R J Cano on Sat, 22 Apr 2017 18:27:22 +0200

 Re: 4 Gigabytes of RAM are not enough!

``` Hi!,

I had a wrong result... some things cannot be grouped inside a single
"while" loop as I shown previously.

But I fixed it... a little bit slower but correct.

Cheers,

Remy

P.S.: Bonus feature, but beware with overflows when using w=1 option.

Notice if 2<=m<n, how the first m!-1 terms given by Ticket1b(m) are
the same corresponding values for Ticket1b(n), but multiplied/divided
by B^(n-m);
```
```/* (PARI) R. J. Cano, Apr 22 2017; Note: For a best performance, use this script with GP2C. */

Ticket1b(n,{B=10},{w=0})={ /* Computes a "final representation" (by default in decimal) for the first n!-1 terms of A217626, storing them as a "t_VECSMALL" object */

n=max(2,n);
my(x0:vecsmall,x:vecsmall,y:vecsmall,i:small,j:small,k:small,u:small,t:small,Q:vecsmall,p:small,bb:small,m:small);
x=Vecsmall(vector(n,i,i-1));
m=n!-1;
Q=Vecsmall(0,m);
k=0;
bb=B;

while(k++<=#Q,

x0=x;

if(w,

/* Implements Schmuel Zak's algorithm -- Start */
u=1;
r=k;
while(u++<k,if(r%u,break,r\=u)); /* These 3 initial lines are an adaptation from the Charles Greathouse IV 's A055881 program. */
j=u;
j\=2;
i=0;
while(i++<=j,
t=x[i];
x[i]=x[1+u-i];
x[1+u-i]=t
)
/* Implements Schmuel Zak's algorithm -- End */

,

/* Implements Narayana Pandit's lex-order algorithm -- Start */
i=#x-1;
while((x[i]>=x[i+1])&&i,i--);
j=#x;
while((x[i]>=x[j])&&(j>i),j--);
t=x[j];
x[j]=x[i];
x[i]=t;
u=#x;
u-=i;
u\=2;
u++;
while(u--,
t=x[i+u];
x[i+u]=x[1+#x-u];
x[1+#x-u]=t;
)
/* Implements Narayana Pandit's lex-order algorithm -- End */

);

y=x;

r=0;
while(r++<#y, y[r  ]-= x0[r] );
r=0;
while(r++<#y, y[r+1]+=  y[r] );
p=1;
r=0;
while(r++<#y, Q[k  ]+=  y[#y-r]*p; p*=bb )

);

return(Q);

}
```