A384147 Array A(n,k) = n*(A(n-1,k)+A(n-2,k)+...+A(n-k,k)), where A(n,k) = n if n <= k, read by antidiagonals with n >= 1 and k >= 1.
1, 1, 2, 1, 2, 3, 1, 8, 3, 4, 1, 20, 3, 4, 5, 1, 56, 27, 4, 5, 6, 1, 152, 99, 4, 5, 6, 7, 1, 416, 387, 64, 5, 6, 7, 8, 1, 1136, 1539, 304, 5, 6, 7, 8, 9, 1, 3104, 6075, 1504, 125, 6, 7, 8, 9, 10, 1, 8480, 24003, 7504, 725, 6, 7, 8, 9, 10, 11, 1, 23168, 94851, 37504, 4325, 216, 7, 8, 9, 10, 11, 12
Offset: 1
Examples
Top left corner of the array: 1 1 1 1 1 1 1 1 1 1 1 2 2 8 20 56 152 416 1136 3104 8480 23168 3 3 3 27 99 387 1539 6075 24003 94851 374787 4 4 4 4 64 304 1504 7504 37504 187264 935104 5 5 5 5 5 125 725 4325 25925 155525 933125 6 6 6 6 6 6 216 1476 10296 72036 504216 7 7 7 7 7 7 7 343 2695 21511 172039 8 8 8 8 8 8 8 8 512 4544 40832 9 9 9 9 9 9 9 9 9 729 7209 10 10 10 10 10 10 10 10 10 10 1000 ...
Links
- Jason Bard, Table of n, a(n) for n = 1..5050
Programs
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Mathematica
nmax = 100; AntiDiagonalFlatten[matrix_] := Module[{n = Length@matrix}, Flatten[Table[matrix[[i, s - i]], {s, 2, 2 n}, {i, Max[1, s - n], Min[n, s - 1]}], 1]]; A384147 = AntiDiagonalFlatten[Table[LinearRecurrence[ConstantArray[n, n], ConstantArray[n, n], {1, nmax}], {n, 1, nmax}]][[;; nmax*(nmax + 1)/2]]
Formula
A(m,m+1) = m^3 for all m >= 1.
A(m,m+2) = m^4 + m^3 - m^2 for all m >= 1.
A(m,m+3) = m^5 + 2m^4 - 2m^2 for all m >= 1.
A(m,m+4) = m^6 + 3m^5 + 2m^4 - 2m^3 - 3m^2 for all m >= 3.
A(m,m+5) = m^7 + 4m^6 + 5m^5 - 5m^3 - 4m^2 for all m >= 4.
...
A(m,m+k) ~ O(m^(k+2)) for all m >= k-1 may be derived similarly.
Comments