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 1-3 of 3 results.

A194805 Number of parts that are visible in one of the three views of the section model of partitions version "tree" with n sections.

Original entry on oeis.org

0, 1, 2, 4, 7, 11, 17, 25, 36, 51, 71, 97, 132, 177, 235, 310, 406, 527, 681, 874, 1116, 1418, 1793, 2256, 2829, 3532, 4393, 5445, 6727, 8282, 10168, 12445, 15190, 18491, 22452, 27192, 32859, 39613, 47651, 57199, 68522, 81920, 97756, 116434, 138435
Offset: 0

Views

Author

Omar E. Pol, Jan 27 2012

Keywords

Comments

The mentioned view of the section model looks like a tree (see example). Note that every column contains the same parts. For more information about the section model of partitions see A135010 and A194803.
Number of partitions of 2n-1 such that n-1 or n is a part, for n >=1. - Clark Kimberling, Mar 01 2014

Examples

			Illustration of one of the three views with seven sections:
.
.                   1
.                 2 1
.                   1 3
.                 2 1
.               4   1
.                   1 3
.                   1   5
.                 2 1
.               4   1
.             3     1
.           6       1
.                     3
.                       5
.                         4
.                           7
.
There are 25 parts that are visible, so a(7) = 25.
Using the formula we have a(7) = p(7) + p(7-1) - 1 = 15 + 11 - 1 = 25, where p(n) is the number of partitions of n.

Links

Crossrefs

Cf. A000041, A000065, A084376, A135010, A138121, A141285, A194550, A194803, A194804.

Programs

  • Mathematica
    Table[Count[IntegerPartitions[2 n - 1],  p_ /; Or[MemberQ[p, n - 1], MemberQ[p, n]]], {n, 50}]  (* Clark Kimberling, Mar 01 2014 *)
Table[PartitionsP[n] + PartitionsP[n-1] - 1, {n, 0, 44}] (* Robert Price, May 12 2020 *)

Formula

a(n) = A084376(n) - 1.
a(n) = A000041(n) + A000041(n-1) - 1, if n >= 1.
a(n) = A000041(n) + A000065(n-1), if n >= 1.

A309267 Expansion of (1 + x) * Product_{k>=1} 1/(1 - x^k)^k.

Original entry on oeis.org

1, 2, 4, 9, 19, 37, 72, 134, 246, 442, 782, 1359, 2338, 3964, 6652, 11046, 18176, 29631, 47935, 76931, 122608, 194072, 305269, 477258, 741977, 1147227, 1764778, 2701403, 4115892, 6242846, 9428575, 14181272, 21245738, 31708402, 47150928, 69867001, 103176007, 151864745, 222821779
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 20 2019

Keywords

Links

Crossrefs

Cf. A000219, A052816, A084376, A091360, A191659, A277963.

Programs

  • Maple
    G:= (1+x)/mul((1-x^k)^k,k=1..100):
S:= series(G,x,101):
seq(coeff(S,x,j),j=0..100); # Robert Israel, Dec 01 2020
  • Mathematica
    nmax = 38; CoefficientList[Series[(1 + x) Product[1/(1 - x^k)^k, {k, 1, nmax}], {x, 0, nmax}], x]
a[0] = 1; a[n_] := a[n] = Sum[DivisorSigma[2, k] a[n - k], {k, 1, n}]/n; Table[a[n] + a[n - 1], {n, 0, 38}]

Formula

a(n) = A000219(n) + A000219(n-1).
a(n) ~ Zeta(3)^(7/36) * 2^(25/36) * exp(3 * Zeta(3)^(1/3) * (n/2)^(2/3) + 1/12) / (A * sqrt(3*Pi) * n^(25/36)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Jul 20 2019

A309266 Expansion of (1 + x) * Product_{k>=1} (1 + x^k)/(1 - x^k).

Original entry on oeis.org

1, 3, 6, 12, 22, 38, 64, 104, 164, 254, 386, 576, 848, 1232, 1768, 2512, 3534, 4926, 6812, 9348, 12736, 17240, 23192, 31016, 41256, 54594, 71890, 94232, 122976, 159816, 206872, 266768, 342756, 438868, 560064, 712448, 903526, 1142478, 1440528, 1811384, 2271720, 2841800, 3546224
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 20 2019

Keywords

Crossrefs

Cf. A001650, A015128, A052816, A084376, A207641, A211971, A277643.

Programs

  • Mathematica
    nmax = 42; CoefficientList[Series[(1 + x) Product[(1 + x^k)/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = Sum[PartitionsP[k] PartitionsQ[n - k], {k, 0, n}]; Table[a[n] + a[n - 1], {n, 0, 42}]

Formula

G.f.: (1 + x)/theta_4(x), where theta_4() is the Jacobi theta function.
a(n) = A015128(n) + A015128(n-1).
a(n) ~ exp(Pi*sqrt(n)) / (4*n) * (1 - (Pi/4 + 1/Pi)/sqrt(n)). - Vaclav Kotesovec, Jul 20 2019
Showing 1-3 of 3 results.

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