Pari/GP, 39 bytes

for(i=1,oo,[print(p[2])|p<-factor(i)~]) 

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A port of @Command Master's 05AB1E answer.

Pari/GP, 47 bytes

for(i=9,oo,[print(d)|d<-digits(i,log(i)^.9\1)]) 

When \$i\$ is larger enough, \$b = \lfloor\log(i)^{0.9}\rfloor\$ will satisfy \$b^b < i\$. So the code will print the base-\$b\$ digits of a sequence of more than \$b^b\$ consecutive integers, which will contain all sequences of integers in \$1,\dots,b-1\$ with length \$<b\$.

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Pari/GP, 50 bytes

for(i=2,oo,for(j=1,i^i,[print(d)|d<-digits(j,i)])) 

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Pari/GP, 39 bytes

for(i=1,oo,[print(p[2])|p<-factor(i)~]) 

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A port of @Command Master's 05AB1E answer.

Pari/GP, 44 bytes (@Polichinelle)

for(i=9,oo,[print(d)|d<-digits(i,log(i)\1)]) 

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Pari/GP, 47 bytes

for(i=9,oo,[print(d)|d<-digits(i,log(i)^.9\1)]) 

When \$i\$ is larger enough, \$b = \lfloor\log(i)^{0.9}\rfloor\$ will satisfy \$b^b < i\$. So the code will print the base-\$b\$ digits of a sequence of more than \$b^b\$ consecutive integers, which will contain all sequences of integers in \$1,\dots,b-1\$ with length \$<b\$.

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Pari/GP, 50 bytes

for(i=2,oo,for(j=1,i^i,[print(d)|d<-digits(j,i)])) 

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Pari/GP, 47 bytes

for(i=9,oo,[print(d)|d<-digits(i,log(i)^.9\1)]) 

When \$i\$ is larger enough, \$b = \lfloor\log(i)^{0.9}\rfloor\$ will satisfy \$b^b < i\$. So the code will print the base-\$b\$ digits of a sequence of more than \$b^b\$ consecutive integers, which will contain all sequences of integers in \$1,\dots,b-1\$ with length \$<b\$.

Try it online!

Pari/GP, 50 bytes

for(i=2,oo,for(j=1,i^i,[print(d)|d<-digits(j,i)])) 

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Pari/GP, 39 bytes

for(i=1,oo,[print(p[2])|p<-factor(i)~]) 

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A port of @Command Master's 05AB1E answer.

Pari/GP, 47 bytes

for(i=9,oo,[print(d)|d<-digits(i,log(i)^.9\1)]) 

When \$i\$ is larger enough, \$b = \lfloor\log(i)^{0.9}\rfloor\$ will satisfy \$b^b < i\$. So the code will print the base-\$b\$ digits of a sequence of more than \$b^b\$ consecutive integers, which will contain all sequences of integers in \$1,\dots,b-1\$ with length \$<b\$.

Try it online!

Pari/GP, 50 bytes

for(i=2,oo,for(j=1,i^i,[print(d)|d<-digits(j,i)])) 

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Pari/GP, 50 bytes

for(i=2,oo,for(j=1,i^i,[print(d)|d<-digits(j,i)])) 

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Pari/GP, 47 bytes

for(i=9,oo,[print(d)|d<-digits(i,log(i)^.9\1)]) 

When \$i\$ is larger enough, \$b = \lfloor\log(i)^{0.9}\rfloor\$ will satisfy \$b^b < i\$. So the code will print the base-\$b\$ digits of a sequence of more than \$b^b\$ consecutive integers, which will contain all sequences of integers in \$1,\dots,b-1\$ with length \$<b\$.

Try it online!

Pari/GP, 50 bytes

for(i=2,oo,for(j=1,i^i,[print(d)|d<-digits(j,i)])) 

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