A294924 Numbers n such that the whole sequence of the first n terms of A293699 is a palindrome.
1, 3, 5, 7, 26, 63, 100, 137, 174, 211, 248, 285, 322, 359, 396, 433, 470, 507, 544, 581, 618, 655, 692, 729, 766, 803, 840, 877, 914, 951, 988, 1025, 1062, 1099, 1136, 1173, 1210, 1247, 1284, 1321, 1358, 1395, 1432, 1469, 1506, 1543, 1580, 1617, 1654, 1691, 1728, 1765, 1802, 1839, 1876, 1913, 1950, 1987, 2024, 2061, 2098, 2135
Offset: 1
Keywords
Examples
The first 7 terms of A293699 are (3, 19, 3, 19, 3, 19, 3) which is a palindromic sequence, so 7 is a term. The first 8 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3) which is not a palindromic sequence, so 8 is not a term. The first 9 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3, 16) which is not a palindromic sequence, so 9 is not a term. The first 25 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 19, 3, 19, 3, 19) which is not a palindromic sequence, so 25 is not a term. The first 26 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 19, 3, 19, 3, 19, 3) which is a palindromic sequence, so 26 is a term.
Links
- V.J. Pohjola, Line plot for n=1..7
- V.J. Pohjola, Line plot for n=1..85
Programs
-
Mathematica
rootsn7 = Flatten[Position[Table[Floor[Tan[-n]], {n, 1, 10^7}], 1]]; difn7 = Differences[rootsn7]; ny = {}; Do[ If[Table[difn7[[i]], {i, 1, n}] == Reverse[Table[difn7[[i]], {i, 1, n}]], AppendTo[ny, n]], {n, 1, Length[difn7]}] ny
Comments