

A141663


Regular triangle T(n,m) = abs(prime(n)  m^2), 0 <= m <= n1.


1



2, 3, 2, 5, 4, 1, 7, 6, 3, 2, 11, 10, 7, 2, 5, 13, 12, 9, 4, 3, 12, 17, 16, 13, 8, 1, 8, 19, 19, 18, 15, 10, 3, 6, 17, 30, 23, 22, 19, 14, 7, 2, 13, 26, 41, 29, 28, 25, 20, 13, 4, 7, 20, 35, 52, 31, 30, 27, 22, 15, 6, 5, 18, 33, 50, 69
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OFFSET

1,1


COMMENTS

Row sums are: {2, 5, 10, 18, 35, 53, 82, 118, 167, 233, 306}.
For p>5, and when only the positive terms of pm^2 are considered, one gets the following irregular triangle:
6, 3,
10, 7, 2,
12, 9, 4,
16, 13, 8, 1,
18, 15, 10, 3,
...
Then in each row, there are 2 integers a,b such that 1<a<b and a divides b (see Problem in links).
(End)


LINKS

David M. Bloom, Problem 1438, Mathematics Magazine, Vol. 67, No. 1 (Feb., 1994), p. 66.


FORMULA

T(n,m) = abs(prime(n)  m^2), 0 <= m <= n1.


EXAMPLE

Triangle begins
{ 2},
{ 3, 2},
{ 5, 4, 1},
{ 7, 6, 3, 2},
{11, 10, 7, 2, 5},
{13, 12, 9, 4, 3, 12},
{17, 16, 13, 8, 1, 8, 19},
{19, 18, 15, 10, 3, 6, 17, 30},
{23, 22, 19, 14, 7, 2, 13, 26, 41},
{29, 28, 25, 20, 13, 4, 7, 20, 35, 52},
{31, 30, 27, 22, 15, 6, 5, 18, 33, 50, 69}


MATHEMATICA

Clear[T, n, m] T[n_, m_] = Abs[Prime[n]  m^2]; Table[Table[T[n, m], {m, 0, n1}], {n, 1, 10}]; Flatten[%]


PROG

(PARI) tabl(nn) = {for (n=1, nn, for (m=0, n1, print1(abs(prime(n)m^2), ", "); ); print(); ); } \\ Michel Marcus, Feb 01 2015


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STATUS

approved



