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 1-9 of 9 results.

A115593 Number of forests of rooted trees with total weight n, where a node at height k has weight 2^k (with root considered to be at height 0).

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 6, 7, 10, 13, 17, 22, 29, 38, 50, 64, 82, 107, 136, 175, 224, 288, 363, 465, 587, 748, 942, 1196, 1503, 1902, 2385, 3004, 3765, 4729, 5911, 7406, 9246, 11549, 14395, 17941, 22326, 27767, 34501, 42821, 53134, 65828, 81546, 100871, 124783
Offset: 0

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Keywords

Comments

The sequence b(2n)=0, b(2n+1)=a(n) is the number of trees of weight n.

Examples

			a(3) = 2; one forest with 3 single-node trees and one with a single two-node tree (root node has weight 1, other node has weight 2).
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    b:= proc(n) local r; `if`(irem(n, 2, 'r')=0, 0, a(r)) end:
    a:= proc(n) option remember; `if`(n=0, 1,
           add(add(d*b(d), d=divisors(j))*a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..50);  # Alois P. Heinz, May 16 2013
  • Mathematica
    b[n_] := b[n] = If[{q, r} = QuotientRemainder[n, 2]; r == 0, 0, a[q]]; a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d*b[d], {d, Divisors[j]}]*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Mar 23 2015, after Alois P. Heinz *)
  • PARI
    {a(n)=my(A=1+x); for(i=1, n, A=exp(sum(m=1, n, subst(A,x,x^(2*m)+x*O(x^n))*x^m/m))); polcoeff(A,n)} /* Paul D. Hanna */

Formula

Euler transform of b(n), where b(2n+1) = a(n) and b(2n) = 0.
From Paul D. Hanna, Oct 26 2011: (Start)
G.f. satisfies: A(x) = exp( Sum_{n>=1} A(x^(2*n)) * x^n/n ).
G.f. satisfies: A(x)*A(-x) = A(x^2). (End)
Let b(n) = a(n-1) for n>=1, then sum(n>=1, b(n)*x^n ) = x / prod(n>=1, (1-x^(2*n-1))^b(n) ); compare to A000081, A004111, and A073075. - Joerg Arndt, Mar 04 2015
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} ( Sum_{d|k and d is odd} d * a(floor(d/2)) ) * a(n-k). - Seiichi Manyama, May 31 2023

A248869 Satisfies Sum_{n>=0} a(n)*x^n = x * Product_{n>=0} (1 + x^n + x^(2*n))^a(n).

Original entry on oeis.org

0, 1, 1, 2, 3, 7, 15, 34, 79, 190, 459, 1136, 2833, 7154, 18206, 46723, 120656, 313514, 818763, 2148434, 5660790, 14972103, 39734107, 105779291, 282403830, 755921733, 2028277115, 5454368549, 14697955778, 39682793675, 107330573239, 290783511134, 789032648219
Offset: 0

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Author

Joerg Arndt, Mar 04 2015

Keywords

Comments

What kind of trees are counted by this sequence (compare with A000081, A004111, A073075, and A115593)?
a(n) is the number of rooted trees of n vertices that have everywhere at most 2 siblings with the same (i.e., isomorphic) subtree below. The g.f. assembles a(n) as a root with child subtrees from among the smaller a(), but takes only 0, 1 or 2 copies of any one of them. Compare asymmetric trees A004111 g.f. which takes 0 or 1 copies. Here the x^(2*n) term allows a 2nd copy. The siblings condition is equivalent to the condition that the tree automorphisms form a 2-group, i.e., group order some power 2^k. 2 same siblings are a swap. 3 same siblings would be an element of order 3 and hence factor 3 in the group order. a(n) >= A213920 since the latter limits same size siblings, whereas here only limits same size plus structure. - Kevin Ryde, Jul 11 2019

Crossrefs

Programs

  • Maple
    h:= proc(n, m, t) option remember; `if`(m=0, binomial(n+t, t),
          `if`(n=0, 0, add(h(n-1, m-j, t+1), j=1..min(2, m))))
        end:
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(b(n-i*j, i-1)*h(a(i), j, 0), j=0..n/i)))
        end:
    a:= n-> `if`(n<2, n, b(n-1$2)):
    seq(a(n), n=0..35);  # Alois P. Heinz, Sep 04 2018
  • Mathematica
    h[n_, m_, t_] := h[n, m, t] = If[m == 0, Binomial[n + t, t], If[n == 0, 0, Sum[h[n - 1, m - j, t + 1], {j, 1, Min[2, m]}]]];
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[b[n - i j, i - 1]* h[a[i], j, 0], {j, 0, n/i}]]];
    a[n_] := If[n < 2, n, b[n - 1, n - 1]];
    a /@ Range[0, 32] (* Jean-François Alcover, Oct 02 2019, after Alois P. Heinz *)

Formula

a(n) ~ c * d^n / n^(3/2), where d = 2.8458470164106425911151048..., c = 0.41694347809945986693376... . - Vaclav Kotesovec, Mar 17 2015
a(n) = A004111(n) + A318859(n). - Kevin Ryde, Jul 11 2019

A306768 G.f. A(x) satisfies: A(x) = x*exp(-A(-x) + A(-x^2)/2 - A(-x^3)/3 + A(-x^4)/4 - A(-x^5)/5 + ...).

