OFFSET
1,1
COMMENTS
All terms m == 1 (mod 3). To prove this, look at gaps of 3 in row 2 of array A322744(n,k). The longest strings of consecutive numbers not in A322744(n,k) can occur only for these 3 numbers. The number following such a gap is A322744(2,e) = (3*2*e)/2 = 3e for some even e. The middle of the string, m = A322744(2,e) - 2 = 3e - 2. Thus m == 1 (mod 3). After the first term, all terms m == 2 (mod 4).- David Lovler, Nov 29 2021
LINKS
David Lovler, Table of n, a(n) for n = 1..852
PROG
(PARI) T319929(n, k) = if (n%2, if (k%2, n+k-1, k), if (k%2, n, 0));
T(n, k) = (3*n*k - T319929(n, k))/2; \\ A322744
list(nn) = {my(list = List()); for (n=2, nn, for (k=2, nn\n, listput(list, T(n, k)); ); ); setminus([1..nn], Set(list)); } \\ A307002
lista(nn) = {my(v=Vec(list(nn))); for (m=4, #v-1, my(x=v[m]); if (vecsearch(v, x-1) && vecsearch(v, x+1), print1(x, ", ")); ); } \\ Michel Marcus, Apr 02 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
David Lovler, Jan 30 2021
STATUS
approved