"Integer partitions"의 두 판 사이의 차이

2020년 12월 28일 (월) 04:18 기준 최신판

background

n:=9

md:=5

n:=12

md:=7

n:=6

md:=11

will be a good choice

\(p(5k+4)\equiv 0 \pmod 5\)

\(p(7k+5)\equiv 0 \pmod 7\)

\(p(11k+6)\equiv 0 \pmod {11}\)

partition rank and crank

(*define a integer you want to investigate*)n := 6 (*choose the proper moduli for the partition statistics*) md := 2 S[n_] := IntegerPartitions[n] (*define the rank of a partition with the name "pr"*) pr[s_] := Max[s] - Length[s] (*define the crank of a partition with the name "crank"*) Om[s_] := Count[s, 1] Mu[s_] := Length[Select[s, # > Om[s] &]] crank[s_] := If[Om[s] == 0, Max[s], Mu[s] - Om[s]] (*modulus distribution of partition rank*) Sort[Tally[Table[Mod[pr[s], md], {s, S[n]}]]] (*modulus distribution of partition crank*) Sort[Tally[Table[Mod[crank[s], md], {s, S[n]}]]] (*list of paritions with rank& crank*) Do[Print[s, ", rank=", pr[s], "\[Congruent]", Mod[pr[s], md], "(mod ", md, ")", ", crank=", crank[s], "\[Congruent]", Mod[crank[s], md], "(mod ", md, ")"], {s, S[n]}] (*you will see the distribution of rank/crank modulus,the partition \ statistics and list of paritions with rank&crank*)

number of partitions with odd and even rank

for theoretical background, see rank of partition and mock theta function

S[n_] := IntegerPartitions[n] pr[s_] := Max[s] - Length[s] PrOd[n_] := Length[Select[S[n], OddQ[pr[#]] &]] PrEv[n_] := Length[Select[S[n], EvenQ[pr[#]] &]] alpha[n_] := PrEv[n] - PrOd[n] Table[alpha[n], {n, 1, 20}]

the generating function is can be shown by Series[(1 + 4 Sum[(-1)^n q^(n (3 n + 1)/2)/(1 + q^n), {n, 1, 10}])/Sum[(-1)^n q^(n (3 n + 1)/2), {n, -8, 8}], {q, 0, 100}]

various partitions

(* partitions with at most 5 parts *) IntegerPartitions[7, 5]

(* partition into exactly three parts *) VS[n_] := IntegerPartitions[n, {3}] VS[11]

(* number of partitions into distinct parts *) PartitionsQ[11]

(* partition into odd parts *) IntegerPartitions[11, All, {1, 3, 5, 7, 9, 11}]

@@ 1번째 줄: / 1번째 줄: @@
+==background==
+n:=9
+md:=5
+n:=12
+md:=7
+n:=6
+md:=11
+will be a good choice
+<math>p(5k+4)\equiv 0 \pmod 5</math>
+<math>p(7k+5)\equiv 0 \pmod 7</math>
+<math>p(11k+6)\equiv 0 \pmod {11}</math>
+==partition rank and crank==
+(*define a integer you want to investigate*)n := 6 (*choose the proper moduli for the partition statistics*) md := 2 S[n_] := IntegerPartitions[n] (*define the rank of a partition with the name "pr"*) pr[s_] := Max[s] - Length[s] (*define the crank of a partition with the name "crank"*) Om[s_] := Count[s, 1] Mu[s_] := Length[Select[s, # > Om[s] &]] crank[s_] := If[Om[s] == 0, Max[s], Mu[s] - Om[s]] (*modulus distribution of partition rank*) Sort[Tally[Table[Mod[pr[s], md], {s, S[n]}]]] (*modulus distribution of partition crank*) Sort[Tally[Table[Mod[crank[s], md], {s, S[n]}]]] (*list of paritions with rank& crank*) Do[Print[s, ", rank=", pr[s], "\[Congruent]", Mod[pr[s], md], "(mod ",    md, ")", ", crank=", crank[s], "\[Congruent]", Mod[crank[s], md],   "(mod ", md, ")"], {s, S[n]}] (*you will see the distribution of rank/crank modulus,the partition \ statistics and list of paritions with rank&crank*)
+==number of partitions with odd and even rank==
+* for theoretical background, see [[rank of partition and mock theta conjecture|rank of partition and mock theta function]]
+S[n_] := IntegerPartitions[n] pr[s_] := Max[s] - Length[s] PrOd[n_] := Length[Select[S[n], OddQ[pr[#]] &]] PrEv[n_] := Length[Select[S[n], EvenQ[pr[#]] &]] alpha[n_] := PrEv[n] - PrOd[n] Table[alpha[n], {n, 1, 20}]
+*  the generating function is can be shown by Series[(1 + 4 Sum[(-1)^n q^(n (3 n + 1)/2)/(1 + q^n), {n, 1, 10}])/Sum[(-1)^n q^(n (3 n + 1)/2), {n, -8, 8}], {q, 0, 100}]
+==various partitions==
+(* partitions with at most 5 parts *) IntegerPartitions[7, 5]
+ (* partition into exactly three parts *) VS[n_] := IntegerPartitions[n, {3}] VS[11]
+ (* number of partitions into distinct parts *) PartitionsQ[11]
+(* partition into odd parts *) IntegerPartitions[11, All, {1, 3, 5, 7, 9, 11}]
+[[분류:math and physics]]
+[[분류:migrate]]