Original entry on oeis.org

0, 1, 1, -1, -2, 2, 6, -5, -18, 15, 59, -54, -215, 199, 813, -744, -3135, 2890, 12394, -11538, -50017, 46806, 204893, -192451, -849681, 800974, 3560927, -3367656, -15058478, 14279426, 64171736, -60992032, -275304665, 262199050, 1188070488, -1133572891, -5153913606
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 14 2019

Keywords

Examples

			G.f.: A(x) = x + x^2 - x^3 - 2*x^4 + 2*x^5 + 6*x^6 - 5*x^7 - 18*x^8 + 15*x^9 + 59*x^10 - 54*x^11 - 215*x^12 + ...
		

Crossrefs

Programs

  • Mathematica
    terms = 36; A[] = 0; Do[A[x] = x Exp[Sum[(-1)^k A[-x^k]/k, {k, 1, terms}]] + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = SeriesCoefficient[x Product[1/(1 + x^k)^((-1)^k a[k]), {k, 1, n - 1}], {x, 0, n}]; a[0] = 0; Table[a[n], {n, 0, 36}]

Formula

G.f.: A(x) = Sum_{n>=1} a(n)*x^n = x * Product_{n>=1} 1/(1 + x^n)^((-1)^n*a(n)).
Recurrence: a(n+1) = (1/n) * Sum_{k=1..n} ( Sum_{d|k} (-1)^(k/d+d)*d*a(d) ) * a(n-k+1).

A248870 Satisfies Sum_{n>=0} a(n)*x^n = x / Product_{n>=0} (1 - x^n/(1 - x^n))^a(n).

Original entry on oeis.org

0, 1, 1, 3, 8, 23, 62, 181, 513, 1513, 4476, 13483, 40933, 125845, 389769, 1217590, 3828775, 12115966, 38546124, 123238296, 395725493, 1275733730, 4127339091, 13396443708, 43610621823, 142354979662, 465838195260, 1527905193504, 5022061115901, 16539625666670, 54571760414658
Offset: 0

Views

Author

Joerg Arndt, Mar 04 2015

Keywords

Comments

Which kind of trees is counted by this sequence (compare to A000081, A004111, A073075, A248869 and A115593)?

Crossrefs

Cf. A248869.

A307538 G.f. A(x) satisfies: A(x) = x*exp(2*A(-x) + 2*A(-x^3)/3 + 2*A(-x^5)/5 + 2*A(-x^7)/7 + 2*A(-x^9)/9 + ...).

Original entry on oeis.org

0, 1, -2, -2, 10, 14, -86, -126, 858, 1302, -9378, -14606, 108954, 172698, -1319966, -2119118, 16489594, 26731542, -210887998, -344490170, 2747510514, 4515757426, -36336187630, -60023827438, 486540793914, 807121753178, -6582918170714, -10959656342678, 89860260268098
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 14 2019

Keywords

Examples

			G.f.: A(x) = x - 2*x^2 - 2*x^3 + 10*x^4 + 14*x^5 - 86*x^6 - 126*x^7 + 858*x^8 + 1302*x^9 - 9378*x^10 - 14606*x^11 + ...
		

Crossrefs

Programs

  • Mathematica
    terms = 28; A[] = 0; Do[A[x] = x Exp[Sum[2 A[-x^(2 k - 1)]/(2 k - 1), {k, 1, terms}]] + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = SeriesCoefficient[x Product[((1 + x^k)/(1 - x^k))^((-1)^k a[k]), {k, 1, n - 1}], {x, 0, n}]; a[0] = 0; Table[a[n], {n, 0, 28}]

Formula

G.f.: A(x) = Sum_{n>=1} a(n)*x^n = x * Product_{n>=1} ((1 + x^n)/(1 - x^n))^((-1)^n*a(n)).
Recurrence: a(n+1) = (2/n) * Sum_{k=1..n} ( Sum_{d|k, k/d odd} (-1)^d*d*a(d) ) * a(n-k+1).

A299021 G.f.: A(x) = Sum_{n>=1} a(n)*x^n = x*Product_{n>=1} (1 + a(n)*x^n)/(1 - a(n)*x^n).

Original entry on oeis.org

1, 2, 6, 22, 86, 358, 1558, 6966, 31894, 148918, 705062, 3380054, 16381158, 80056550, 394266950, 1955139942, 9749771926, 48873487942, 246160229782, 1244801094742, 6318514387638, 32184084454166, 164425969781062, 842429440124854, 4327629345403078, 22283328480744070
Offset: 1

Views

Author

Ilya Gutkovskiy, Jun 18 2018

Keywords

Examples

			G.f.: A(x) = x + 2*x^2 + 6*x^3 + 22*x^4 + 86*x^5 + ... = x * ((1 + x) * (1 + 2*x^2) * (1 + 6*x^3) * (1 + 22*x^4) * (1 + 86*x^5) * ...) / ((1 - x) * (1 - 2*x^2) * (1 - 6*x^3) * (1 - 22*x^4) * (1 - 86*x^5) * ...).
		

