A165195 -id:A165195 - cp's OEIS frontend

A165196 Partial sums of A165195.

1, 2, 4, 5, 7, 12, 14, 15, 17, 22, 37, 42, 44, 49, 51, 52, 54, 59, 74, 79, 94, 146, 161, 166, 165, 173, 188, 193, 195, 200, 202, 203

Offset: 1

Gary W. Adamson, Sep 06 2009

nonn

a(2^n) = A000110(n-1); e.g. a(8) = a(2^3) = A000110(4), where A000110 =

the Bell sequence.

			a(8) = 15 = sum of the first 8 terms of A165195: (1, 1, 2, 1, 2, 5, 2, 1...).

A000110, A165194, A165195

Partial sums of sequence A165195

A165194 Triangle of 2^n terms by rows, left half of (n+1)-th row = row n; right half = "reverse and increment" row n; using terms in A000110.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 1, 2, 5, 15, 5, 2, 5, 2, 1

Offset: 1

Gary W. Adamson, Sep 06 2009

Row sums = A000110, the Bell sequence starting with offset 1; (1, 2, 5, 15,...).

Rows tend to A165195.

			First few rows of the triangle =
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 \kQ halves, we obtain row 4: (1, 1, 2, 1, 2, 5, 2, 1), sum = 15 = A000110(4).

A000110, A165195, A165196

Given the Bell sequence, A000110: (1, 1, 2, 5, 15,...); row 1 = 1, row 2 =
(1, 1);...where left half of row (n+1) = row n. Right half of row (n+1)
= reversal of row n, replacing terms with the next Bell number.

cp's OEIS Frontend

A165196 Partial sums of A165195.

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