A centered decagonal number is a centered figurate number that represents a decagon with a dot in the center and all other dots surrounding the center dot in successive decagonal layers. The centered decagonal number for n is given by the formula
Thus, the first few centered decagonal numbers are
 1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, ... (sequence A062786 in the OEIS)
Like any other centered kgonal number, the nth centered decagonal number can reckoned by multiplying the (n − 1)th triangular number by k, 10 in this case, then adding 1. As a consequence of performing the calculation in base 10, the centered decagonal numbers can be obtained by simply adding a 1 to the right of each triangular number. Therefore, all centered decagonal numbers are odd and in base 10 always end in 1.
Another consequence of this relation to triangular numbers is the simple recurrence relation for centered decagonal numbers:
where
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Método general para dibujar polígonos inscritos en circunferencias
Transcription
Let's apply the general method to draw regular polygons inscribed in circumferences. The fist step we must do is to draw a vertical diameter. This vertical diameter will intersect with the circumference in two ponts, "A" and "P". Then, centered at "A", with radius "AP", let's draw an arc to the right with the same radius, but in this case centered at "P", we will draw another arc that will intersect with the fist one at a point that we will call "Q". In the next step, we have to divide the diameter we drew previously into as many parts as we want the polygon to have sides. In this case we will draw a decagon, I mean, a tensided polygon. Let's divide the diameter into ten equal parts. For this we will use Thales' theorem. First we draw a ray that has any angle and starts at point "A". After that we will choose any given measurement and place it on the ray ten times, starting at point "A". In this case we are taking that measurement with compass. Also you can take it with the rule One centimeter, two centimeters,... always considering it is not too far from the other end of the segment we want to divide, in this case, "P" The eighth measurement, here would be the ninth. The accuracy is very important in this part of the exercise, and this is the tenth line mark, the tenth time I put this same measurement. At this tenth line mark, this point, we will join it with "P". And from now on, we will draw parallel segments to the previous junction, through successive marks drawn above previosly, through the ninth one, through the eighth one. We use this square with a kind of attached protractor to do this with the software we use to record this videos, but you should do it with a square and bevel. This would be the fifth line mark, the fifth measurement you can see it pass through "O" This means that we're doing the things correctly. And it should always happen you draw an even number of sides. This would be the line mark number "3" and then comes the most important of all, number "2". We wiil draw this line thicker because of its importance. This would be the line mark number "2" in the diameter. Regardless of the number of sides of the polygon we are looking for line mark number "2", which will always be the key. We'll always search this line mark number "2" We number it. The next step will be to draw a ray which starts at ponit "Q" through point "2" continues until it cuts the circumference at a point we will call "B" OK, so we already have the measurement of the side we are looking for, the decagon side. This measurement is "AB". If we have drawn it correctly, the measurement "AB" will fit in the circumference just ten times. The third time the fourth, the fifth time that you can check it concurs with point "P". This means that we are working with accuracy. Let's put the side's measurement successively obtaining the polygon's vertex that we are looking for. Remind that it is a general method to draw polygons of any number os sides. The only condition is that they are inscribed in a previous drawn circumference. The only difference to do the exercise will be the division of the diameter into as many parts as sides the polygon has. Whatever the number of parts, the key division will always be number "2". Here you can check how we get to point "A". It means that everyside has the same dimension. We give each vertex a letter and we just join them. We join "D" to "E". "E" to point "P", "P" to point "F", "G" to "H", "H" to "I" and finally, closing the polygon joining "I" to the vertex "A". Here we have the decagon we are looking for.
See also
 [ordinary] decagonal number