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = SeriesCoefficient[x Product[(1 + a[k] x^k)/(1 - a[k] x^k), {k, 1, n - 1}], {x, 0, n}]; a[1] = 1; Table[a[n], {n, 26}]
    a[n_] := a[n] = SeriesCoefficient[x Exp[Sum[Sum[(1 + (-1)^(k + 1)) a[j]^k x^(j k)/k, {j, 1, n - 1}], {k, 1, n - 1}]], {x, 0, n}]; a[1] = 1; Table[a[n], {n, 26}]

Formula

G.f.: A(x) = Sum_{n>=1} a(n)*x^n = x*exp(Sum_{k>=1} Sum_{n>=1} (1 + (-1)^(k+1))*a(n)^k*x^(n*k)/k).

A307722 G.f. A(x) satisfies: A(x) = x*exp(2*Sum_{n>=1} Sum_{k>=1} n*a(n)*x^(n*(2*k-1))/(2*k - 1)).

Original entry on oeis.org

0, 1, 2, 10, 78, 794, 9870, 143610, 2382350, 44266538, 909575170, 20468012850, 500542618118, 13218631046786, 374965272837542, 11372416113131346, 367296622702990270, 12587154399475110546, 456238999451039779510, 17440439387336903608866, 701272672299320517560470
Offset: 0

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Author

Ilya Gutkovskiy, Apr 24 2019

Keywords

Examples

			G.f.: A(x) = x + 2*x^2 + 10*x^3 + 78*x^4 + 794*x^5 + 9870*x^6 + 143610*x^7 + 2382350*x^8 + 44266538*x^9 + ...
		

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = SeriesCoefficient[x Exp[2 Sum[Sum[j a[j] x^(j (2 k - 1))/(2 k - 1), {k, 1, n - 1}], {j, 1, n - 1}]], {x, 0, n}]; a[1] = 1; Table[a[n], {n, 0, 20}]
    a[n_] := a[n] = SeriesCoefficient[x Product[((1 + x^k)/(1 - x^k))^(k a[k]), {k, 1, n - 1}], {x, 0, n}]; a[1] = 1; Table[a[n], {n, 0, 20}]

Formula

G.f.: A(x) = Sum_{n>=1} a(n)*x^n = x * Product_{n>=1} ((1 + x^n)/(1 - x^n))^(n*a(n)).

A308152 G.f.: x * Product_{j>=1, k>=1} ((1 + x^(j*k))/(1 - x^(j*k)))^a(j).

Original entry on oeis.org

1, 2, 8, 32, 138, 612, 2864, 13712, 67416, 337482, 1716208, 8837392, 45997032, 241571408, 1278625480, 6813568656, 36524390042, 196820310100, 1065583770168, 5793299764208, 31615962617272, 173131117881312, 951040865156928, 5239171609158304, 28937688613453048
Offset: 1

Views

Author

Ilya Gutkovskiy, May 14 2019

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = SeriesCoefficient[x Product[Product[((1 + x^(j k))/(1 - x^(j k)))^a[j], {k, 1, n - 1}], {j, 1, n - 1}], {x, 0, n}]; Table[a[n], {n, 1, 25}]
    terms = 25; A[] = 0; Do[A[x] = x Exp[2 Sum[Sum[A[x^(i (2 j - 1))]/(2 j - 1), {j, 1, terms}], {i, 1,terms}]] + O[x]^(terms + 1) // Normal, terms + 1]; Rest[CoefficientList[A[x], x]]

Formula

G.f. A(x) satisfies: A(x) = x * exp(2 * Sum_{i>=1} Sum_{j>=1} A(x^(i*(2*j-1)))/(2*j - 1)).

A308227 G.f.: (x/(1 - x)) * Product_{k>=1} ((1 + x^k)/(1 - x^k))^a(k).

Original entry on oeis.org

1, 3, 11, 47, 217, 1065, 5453, 28789, 155633, 857207, 4793103, 27136555, 155249971, 896133487, 5212477023, 30522169103, 179777122393, 1064411910393, 6331361864657, 37817265028841, 226731778956181, 1363993567341257, 8231111557650837, 49812263080757845
Offset: 1

Views

Author

Ilya Gutkovskiy, May 16 2019

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = SeriesCoefficient[x/(1 - x) Product[((1 + x^k)/(1 - x^k))^a[k], {k, 1, n - 1}], {x, 0, n}]; Table[a[n], {n, 1, 24}]
    terms = 24; A[] = 0; Do[A[x] = x Exp[Sum[2 A[x^(2 k - 1)]/(2 k - 1) + x^k/k, {k, 1, terms}]] + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x] // Rest

Formula

G.f. A(x) satisfies: A(x) = x * exp(Sum_{k>=1} 2*A(x^(2*k-1))/(2*k - 1) + x^k/k).
Showing 1-9 of 9 results.