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%I #12 Nov 24 2018 01:34:39
%S 1,121,12321,1234321,123454321,12345654321,1234567654321,
%T 123456787654321,12345678987654321,1234567898987654321,
%U 123456789878987654321,12345678987678987654321,1234567898765678987654321,123456789876545678987654321,12345678987654345678987654321
%N a(n) = (2*n+1)-digit palindrome formed by first concatenating A158289(i) for i=1..n+1 and then A158289(n-j) for j=1..n.
%e 1
%e 121
%e 12321
%e 1234321
%e 123454321
%e 12345654321
%e 1234567654321
%e 123456787654321
%e 12345678987654321
%e 1234567898987654321
%e 123456789878987654321
%e 12345678987678987654321
%e 1234567898765678987654321
%e 123456789876545678987654321
%e 12345678987654345678987654321
%e 1234567898765432345678987654321
%e 123456789876543212345678987654321
%e 12345678987654321012345678987654321
%e 1234567898765432101012345678987654321
%e 123456789876543210121012345678987654321
%o (PARI) a158289(n)=abs(n-round(n/18)*18)
%o eva(n) = subst(Pol(n), x, 10)
%o a(n) = if(n==0, return(1)); my(v=vector(2*n+1), x=#v); for(k=1, ceil(#v/2), v[k]=a158289(k)); while(1, v[x]=v[#v-x+1]; x--; if(x==ceil(#v/2), return(eva(v))))
%Y Cf. A158289.
%K nonn,base,easy
%O 0,2
%A _Felix Fröhlich_, Nov 17 2018