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A275605
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Number of set partitions of [n] such that j is member of block b only if b = 1 or at least one of j-1, j-2 is member of a block >= b-1.
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2
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1, 1, 2, 5, 15, 51, 191, 773, 3336, 15207, 72697, 362447, 1876392, 10051083, 55544661, 315899245, 1845139684, 11048651523, 67719859612, 424287619507, 2714074517843, 17706680249505, 117704101959444, 796546613501759, 5483490237025393, 38372546811580251
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OFFSET
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0,3
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COMMENTS
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Original name was: The 'AND' Motzkin Numbers.
This sequence consists of the place values from counting in a pattern where the digit is carried if the current place exceeds both the next place plus one and the place after that plus one. (Note that the place "after" a digit is equally described as the digit preceding it, since we write high-order digits first.)
If the "and" logical comparison is changed to "or", then that modified definition produces the Motzkin numbers A001006.
If the definition looks only at the next term, this generates the Catalan numbers A000108.
This is the case k = 2 of a class of sequences, counting sequences where the k-th term is not more than one more than the maximum of the previous k values. The case k = 1 is the Catalan numbers. The limit as k goes to infinity is the Bell numbers A000110. A similar series limiting terms to no more than one more than the minimum of the previous k values has again the Catalan numbers for k = 1, the Motzkin numbers for k = 2, and continues from there. In this case the limit is the all-ones sequence. - Franklin T. Adams-Watters, Mar 14 2017
To get all the sequences of numerals of length n, take all the numerals of length strictly less than n, and pad them on the left with zeros to length n. - Franklin T. Adams-Watters, May 26 2017
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LINKS
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EXAMPLE
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The sequence of numerals starts 0, 1, 10, 11, 12, 100, 101, 102, 110, 111, 112, 120, 121, 122, 123.
To get the numeral following 12, we first increment the final digit: 13. But the digits before the 3 are 0 (implied) and 1, and 3 is greater than either of those by more than 1. So we set the last digit to 0, and increment the previous one: 20. Again, 2 is too large for the two implicit zeros in front of it, so we set it to 0 and increment the preceding digit, an implicit zero; so we get 100, which presents no problems.
The length 3 numerals come from the numerals less than 100: 0, 1, 10, 11, 12. Inserting leading zeros to length 3 gives 000, 001, 010, 011, 012.
The values of 1, 10, 100, 1000, etc. make up the sequence.
a(5) = 51 = 52 - 1 = A000110(5) - 1 counts all set partitions of [5] except: 134|2|5. - Alois P. Heinz, May 27 2017
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MAPLE
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b:= proc(n, i, j) option remember; `if`(n=0, 1,
add(b(n-1, max(j, k), k), k=1..i+1))
end:
a:= n-> b(n, 0$2):
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MATHEMATICA
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SIZ = 30; MAX = 100000;
M = Table[0, {n, 1, SIZ + 2}];
For[i = 0, i <= MAX, i++, sum = 0; For[j = 1, j <= SIZ, j++, sum += M[[j]]; ]
If[sum == 1, Print[i]]M[[1]]++;
For[j = 1, j <= SIZ, j++, If[M[[j]] > M[[j + 1]] + 1 && M[[j]] > M[[j + 2]] + 1, M[[j]] = 0; M[[j + 1]]++]]] (* Benedict W. J. Irwin, Nov 14 2016 *)
b[n_, i_, j_] := b[n, i, j] = If[n==0, 1, Sum[b[n-1, Max[j, k], k], {k, 1, i+1} ] ];
a[n_] := b[n, 0, 0];
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PROG
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#include <stdio.h>
int main(){
int N=25; //max digits
int j, s[N], sum;
long long int i=0;
for(j=0; j<N; j++) s[j]=0;
while(1){
sum=0;
for(j=0; j<N; j++) sum+=s[j];
if(sum==1){ printf("%lld, ", i); fflush(stdout); }
s[0]++;
for(j=0; j<N-2; j++){
if( (s[j]>s[j+1]+1) && (s[j]>s[j+2]+1) ){
s[j]=0;
s[j+1]+=1;
}
}
i++;
}
return 0;
}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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