0,            0, ... A000007
  k=1| 1,  2,   6,    20,      70,       252,          924, ... A000984
  k=2| 1,  4,  36,   400,    4900,     63504,       853776, ... A002894
  k=3| 1,  6,  90,  1860,   44730,   1172556,     32496156, ... A002896
  k=4| 1,  8, 168,  5120,  190120,   7939008,    357713664, ... A039699
  k=5| 1, 10, 270, 10900,  551950,  32232060,   2070891900, ... A287317
  k=6| 1, 12, 396, 19920, 1281420,  96807312,   8175770064, ... A356258
  k=7| 1, 14, 546, 32900, 2570050, 238935564,  25142196156, ...
  k=8| 1, 16, 720, 50560, 4649680, 514031616,  64941883776, ...
  k=9| 1, 18, 918, 73620, 7792470, 999283068, 147563170524, ...
		

Crossrefs

Rows: A000007 (k=0), A000984 (k=1), A002894 (k=2), A002896 (k=3), A039699 (k=4), A287317 (k=5), A356258 (k=6).
Columns: A005843 (n=1), A152746 (n=2), 20*A169711 (n=3), 70*A169712 (n=4), 252*A169713 (n=5).
Main diagonal gives A303503.
Cf. A287316.

Programs

  • Maple
    A287318_row := proc(k, len) local b, ser;
    b := k -> BesselI(0, 2*sqrt(x))^k: ser := series(b(k), x, len);
    seq((2*i)!*coeff(ser,x,i), i=0..len-1) end:
    for k from 0 to 6 do A287318_row(k, 9) od;
  • Mathematica
    Table[Table[SeriesCoefficient[BesselI[0, 2 Sqrt[x]]^k, {x, 0, n}] (2 n)!, {n, 0, 6}], {k, 0, 6}]

Formula

A(n,k) = A287316(n,k) * binomial(2*n,n).

A303503 a(n) = (2*n)! * [x^(2*n)] BesselI(0,2*x)^n.

Original entry on oeis.org

1, 2, 36, 1860, 190120, 32232060, 8175770064, 2898980908824, 1369263687414480, 830988068906518380, 630109741730668410640, 583773362067938664133512, 648851848280206013365243776, 852146184628067383511375555000, 1305460597778526044143501996708800, 2307324514460203126471248458864413200
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 25 2018

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
          add(b(n-j, i-1)*binomial(n, j)^2, j=0..n))
        end:
    a:= n-> (2*n)!*b(n$2)/n!^2:
    seq(a(n), n=0..17);  # Alois P. Heinz, Jan 29 2023
  • Mathematica
    Table[(2 n)! SeriesCoefficient[BesselI[0, 2 x]^n, {x, 0, 2 n}], {n, 0, 15}]

Formula

a(n) = A287318(n,n).
a(n) ~ c * d^n * n^(2*n), where c = 1.72802011936236389522137050964080... and d = 1.1381284656425793765251319541847869000364101065484286935... - Vaclav Kotesovec, Apr 26 2018
a(n) = A000984(n)*A033935(n). - Alois P. Heinz, Jan 30 2023

A359801 Number of 4-dimensional cubic lattice walks that start and end at origin after 2n steps, not touching origin at intermediate stages.

Original entry on oeis.org

1, 8, 104, 2944, 108136, 4525888, 204981888, 9792786432, 486323201640, 24874892400064, 1302278744460352, 69474942954714112, 3764568243058030208, 206675027529594291200, 11473858525271117889536, 643154944963894079717376, 36355546411928157876528744, 2070313613815122857027563200
Offset: 0

Views

Author

Shel Kaphan, Mar 08 2023

Keywords

Comments

In Novak's note it is mentioned that if P(z) and Q(z) are the g.f.s for the probabilities of indecomposable and decomposable loops, respectively, that P(z) = 1 - 1/Q(z). This works equally well using loop counts rather than probabilities. The g.f.s may be expressed by the series constructed from the sequences of counts of loops of length 2*n. Q(z) for the 4-d case is the series corresponding to A039699.
To obtain the probability of returning to the point of origin for the first time after 2*n steps, divide a(n) by the total number of walks of length 2*n in d dimensions: (2*d)^(2*n) = 64^n.

Crossrefs

Cf. A039699, A287317 (number of walks that return to the origin in 2n steps).
Number of walks that return to the origin for the first time in 2n steps, in 1..3 dimensions: |A002420|, A054474, A049037.
Column k=4 of A361397.

Programs

  • Mathematica
    walk4d[n_] :=
     Sum[(2 n)!/(i! j! k! (n - i - j - k)!)^2, {i, 0, n}, {j, 0,
       n - i}, {k, 0, n - i - j}]; invertSeq[seq_] :=
      CoefficientList[1 - 1/SeriesData[x, 0, seq, 0, Length[seq], 1], x]; invertSeq[Table[walk4d[n], {n, 0, 17}]]
  • PARI
    seq(n) = {my(v=Vec(2 - 1/serlaplace(besseli(0, 2*x + O(x^(2*n+1)))^4))); vector(n+1, i, v[2*i-1])} \\ Andrew Howroyd, Mar 08 2023

Formula

G.f.: 2 - 1/Q(x) where Q(x) is the g.f. of A039699.
G.f.: 2 - 1/Integral_{t=0..oo} exp(-t)*BesselI(0,2*t*sqrt(x))^4 dt.
INVERTi transform of A039699.

A356258 Number of 6-dimensional cubic lattice walks that start and end at origin after 2n steps, free to pass through origin at intermediate stages.

