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A379260 Index of first appearance of n in sequence A379049.

%I A379260 #10 Dec 21 2024 01:07:49
%S A379260 0,1,3,2,9,4,6,26,24,5,18,7,78,28,12,11,54,19,216,71,15,29,162,53,21,
%T A379260 73,231,16,486,13,51,217,84,83,36,14,33,647,57,32,4374,31,237,649,45,
%U A379260 22,207,236,165,1945,693,50,2151,212,90,46,87,160,39366,86,63
%N A379260 Index of first appearance of n in sequence A379049.
%C A379260 a(n) is the smallest integer that makes A379049(a(n)) = n.
%C A379260 Conjecture: a(n) is defined for all integer n > 1.
%e A379260 For n = 2, A379049(0) = 1 + 1 = 2.  Thus a(2) = 0;
%e A379260 For n = 3, A379049(1) = 2 + 1 = 3, since 1's balanced ternary representation is 1.  Thus a(3) = 1;
%e A379260 For n = 4, A379049(3) = 3 + 1 = 4, since 3's balanced ternary representation is 10.  Thus a(4) = 3;
%e A379260 ...
%e A379260 For n = 60, A379049(39366) = 31 + 29 = 60, since 39366's balanced ternary representation is 1T000000000, where the 11's digit is 1 represents the 11's prime 31 in the term before the plus sign, and the 10's digit is T representing the 10's prime 29 in the term after the plus sign. And evaluation of A379049 found no number i smaller than 39366 can make A379049(i) = 60.  Thus a(60) = 39366.
%t A379260 BTDigits[m_Integer, g_] :=  Module[{n = m, d, sign, t = g}, If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n]; d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];    While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign; t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
%t A379260 goal = 62; res = {}; ct = 1;
%t A379260 Do[AppendTo[res, 0], {i, 2, goal}]; i = -1; While[ct < goal, i++; BT = BTDigits[i, {0}]; BTl = Length[BT]; f = 1; b = 1;  Do[If[BT[[j]] == 1, f = f*Prime[BTl - j + 1]];   If[BT[[j]] == -1, b = b*Prime[BTl - j + 1]], {j, 1, BTl}]; d = f + b; If[(d <= goal) && (res[[d - 1]] == 0), res[[d - 1]] = i; ct++]];
%t A379260 res
%Y A379260 Cf. A379049, A345128, A005812, A134022, A134023, A140267.
%K A379260 base,nonn
%O A379260 2,3
%A A379260 _Lei Zhou_, Dec 19 2024