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A204539
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a(n) is the number of integers N=4k whose "basin" sequence (cf. comment) ends in n^2.
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4
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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
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OFFSET
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2,4
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COMMENTS
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The "basin" (analogous to river basins, for reasons set out below) is the number of positive integers N=4k which end in the "sea" at n^2. The "sea" of N is found as follows:
Starting out with N, in step i=1,2,3,..., stop if you have reached N=(i+1)^2 (the "sea" of N), otherwise set N to the next higher, odd or even (according to the parity of i), multiple of i+2, and go to step i+1.
Partial sums of this sequence appear to be A104738 (with a shift in offset). This has been confirmed for at least the first 4000 terms, but it is not at all clear why this is the case. - Ray Chandler, Jan 20 2012
After the first term, this sequence agrees with A028914 except for offset. Therefore this sequence is related to A028913, A007952, A002491 and A108696 dealing with the sieve of Tchoukaillon (or Mancala, or Kalahari). - Ray Chandler, Jan 20 2012
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LINKS
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EXAMPLE
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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}
etc.
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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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