

A350811


a(n) = ceiling((nR(n))^2/(n+R(n))), where R(n) is the digit reversal of n.


1



0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 3, 8, 14, 20, 27, 34, 41, 48, 15, 3, 0, 2, 5, 10, 15, 21, 27, 33, 23, 8, 2, 0, 2, 4, 8, 12, 17, 23, 30, 14, 5, 2, 0, 1, 3, 7, 10, 15, 37, 20, 10, 4, 1, 0, 1, 3, 6, 9, 45, 27, 15, 8, 3, 1, 0, 1, 3, 5, 52, 34, 21, 12, 7, 3, 1, 0, 1, 2, 59, 41
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OFFSET

1,10


COMMENTS

If n is a palindrome, R(n) = n and nR(n) = 0, thus a(n) = 0. So, for a given length of the sequence, the number of zeros is equal to the number of palindromic n's. For example, in the interval {0, 100} there are 18 zeros corresponding to the 18 palindromic n's: 1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88 and 99.
From the 11th term, the sequence can be seen as a series of subsequences separated by a zero. Between two zeros, the terms start by increasing, then go through a maximum and finally decrease as shown for example by the zero separated subsets: [3,8,14,20,27,34,41,48,15,3]; [2,5,10,15,21,27,33,23,8,2]; [2,4,8,12,17,23,30,14,5,2].
The above described characteristics of the sequence give wellstructured patterns to nice graphs.


LINKS



EXAMPLE

For n = 1, R(n) = 1, thus a(1) = ceiling((11)^2/(1+1)) = 0.
For n = 10, R(n) = 1, thus a(10) = ceiling((101)^2/(10+1)) = 8.
For n = 22, R(n) = 22, thus a(22) = ceiling((2222)^2/(22+22)) = 0.


MATHEMATICA

Table[Ceiling[(nFromDigits[Reverse[IntegerDigits[n]]])^2/(n+FromDigits[Reverse[IntegerDigits[n]]])], {n, 81}] (* Stefano Spezia, Jan 18 2022 *)


PROG

(PARI) a(n)=my(x=fromdigits(Vecrev(digits(n)))); r=ceil((nx)^2/(n+x));
for(n=1, 2000, print1(a(n)", "))
(Python)
def R(n): return int(str(n)[::1])
def a(n):
Rn = R(n)
q, r = divmod((nRn)**2, n+Rn)
return q if r == 0 else q + 1


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



