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A209260
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Irregular triangular array read by rows, resulting in a permutation of the natural numbers.
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5
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1, 2, 3, 5, 6, 4, 7, 9, 10, 12, 14, 15, 11, 18, 20, 21, 13, 22, 25, 27, 28, 8, 26, 30, 33, 35, 36, 17, 24, 39, 42, 44, 45, 19, 34, 40, 49, 52, 54, 55, 38, 51, 56, 60, 63, 65, 66, 23, 50, 57, 68, 72, 75, 77, 78
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OFFSET
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1,2
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COMMENTS
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One way to derive this sequence row by row is by eliminating from row n of A141419 any integer already present in row m < n of A141419, with the remaining entries comprising row n of this triangle. The resulting irregular triangle then begins as {1}; {2, 3}; {5, 6}; {4, 7, 9, 10}; ... (see the examples below).
If 2*n+1 is a prime, then 2*n+1 appears in row n+1.
Conjecture 1. 2*n+1 is in the set S_n = {union of first n rows of A209260} if and only if 2*n+1 is composite. (Verified for all n <= 10^6 by Charles R Greathouse IV.) This conjecture has been proved (see [Jeffery]).
Conjecture 2. Let N be an integer, N>0, and let D_N be the set of all positive integral divisors of N. Let B_N be the set of all integral solutions of d + (h-1)/2 for which d in D_N, with h=N/d and N=d*h (some solutions will not be integers and so are not in B_N). Then N appears in row r = min(B_N) of A209260. Also, N appears in row b of A141419, for each b in B_N. - L. Edson Jeffery, Feb 13 2013
For a proof of Conjecture 2 see the link.
The row-column index pair in A141419 for number v having factorization v = d*h with h odd is: (d+(h-1)/2, h) if h+1 <= 2*d, and (d+(h-1)/2, 2*d) if h+1 > 2*d. The position of number v in this triangle occurs for the minimum value of d+(h-1)/2 among all divisor pairs d,h with v = d*h and h odd. (END)
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LINKS
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EXAMPLE
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As an irregular triangle:
{1};
{2, 3};
{5, 6};
{4, 7, 9, 10};
{12, 14, 15};
{11, 18, 20, 21};
{13, 22, 25, 27, 28};
{8, 26, 30, 33, 35, 36};
{17, 24, 39, 42, 44, 45};
{19, 34, 40, 49, 52, 54, 55};
{38, 51, 56, 60, 63, 65, 66};
{23, 50, 57, 68, 72, 75, 77, 78};
and as the terms appear in their correct positions in A141419:
{1 };
{2, 3 };
{ 5, 6 };
{4, 7, 9, 10 };
{ 12, 14, 15 };
{ 11, 18, 20, 21 };
{ 13, 22, 25, 27, 28 };
{8, 26, 30, 33, 35, 36 };
{ 17, 24, 39, 42, 44, 45 };
{ 19, 34, 40, 49, 52, 54, 55 };
{ 38, 51, 56, 60, 63, 65, 66 };
{ 23, 50, 57, 68, 72, 75, 77, 78};
The row-column index pair for a hole in the triangle can be computed from the expressions in the Formula section and the triangle produced with the Mathematica code. - Hartmut F. W. Hoft, Apr 14 2016
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MATHEMATICA
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oddDivs[v_] := Module[{d=Divisors[v]}, Select[Transpose[{d, Reverse[d]}], OddQ[#[[2]]]&]]
holes[v_] := Drop[Sort[Map[{#[[1]]+(#[[2]]-1)/2, If[#[[2]]+1<=2*#[[1]], #[[2]], 2*#[[1]]]}&, oddDivs[v]], #1[[1]]<#2[[1]]&], 1]
a141419[i_, j_] := j*(2*i-j+1)/2
triangle141419[r_] := Table[a141419[i, j], {i, 1, r}, {j, 1, i}]
holes209260[r_] := Select[Flatten[Map[holes, Union[Flatten[triangle141419[r]]]], 1], #[[1]]<=r&]
triangle209260[r_] := Module[{rT=triangle141419[r], rH=holes209260[r], k, i, j}, For[k=1, k<=Length[rH], k++, {i, j}=rH[[k]]; rT[[i, j]]=" "]; rT]
a209260[r_] := Select[Flatten[triangle209260[r]], #!=" "&]
a209260[12] (* data *)
TableForm[triangle209260[12], TableDepth->2] (* triangle with holes *)
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PROG
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(PARI) try(p)={ \\ Test conjecture at prime p
my(n=p\2);
my(v=vectorsmall(n, i, i), v1=vectorsmall(n-1, i, 2*i+1));
while(1,
my(t=#v1-1);
while(t && v1[t]>p, t--);
if(t<1, return(0));
[v1, v]=[vectorsmall(t, i, v1[i]+v1[i+1]-v[i+1]), v1];
for(i=1, #v1,
if(v1[i]==p && isprime(p), return("Conjecture fails at "p))
)
)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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