OFFSET
1,2
FORMULA
For odd rows:
a(1, k) = a(1, k-1) - a(1, k-2)
a(2, k) = a(1, k-1) + [ a(2, k-1) - a(2, k-2) ]
a(3, k) = a(2, k-1) + [ a(3, k-1) - a(3, k-2) ]
...
a((k-1)/2, k) = a((k-3)/2, k-1) + [ a((k-1)/2, k-1) - a((k-1)/2, k-2) ]
a((k+1)/2, k) = Prime(k) - 2
and a((k-1)/2, k-1) = Prime(k-1) - 2
a((k-1)/2, k-2) = Prime(k-2) - 2
For even rows:
a(1, k) = a(1, k-1) + [ a(2, k-1) - a(1, k-2) ]
a(2, k) = a(2, k-1) + [ a(3, k-1) - a(2, k-2) ]
a(3, k) = a(3, k-1) + [ a(4, k-1) - a(3, k-2) ]
...
a((k-2)/2, k) = a((k-2)/2, k-1) + [ a(k/2, k-1) - a((k-2)/2, k-2) ]
a(k/2, k) = Prime(k) - 2
and a(k/2, k-1) = Prime(k-1) - 2
a((k-2)/2, k-2) = Prime(k-2) - 2
The recurrent prime formulas for odd and even rows are the following : prime(k_odd) = A000040(k_odd) = A115298(k) + Sum_{n=1..(k-3)/2} [ a(n,k-2) -2*a(n,k-1) ] + A000040(k-2) - A000040(k-1) +2; prime(k_even) = A000040(k_even) = A115298(k) + Sum_{n=1..(k-2)/2} [ a(n,k-2) -a((k-2)/2,k-2) -2*a(n,k-1) +a(1,k-1) ] + A000040(k-2) - A000040(k-1) + 2
EXAMPLE
The computation for obtaining the coefficients of each row of the Treillis triangle are the paired differences between primes ascending and those descending. Only half-rows are to be considered for deducing such terms.
For the 13th row:
...................19-17,.23-13,.29-11,.31-7,.37-5,.41-3,.43-2
.....................2,.....10,....18,....24,...32,...38,...41
For the 14th row:
...................19-19,.23-17,.29-13,.31-11,.37-7,.41-5,.43-3,.47-2
.....................0,.....6,.....16,....20,....30,...36,...40,...45
From Michael Somos, Oct 17 2016: (Start)
Triangle:
1: 1,
2: 3,
3: 2, 5,
4: 4, 9,
5: 2, 8, 11,
6: 6, 10, 15,
7: 4, 8, 14, 17,
8: 6, 12, 16, 21,
... (End)
CROSSREFS
KEYWORD
easy,nonn,tabf,uned
AUTHOR
André F. Labossière, Jan 19 2006
STATUS
approved