A333429 A(n,k) is the n-th number m that divides k^m + 1 (or 0 if m does not exist); square array A(n,k), n>=1, k>=1, read by antidiagonals.
1, 1, 2, 1, 3, 0, 1, 2, 9, 0, 1, 5, 10, 27, 0, 1, 2, 25, 50, 81, 0, 1, 7, 3, 125, 250, 171, 0, 1, 2, 49, 9, 205, 1250, 243, 0, 1, 3, 10, 203, 21, 625, 5050, 513, 0, 1, 2, 9, 50, 343, 26, 1025, 6250, 729, 0, 1, 11, 5, 27, 250, 1379, 27, 2525, 11810, 1539, 0
Offset: 1
Examples
Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... 2, 3, 2, 5, 2, 7, 2, 3, 2, 11, ... 0, 9, 10, 25, 3, 49, 10, 9, 5, 121, ... 0, 27, 50, 125, 9, 203, 50, 27, 25, 253, ... 0, 81, 250, 205, 21, 343, 250, 57, 82, 1331, ... 0, 171, 1250, 625, 26, 1379, 1250, 81, 125, 2783, ... 0, 243, 5050, 1025, 27, 1421, 2810, 171, 625, 5819, ... 0, 513, 6250, 2525, 63, 2401, 5050, 243, 2525, 11891, ... 0, 729, 11810, 3125, 81, 5887, 6250, 513, 3125, 14641, ... 0, 1539, 25250, 5125, 147, 9653, 14050, 729, 3362, 30613, ...
Links
- Alois P. Heinz, Antidiagonals n = 1..20, flattened
Crossrefs
Programs
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Maple
A:= proc() local h, p; p:= proc() [1] end; proc(n, k) if k=1 then `if`(n<3, n, 0) else while nops(p(k))0 do od; p(k):= [p(k)[], h] od; p(k)[n] fi end end(): seq(seq(A(n, 1+d-n), n=1..d), d=1..12); -
Mathematica
dmax = 12; mmax = 2^(dmax+3); col[k_] := col[k] = Select[Range[mmax], Divisible[k^#+1, #]&]; A[n_, k_] := If[n>2 && k==1, 0, col[k][[n]]]; Table[A[n, d-n+1], {d, 1, dmax}, {n, 1, d}] // Flatten (* Jean-François Alcover, Jan 05 2021 *)