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%I #17 Aug 07 2017 12:25:12
%S 1,11,19,67,71,263,271,781,1273,1349,2981,4757,5041,18157,18673,19241,
%T 55451,71273,73441,95779,211651,337747,357911,1289147,1325783,1366111,
%U 3937021,5060383,5214311,6800309,15027221,19314983,19902511,23980037,25411681
%N Where the ratio A235027(n)/n obtains record values.
%C Because A056539(n)/n < 2 for all n, and already for the tenth term of this sequence 1349 we have A235027(1349)/1349 = 2.094... it follows that the only primes present are terms a(2) .. a(7): 11, 19, 67, 71, 263, 271. Conjecture: every term after that is a product of some of those six primes. For example: 781 = 11*71, 1273 = 19*67, 1349 = 19*71, 2981 = 11*271, 4757 = 67*71, 5041 = 71*71.
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%o (PARI)
%o revbits(n) = fromdigits(Vecrev(binary(n)), 2);
%o A235027(n) = {my(f = factor(n)); for (k=1, #f~, if (f[k, 1] != 2, f[k, 1] = revbits(f[k, 1]); ); ); factorback(f); } \\ This function from _Michel Marcus_, Aug 05 2017
%o m=0; i=0; n=0; while(i<35, n++; if((A235027(n)/n) > m, m = A235027(n)/n; i++; print1(n,",")));
%Y Cf. A235027.
%K nonn,base
%O 1,2
%A _Antti Karttunen_, Aug 07 2017