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URL: https://oeis.org/A165194

⇱ A165194 - OEIS

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A165194
Triangle of 2^(n-1) terms by rows, left half of (n+1)-th row = row n; right half = "reverse and increment" row n; using terms in A000110.
4
1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1, 2, 5, 15, 5, 15, 52, 15, 5, 2, 5, 15, 5, 2, 5, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1, 2, 5, 15, 5, 15, 52
OFFSET
1,6
COMMENTS
Given the Bell sequence, A000110: row 1 = [1]; thereafter: The left half of row (n+1) = row n. The right half of row (n+1) = the reversal of row n, replacing terms with the next Bell number.
Rows tend to A165195.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..8191 (rows 1..13)
FORMULA
T(n,k) = A165195(k). - Andrew Howroyd, Sep 24 2025
EXAMPLE
Triangle T(n,k) begins (n >= 1, 1 <= k <= 2^(n-1)):
1;
1, 1;
1, 1, 2, 1;
1, 1, 2, 1, 2, 5, 2, 1;
1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1;
...
For example: row 4, left half = (1, 1, 2, 1); right half = (1, 2, 1, 1), replaced with the next higher Bell numbers: (2, 5, 2, 1). Appending the two halves, we obtain row 4: (1, 1, 2, 1, 2, 5, 2, 1), sum = 15 = A000110(4).
PROG
(PARI) row(n)={my(v=vector(2^(n-1), i, i--; hammingweight(bitxor(i, i>>1))), b=Vec(serlaplace(exp(exp(x+O(x^n)) - 1)))); vector(#v, i, b[1+v[i]])} \\ Andrew Howroyd, Sep 24 2025
CROSSREFS
Row sums are A000110.
Sequence in context: A331310 A241597 A374164 * A002951 A331287 A365105
KEYWORD
nonn,tabf
AUTHOR
Gary W. Adamson, Sep 06 2009
EXTENSIONS
Duplicate term removed and a(32) onwards from Andrew Howroyd, Sep 24 2025
STATUS
approved