This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
For n=4, s(0)=4, 4 ->4-4 mod 1=4 ->4-4 mod 2=4 ->4-4 mod 3=3 ->3-3 mod 4=0, hence s(4)=0 and a(4)=4. For n=6, s(0)=6, s(1)=6-6 mod 2=6, s(2)=6-6 mod 3=6, s(3)=6-6 mod 4=6-2=4, s(4)=4-4 mod 5=0, hence a(6)=4.
Flatten@Table[First@Position[Rest@FoldList[#1-Mod[#1,#2]&,i,Range[2,i+1]],0], {i,30}] (* Birkas Gyorgy, Feb 26 2011 *)
a(n)=if(n<1, 0, s=n; c=1; while(s-s%c>0, s=s-s%c; c++); c--) \\ corrected by Dan Dima, Jan 18 2025
If x(1)=4, x(2)= 2*floor(4/2)=4, x(3)=3*floor(4/3)=3; x(4)=4*floor(3/4)=0 hence a(4)=4.
f:= proc(n,k) option remember; if n = 0 then return k-1 fi; procname(k*floor(n/k),k+1) end proc: map(f, [$1..100], 1); # Robert Israel, Jul 25 2019
a[n_] := Module[{x}, x[1] = n; x[k_] := x[k] = k Floor[x[k-1]/k]; For[k = 1, True, k++, If[x[k] == 0, Return[k]]]];
Array[a, 100] (* Jean-François Alcover, Jun 07 2020 *)
a(n)=if(n<0,0,s=n; c=1; while(s-s%c>0,s=s-s%c; c++); c)
For integers N=4,8,12,16,... we have the following sequences:
{4}
{8, 9} (8 -> the next higher odd multiple of 3, which is 9 -> STOP)
{12, 15, 16} (12 -> 3*5=15 -> 4*4=16 -> STOP)
{16, 21, 24, 25}
{20, 21, 24, 25}
{24, 27, 32, 35, 36}
{28, 33, 40, 45, 48, 49}
{32, 33, 40, 45, 48, 49}
{36, 39, 40, 45, 48, 49}
...
Thus there is 1 integer N=4k ending in the sea at 2^2, whence basin a(2)=1, and idem for 3 and 4.
The two integers 16 and 20 end at 5^2, so the basin of 5 is a(5)=2.
There is again a(6)=1 integer ending in 6^2, while the basin of 7 are the 3 integers 28, 32, and 36, which all merge into the "river" that enters the "sea" in 7^2=49.
Thus the first 6 terms in the sequence are 1, 1, 1, 2, 1, 3.
Take N=100 as an example: the next integer on the same line is the next higher odd multiple of 3, i.e., smallest 3*(2m+1) > 100, which is 105. The next number is the least even multiple of 4, 4*(2m) = 112, etc., leading to 115 = 5*(2m+1), followed by 120 = 6*(2m), 133 = 7*(2m+1), 144 = 8*2m (where we have a square, but not the square of 8), 153 =9*(2m+1), 160 = 10*2m, 165 = 11*(2m+1), 168 = 12*(2m) and finally 169 = 13*13.
cumul[n_Integer] := Module[{den1 = n, num = n^2, den2}, While[num > 4 && den1 != 2, num = num - 1; den1 = den1 - 1; den2 = Floor[num/den1]; If[Not[EvenQ[den1 + den2]], den2 = den2 - 1]; num = den1 den2]; Return[num/4]]; basin[2] := 1; basin[n_Integer] := cumul[n] - cumul[n - 1]; Table[basin[n], {n, 2, 75}] (* Alonso del Arte, Jan 19 2012 *)
bs(n,s,m=2)={while(n>m^2,n=(n\m+++2-bittest(n\m-m,0))*m; s & print1(n","));n}
n=4; for(c=2,50, for(k=1,9e9, bs(n+=4)==c^2 || print1(k",")||break)) \\ M. F. Hasler, Jan 20 2012
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