Thotsaporn Thanatipanonda has authored 4 sequences.
A207818
The sum of the second powers of the numbers of solid standard Young tableaux over all planar partitions of n.
Original entry on oeis.org
1, 3, 15, 105, 981, 11775, 174309, 3070293, 62728749, 1462046235, 38446605975, 1131448219263, 37011386961477, 1337106744965847, 53005310776229283, 2290750722850361415, 107258522028595441911, 5410293190992424617243, 292567014870308385871989
Offset: 1
a(2) = 3 since the planar partitions [[2]], [[1,1]], [[1],[1]] in which each of these has only one tableaux.
- Shalosh B. Ekhad and Doron Zeilberger, Computational and Theoretical Challenges On Counting Solid Standard Young Tableaux.
A207816
Number of distinct necklaces with n red, n green, n blue and n white beads.
Original entry on oeis.org
1, 6, 318, 30804, 3941598, 586637256, 96197661156, 16875655269948, 3111284141045598, 595909785174057204, 117634021777132574568, 23797087019979071174580, 4912693780461352534397604, 1031629572413246016139181544, 219809927417367534490107035244, 47426945434432859336092700072304
Offset: 0
For n=1, a(1)=6 since for four beads necklaces with each bead from each of the four colors say (R,G,B,W), we can arrange as following, [R,G,B,W], [R,G,W,B], [R,B,G,W], [R,B,W,G], [R,W,G,B] and [R,W,B,G].
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with(combinat): with(numtheory):
# This formula comes from Polya Counting Theorem:
# Z(C_n) = add(phi(d)*(a_d)^(n/d), d in divisors(n))/n;
PolyaBrace:= proc(S) option remember; local n, s, d;
n:= add(s, s=S);
add(phi(d) *PolyaCoeff(d, S), d=divisors(n))/n
end:
# Find coeff of prod(a[i]^s[i], i=1..n) of a_d^(n/d) (symmetric function)
PolyaCoeff:= proc(d, S) option remember; local n, pow, s;
n:= add(s, s=S);
pow:= n/d;
if {seq(s mod d, s = S)} = {0}
then multinomial(pow, seq(s/d, s = S))
else 0
fi:
end:
a:= n-> `if`(n=0, 1, PolyaBrace([n$4])):
seq(a(n), n=0..20);
-
a[n_] := DivisorSum[n, EulerPhi[n/#] (4#)!/(#!^4 * 4n)&]; a[0]=1;
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 24 2017, after Alois P. Heinz *)
A207817
a(n) = (4*n)! / (n!^4 * (n+1)).
Original entry on oeis.org
1, 12, 840, 92400, 12612600, 1955457504, 329820499008, 59064793444800, 11062343605599000, 2145275226626532000, 427760079188506384320, 87255985739923260973440, 18139177035549431752363200, 3831766983249199488516960000, 820623729024838763928509760000
Offset: 0
-
with(combinat, multinomial): seq(multinomial(4*n, n$4)/(n+1), n=0..20);
-
CoefficientList[Series[HypergeometricPFQ[{1/4, 1/2, 3/4}, {1, 2}, 256 x], {x, 0, 20}], x] (* Benedict W. J. Irwin, Jul 13 2016 *)
A175707
Number of ways to put n copies of 1,2,3,4 into sets.
Original entry on oeis.org
1, 15, 139, 862, 4079, 15791, 52450, 154279, 411180, 1009741, 2314278, 5000125, 10264997, 20152950, 38037517, 69323949, 122448455, 210271756, 351989816, 575711716, 921889652, 1447822620, 2233501928, 3389114724, 5064582169, 7461570579, 10848490675, 15579077786, 22115241763, 31054971635, 43166197978, 59427633555, 81077755892, 109673237289, 147158299390, 195946638641
Offset: 0
For n=1, the solution is the fourth term of Bell numbers A000110.
For n=2, one way to partition 2 copies of 1, 2 copies of 2, 2 copies of 3 and 2 copies of 4 is {1}{2}{34}{12}{34}. On the other hand {112}{34}{23}{4} is not allowed since the same numbers are in the same set {112}.
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Doron Zeilberger, In How Many Ways Can You Reassemble Several Russian Dolls? (2009).
- Doron Zeilberger, In How many ways can you reassemble several russian dolls?, arXiv:0909.3453 [math.CO], 2009.
- Doron Zeilberger, BABUSHKAS; Local copy
- Index entries for linear recurrences with constant coefficients, signature (7, -17, 8, 36, -60, 4, 56, -22, -22, -22, 56, 4, -60, 36, 8, -17, 7, -1).
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a:= n-> (5382*n^11 +236808*n^10 +4643760*n^9 +53507520*n^8 +402098796*n^7 +2067612624*n^6 +7421736960*n^5 +18616942080*n^4 +32101468047*n^3 +36555545268*n^2 +25131098880*n +8024016000 +7016625*(-1)^n*n^3 +84199500*(-1)^n*n^2 +359251200*(-1)^n*n +538876800*(-1)^n) /(2^11*3^7*5^2*7*11) +5/3^6*(-1)^n * (sin(n*Pi/3)/sqrt(3)+ cos(n*Pi/3));
seq(a(n), n=0..40);
seq(SeqBrnDJ(n,4)[5], n=1..6); # using the Maple package BABUSHKAS (see links)
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LinearRecurrence[{7, -17, 8, 36, -60, 4, 56, -22, -22, -22, 56, 4, -60, 36, 8, -17, 7, -1}, {1, 15, 139, 862, 4079, 15791, 52450, 154279, 411180, 1009741, 2314278, 5000125, 10264997, 20152950, 38037517, 69323949, 122448455, 210271756}, 36] (* Jean-François Alcover, Nov 13 2018 *)
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