A179298
a(n)=n^3-(n-1)^3-(n-2)^3-...-1.
Original entry on oeis.org
1, 7, 18, 28, 25, -9, -98, -272, -567, -1025, -1694, -2628, -3887, -5537, -7650, -10304, -13583, -17577, -22382, -28100, -34839, -42713, -51842, -62352, -74375, -88049, -103518, -120932, -140447, -162225, -186434, -213248, -242847, -275417
Offset: 1
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f[n_]:=Module[{k=n-1,x=n^3},While[k>0,x-=k^3;k--;];x];lst={};Do[AppendTo[lst,f[n]],{n,5!}];lst
A341331
a(n) = n^n - (n-1)^n - (n-2)^n - ... - 1^n.
Original entry on oeis.org
1, 3, 18, 158, 1825, 26141, 446782, 8869820, 200535993, 5085658075, 142947350986, 4410243535402, 148156328308105, 5382924338773177, 210309307208574750, 8791961076113491704, 391581231268402937041, 18510377905675629883959, 925555262359725659407258
Offset: 1
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f:= proc(n) local k; n^n - add(k^n,k=1..n-1) end proc:
map(f, [$1..30]); # Robert Israel, Feb 10 2021
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a[n_] := n^n - Sum[k^n, {k, 0, n - 1}]; Array[a, 20] (* Amiram Eldar, Apr 28 2021 *)
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a(n) = n^n-sum(k=0, n-1, k^n);
A360665
Square array T(n, k) = k*((2*n-1)*k+1)/2 read by rising antidiagonals.
Original entry on oeis.org
0, 0, 0, 0, 1, -1, 0, 2, 3, -3, 0, 3, 7, 6, -6, 0, 4, 11, 15, 10, -10, 0, 5, 15, 24, 26, 15, -15, 0, 6, 19, 33, 42, 40, 21, -21, 0, 7, 23, 42, 58, 65, 57, 28, -28, 0, 8, 27, 51, 74, 90, 93, 77, 36, -36, 0, 9, 31, 60, 90, 115, 129, 126, 100, 45, -45
Offset: 0
By rows:
0, 0, -1, -3, -6, -10, -15, -21, -28, ... = -A161680
0, 1, 3, 6, 10, 15, 21, 28, 36, ... = A000217
0, 2, 7, 15, 26, 40, 57, 77, 100, ... = A005449
0, 3, 11, 24, 42, 65, 93, 126, 164, ... = A005475
0, 4, 15, 33, 58, 90, 129, 175, 228, ... = A022265
0, 5, 19, 42, 74, 115, 165, 224, 292, ... = A022267
0, 6, 23, 51, 90, 140, 201, 273, 356, ... = A022269
... .
Cf. Antidiagonal sums:
A034827(n+1).
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T[n_, k_] := ((2*n - 1)*k^2 + k)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 31 2023 *)
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T(n, k) = ((2*n-1)*k^2+k)/2 \\ Thomas Scheuerle, Mar 17 2023
Showing 1-3 of 3 results.
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