A379260 Index of first appearance of n in sequence A379049.
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, 73, 231, 16, 486, 13, 51, 217, 84, 83, 36, 14, 33, 647, 57, 32, 4374, 31, 237, 649, 45, 22, 207, 236, 165, 1945, 693, 50, 2151, 212, 90, 46, 87, 160, 39366, 86, 63
Offset: 2
Examples
For n = 2, A379049(0) = 1 + 1 = 2. Thus a(2) = 0; For n = 3, A379049(1) = 2 + 1 = 3, since 1's balanced ternary representation is 1. Thus a(3) = 1; For n = 4, A379049(3) = 3 + 1 = 4, since 3's balanced ternary representation is 10. Thus a(4) = 3; ... 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.
Programs
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Mathematica
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]; goal = 62; res = {}; ct = 1; 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++]]; res
Comments