A204539 a(n) is the number of integers N=4k whose "basin" sequence (cf. comment) ends in n^2.
1, 1, 1, 2, 1, 3, 2, 4, 2, 4, 3, 5, 1, 9, 2, 10, 3, 5, 7, 9, 2, 10, 9, 9, 2, 13, 9, 8, 4, 20, 4, 15, 6, 15, 8, 12, 6, 22, 6, 15, 15, 21, 5, 13, 12, 23, 7, 24, 11, 19, 15, 24, 6, 30, 6, 26, 7, 27, 26, 13, 6, 33, 27, 30, 5, 13, 30, 30, 5, 37, 15, 26, 28, 32, 7, 17, 25, 54, 9, 30, 21, 41, 25
Offset: 2
Keywords
Examples
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.
Links
- Ray Chandler, Table of n, a(n) for n = 2..10001
- Mark Dukes, Fagan's Construction, Strange Roots, and Tchoukaillon Solitaire, Journal of Integer Sequences, Vol. 24 (2021), Article 21.7.1.
- Mark Dukes, Fagan's Construction, Strange Roots, and Tchoukaillon Solitaire, arXiv:2202.02381 [math.NT], 2022.
Crossrefs
Programs
-
Mathematica
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 *)
-
PARI
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
Comments