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%I #23 Nov 13 2023 17:54:07
%S 1,0,0,0,1,1,0,1,1,1,0,1,1,2,1,2,1,1,1,2,0,2,2,1,1,2,1,0,2,2,0,1,1,2,
%T 2,3,1,2,1,1,3,1,1,1,2,1,1,1,1,1,3,1,3,1,0,2,2,3,1,1,2,2,2,3,0,1,3,1,
%U 3,1,2,2,1,1,2,1,2,3,1,2,1,3,1,2,2,0,2,3,2,1,1,1,1,1,2,1,1,1,2,3
%N a(n) = minimal number of 0's that must be changed to 1's in the binary expansion of the n-th prime in order to make it into a palindrome.
%H Paolo Xausa, <a href="/A144477/b144477.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) is half the Hamming distance between the binary expansion of prime(n) and its reversal.
%e a(5) = 1 since prime(5) = 11 = 1011_2 becomes a palindrome if we change the third bit to 0.
%t A144477[n_]:=With[{p=IntegerDigits[Prime[n],2]},HammingDistance[p,Reverse[p]]/2];Array[A144477,100] (* _Paolo Xausa_, Nov 13 2023 *)
%o (PARI)
%o HD(p)=
%o {
%o v=binary(p); H=0; j=#v;
%o for(k=1,#v, H+=abs(v[k]-v[j]); j--);
%o return(H)
%o };
%o for(n=1,100, p=prime(n); an=HD(p)/2; print1(an,", "))
%Y Subsequence of A037888.
%K nonn,base
%O 1,14
%A _Washington Bomfim_, Jan 15 2011, following a suggestion from _Joerg Arndt_.
%E Edited by _N. J. A. Sloane_, Apr 23 2020 at the suggestion of _Harvey P. Dale_
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