Original entry on oeis.org

1, 12, 396, 19920, 1281420, 96807312, 8175770064, 748315668672, 72729762868620, 7402621930738320, 781429888276676496, 84955810313787521472, 9463540456205136873936, 1075903653146632508721600, 124461755084172965028753600, 14615050011682746903615601920
Offset: 0

Views

Author

Dave R.M. Langers, Oct 12 2022

Keywords

Examples

			a(1)=12, because twelve paths start at the origin, visit one of the adjacent vertices, and immediately return to the origin, resulting in 12 different paths of length 2n=2*1=2.
		

Crossrefs

Row k=6 of A287318.
1-5 dimensional analogs are A000984, A002894, A002896, A039699, A287317.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
          add(b(n-j, i-1)*binomial(n, j)^2, j=0..n))
        end:
    a:= n-> (2*n)!*b(n, 6)/n!^2:
    seq(a(n), n=0..15);  # Alois P. Heinz, Jan 30 2023

Formula

E.g.f.: Sum_{n>=0} a(2*n) * x^(2*n)/(2*n)! = I_0(2*x)^6. (I = Modified Bessel function first kind).
a(n) = Sum_{h+i+j+k+l+m=n, 0<=h,i,j,k,l,m<=n} multinomial(2n [h,h,i,i,j,j,k,k,l,l,m,m]). - Shel Kaphan, Jan 29 2023

A361364 Number of 5-dimensional cubic lattice walks that start and end at origin after 2n steps, not touching origin at intermediate stages.

Original entry on oeis.org

1, 10, 170, 6500, 332050, 19784060, 1296395700, 90616189800, 6637652225250, 503852804991500, 39337349077483420, 3142010167321271000, 255747325678297576100, 21150729618673827139000, 1773152567858996728205000, 150409554094012703302602000, 12890454660664800562838261250
Offset: 0

Views

Author

Shel Kaphan, Mar 09 2023

Keywords

Comments

In Novak's note it is mentioned that if P(z) and Q(z) are the g.f.s for the probabilities of indecomposable and decomposable loops, respectively, that P(z) = 1 - 1/Q(z). This works equally well using loop counts rather than probabilities. The g.f.s may be expressed by the series constructed from the sequences of counts of loops of length 2*n. Q(z) for the 5-d case is the series corresponding to A287317.
To satisfy this g.f. equation, a(0) should be 0, but we give it as 1 since there is one trivial loop of 0 steps, and for consistency with related sequences.
To obtain the probability of returning to the point of origin for the first time after 2*n steps, divide a(n) by the total number of walks of length 2*n in d dimensions: (2*d)^(2*n) = 100^n.

Crossrefs

Cf. A287317, A039699 (number of walks that return to the origin in 2n steps).
Number of walks that return to the origin for the first time in 2n steps, in 1..4 dimensions: |A002420|, A054474, A049037, A359801.
Column k=5 of A361397.
Cf. A169714.

Programs

  • Mathematica
    walk5d[n_] :=
     Sum[(2 n)!/(i! j! k! l! (n - i - j - k - l)!)^2, {i, 0, n}, {j, 0,
       n - i}, {k, 0, n - i - j}, {l, 0, n - i - j - k}]; invertSeq[seq_] :=
     CoefficientList[1 - 1/SeriesData[x, 0, seq, 0, Length[seq], 1], x]; invertSeq[Table[walk5d[n], {n, 0, 15}]]

Formula

G.f.: 2 - 1/Integral_{t=0..oo} exp(-t)*BesselI(0,2*t*sqrt(x))^5 dt.
INVERTi transform of A169714.

A328918 a(n) is the number of ordered pairs of positive integers (x, y) with x + y = 10^n, where x and y each have exactly n-digits but with initial zero digits allowed, and as strings, x and y are permutations of each other.

Original entry on oeis.org

1, 1, 11, 11, 281, 281, 11181, 11181, 563131, 563131, 32795191, 32795191, 2103687091, 2103687091, 144420919291, 144420919291, 10421915468041, 10421915468041, 781300466839541, 781300466839541, 60358948031151561, 60358948031151561, 4777791013174712961
Offset: 1

Views

Author

Dr. Michael W. Ecker, Oct 30 2019

Keywords

Comments

Published with slightly different wording in Mathematics Magazine, Problem 1016, Dec. 1977.
Analyzed for n = 1, 2, 3; computer-verified for n up to 8.
All solutions consist of an even number of digits followed by the digit 5 followed by zero or more 0's. This pattern means that a(2*n-1) = a(2*n). The initial segment consists of pairs of digits that add to 9 (0 with 9, 1 with 8, etc) arranged in arbitrary order and in particular leading 0's are permitted by the definition of the problem. A287317(k) gives the number of such arrangement with k pairs. For example, 339606500 + 660393500 is a solution. - Andrew Howroyd, Nov 03 2019

Examples

			For n = 3, solutions are (095, 905), (185, 815), (275, 725), (365, 635), (455, 545), (500, 500), (545, 455), (635, 365), (725, 275), (815, 185), (905, 095).
		

Crossrefs

Cf. A287317.

Programs

  • PARI
    seq(n)={Vec(serlaplace(besseli(0,2*x + O(x*x^n))^5)/(1-x))} \\ Andrew Howroyd, Nov 03 2019

Formula

a(n) = Sum_{k=0..floor((n-1)/2)} A287317(k). - Andrew Howroyd, Nov 03 2019

Extensions

Terms a(9) and beyond from Andrew Howroyd, Nov 03 2019
Showing 1-6 of 6 results.