Metamath Proof Explorer


Theorem pcbc

Description: Calculate the prime count of a binomial coefficient. (Contributed by Mario Carneiro, 11-Mar-2014) (Revised by Mario Carneiro, 21-May-2014)

Ref Expression
Assertion pcbc NK0NPPpCnt(NK)=k=1NNPkNKPk+KPk

Proof

Step Hyp Ref Expression
1 simp3 NK0NPP
2 nnnn0 NN0
3 2 3ad2ant1 NK0NPN0
4 3 faccld NK0NPN!
5 4 nnzd NK0NPN!
6 4 nnne0d NK0NPN!0
7 fznn0sub K0NNK0
8 7 3ad2ant2 NK0NPNK0
9 8 faccld NK0NPNK!
10 elfznn0 K0NK0
11 10 3ad2ant2 NK0NPK0
12 11 faccld NK0NPK!
13 9 12 nnmulcld NK0NPNK!K!
14 pcdiv PN!N!0NK!K!PpCntN!NK!K!=PpCntN!PpCntNK!K!
15 1 5 6 13 14 syl121anc NK0NPPpCntN!NK!K!=PpCntN!PpCntNK!K!
16 bcval2 K0N(NK)=N!NK!K!
17 16 3ad2ant2 NK0NP(NK)=N!NK!K!
18 17 oveq2d NK0NPPpCnt(NK)=PpCntN!NK!K!
19 fzfid NK0NP1NFin
20 nnre NN
21 20 3ad2ant1 NK0NPN
22 21 adantr NK0NPk1NN
23 simpl3 NK0NPk1NP
24 prmnn PP
25 23 24 syl NK0NPk1NP
26 elfznn k1Nk
27 26 nnnn0d k1Nk0
28 27 adantl NK0NPk1Nk0
29 25 28 nnexpcld NK0NPk1NPk
30 22 29 nndivred NK0NPk1NNPk
31 30 flcld NK0NPk1NNPk
32 31 zcnd NK0NPk1NNPk
33 11 nn0red NK0NPK
34 21 33 resubcld NK0NPNK
35 34 adantr NK0NPk1NNK
36 35 29 nndivred NK0NPk1NNKPk
37 36 flcld NK0NPk1NNKPk
38 37 zcnd NK0NPk1NNKPk
39 33 adantr NK0NPk1NK
40 39 29 nndivred NK0NPk1NKPk
41 40 flcld NK0NPk1NKPk
42 41 zcnd NK0NPk1NKPk
43 38 42 addcld NK0NPk1NNKPk+KPk
44 19 32 43 fsumsub NK0NPk=1NNPkNKPk+KPk=k=1NNPkk=1NNKPk+KPk
45 3 nn0zd NK0NPN
46 uzid NNN
47 45 46 syl NK0NPNN
48 pcfac N0NNPPpCntN!=k=1NNPk
49 3 47 1 48 syl3anc NK0NPPpCntN!=k=1NNPk
50 11 nn0ge0d NK0NP0K
51 21 33 subge02d NK0NP0KNKN
52 50 51 mpbid NK0NPNKN
53 11 nn0zd NK0NPK
54 45 53 zsubcld NK0NPNK
55 eluz NKNNNKNKN
56 54 45 55 syl2anc NK0NPNNKNKN
57 52 56 mpbird NK0NPNNK
58 pcfac NK0NNKPPpCntNK!=k=1NNKPk
59 8 57 1 58 syl3anc NK0NPPpCntNK!=k=1NNKPk
60 elfzuz3 K0NNK
61 60 3ad2ant2 NK0NPNK
62 pcfac K0NKPPpCntK!=k=1NKPk
63 11 61 1 62 syl3anc NK0NPPpCntK!=k=1NKPk
64 59 63 oveq12d NK0NPPpCntNK!+PpCntK!=k=1NNKPk+k=1NKPk
65 9 nnzd NK0NPNK!
66 9 nnne0d NK0NPNK!0
67 12 nnzd NK0NPK!
68 12 nnne0d NK0NPK!0
69 pcmul PNK!NK!0K!K!0PpCntNK!K!=PpCntNK!+PpCntK!
70 1 65 66 67 68 69 syl122anc NK0NPPpCntNK!K!=PpCntNK!+PpCntK!
71 19 38 42 fsumadd NK0NPk=1NNKPk+KPk=k=1NNKPk+k=1NKPk
72 64 70 71 3eqtr4d NK0NPPpCntNK!K!=k=1NNKPk+KPk
73 49 72 oveq12d NK0NPPpCntN!PpCntNK!K!=k=1NNPkk=1NNKPk+KPk
74 44 73 eqtr4d NK0NPk=1NNPkNKPk+KPk=PpCntN!PpCntNK!K!
75 15 18 74 3eqtr4d NK0NPPpCnt(NK)=k=1NNPkNKPk+KPk