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A065157 Table of binary string substitutions: a(i,j) is obtained by substituting i for each 1-bit in j.

%I A065157 #13 Feb 11 2023 09:10:15
%S A065157 1,2,2,3,4,3,4,6,10,4,5,8,15,8,5,6,10,36,12,18,6,7,12,45,16,27,20,7,8,
%T A065157 14,54,20,68,30,42,8,9,16,63,24,85,72,63,16,9,10,18,136,28,102,90,292,
%U A065157 24,34,10,11,20,153,32,119,108,365,32,51,36,11,12,22,170,36,264,126,438,40,132,54,74,12
%N A065157 Table of binary string substitutions: a(i,j) is obtained by substituting i for each 1-bit in j.
%F A065157 Table origin is a(1,1).
%F A065157 a(1,n) = a(n,1) = n.
%F A065157 a(0,n) = a(n,0) = 0.
%F A065157 a(i,j) = A065158(i,j)*i.
%F A065157 a(n,n) = A065159(n) = A065160(n)*n.
%e A065157 a(3,5): 5 = 101_2 -> (3)0(3) = (11)0(11)_2 = 11011_2 = 27.
%e A065157 a(5,3): 3 = 11_2 -> (5)(5) = (101)(101)_2 = 101101_2 = 45.
%o A065157 (PARI) T(n,k) = my(bk=binary(k), sn=Str(fromdigits(binary(n))), s=""); for (i=1, #bk, if (bk[i] == 1, s=concat(s, sn), s=concat(s, "0"))); fromdigits(apply(eval, Vec(s)), 2); \\ _Michel Marcus_, Feb 11 2023
%Y A065157 Cf. A065158, A065159, A065160.
%K A065157 base,easy,nonn,tabl
%O A065157 1,2
%A A065157 _Marc LeBrun_, Oct 18 2001