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A099964
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Triangle read by rows: The n-th row is constructed by forming the partial sums of the previous row, reading from the right and if n is a triangular number repeating the final term.
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3
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1, 1, 1, 1, 2, 2, 3, 3, 3, 6, 8, 8, 14, 17, 17, 31, 39, 39, 39, 78, 109, 126, 126, 235, 313, 352, 352, 665, 900, 1026, 1026, 1926, 2591, 2943, 2943, 2943, 5886, 8477, 10403, 11429, 11429, 21832, 30309, 36195, 39138, 39138, 75333, 105642, 127474, 138903
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| ...
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EXAMPLE
| Triangle begins
1
1 1
1 2
2 3 3
3 6 8
8 14 17
17 31 39 39
39 78 109 126
126 235 313 352
352 665 900 1026
1026 1926 2591 2943 2943
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MAPLE
| with(linalg):rev:=proc(a) local n, p; n:=vectdim(a): p:=i->a[n+1-i]: vector(n, p) end: ps:=proc(a) local n, q; n:=vectdim(a): q:=i->sum(a[j], j=1..i): vector(n, q) end: pss:=proc(a) local n, q; n:=vectdim(a): q:=proc(i) if i<=n then sum(a[j], j=1..i) else sum(a[j], j=1..n) fi end: vector(n+1, q) end: tr:={seq(n*(n+1)/2, n=1..30)}: R[0]:=vector(1, 1): for n from 1 to 15 do if member(n, tr)=false then R[n]:=ps(rev(R[n-1])) else R[n]:=pss(rev(R[n-1])) fi od: for n from 0 to 15 do evalm(R[n]) od; # (Deutsch)
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MATHEMATICA
| triQ[n_] := Reduce[ n == k(k+1)/2, k, Integers] =!= False; row[0] = {1}; row[1] = {1, 1}; row[n_] := row[n] = (ro = Accumulate[ Reverse[ row[n-1]]]; If[triQ[n], Append[ ro, Last[ro] ], ro]); Flatten[ Table[ row[n], {n, 0, 13}]](* From Jean-François Alcover, Nov 24 2011 *)
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CROSSREFS
| First column (and row sums) gives A099965. Cf. A099966, A099968.
If an extra term is added to /every/ row we get A008282. Cf. A099959, A099961.
Sequence in context: A115733 A025496 A099959 * A094440 A093736 A014589
Adjacent sequences: A099961 A099962 A099963 * A099965 A099966 A099967
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KEYWORD
| nonn,tabf,nice,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Nov 13 2004, following a suggestion made by Douglas G. Rogers, Mar 10, 2003
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EXTENSIONS
| More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Nov 16 2004
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