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A035313
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(Largest) diagonal of the Zorach additive triangle A035312.
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11
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1, 3, 9, 26, 66, 154, 346, 771, 1726, 3887, 8768, 19700, 43890, 96717, 210665, 453893, 968903, 2053260, 4328489, 9093971, 19068611, 39943689, 83628399, 175018523, 366081209, 765102907, 1597315656, 3330380593, 6933810145
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graph;
refs;
listen;
history;
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OFFSET
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0,2
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COMMENTS
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From Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Apr 22 2007: (Start)
Starting with 1, smallest sequence for which:
all its terms a1(n).............................. 1,3,9,26,66
all terms of first differences a2(n)=a1(n+1)-a1(n) 2,6,17,40
all terms of second differences a3(n)=a2(n+1)-a2(n) 4,11,23
...
all terms of (1+i)th differences ai(n)=ai-1(n+1)-ai-1(n)
are different for any n and any i (End)
Which is to say, this sequence is the lexicographically earliest sequence of positive integers such that the sequence itself and its n-th differences for n >= 1 are pairwise disjoint. - David W. Wilson, Feb 26 2012
Conjecturally, every positive integer occurs in the sequence or one of its n-th differences, which would imply that the sequence and its n-th differences partition the positive integers. - David W. Wilson, Feb 26 2012
Conjecture: lim(n->infinity, a(n+1)/a(n)) = 2. - David W. Wilson, Feb 26 2012
Note that the n-th differences yield the n-th subdiagonals (parallels to the right edge) in the triangle A035312. Therefore Lallouet's statement and Wilson's 1st comment above are just rephrasing the definition of that triangle. - M. F. Hasler, May 09 2013
Binomial transform of A035311. Hence, from the observed asymptotic equality A035311(n) ~ 2*n, a stronger statement than the one given above follows: a(n) ~ n*2^n. - Andrey Zabolotskiy, Feb 08 2017
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LINKS
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EXAMPLE
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Start with 1; 2 is the next, then add 1+2 to get 3, then 4 is next, then 4+2=6 and 6+3 is 9, then 5 is not next because 5+4=9 and 9 was already used, so 7 is next...which ultimately generates 26 in the final column...
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MATHEMATICA
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(* Assuming n <= t(n, 1) <= 3n *) rows = 29; uniqueQ[t1_, n_] := (t[n, 1] = t1; Do[t[n, k] = t[n, k-1] + t[n-1, k-1], {k, 2, n}]; n*(n+1)/2 == Length[ Union[ Flatten[ Table[t[m, k], {m, 1, n}, {k, 1, m}]]]]); t[n_, 1] := t[n, 1] = Select[ Complement[Range[n, 3 n], Flatten[ Table[t[m, k], {m, 1, n-1}, {k, 1, m}]]], uniqueQ[#, n] &, 1][[1]]; Last /@ Table[t[n, k], {n, 1, rows}, {k, 1, n}] (* Jean-François Alcover, Jun 05 2012 *)
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PROG
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See link for Haskell program.
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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