A165195 Rows of triangle A165194 tend to this sequence; generated from A000110.

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, 2, 5, 15, 5, 15, 52, 15, 5, 15, 52, 203, 52, 15, 52, 15, 5, 2, 5, 15, 5, 15, 52, 15, 5, 2, 5, 15, 5, 2, 5, 2, 1, 2, 5, 15, 5, 15, 52, 15, 5, 15, 52, 203

Offset

1,3

Comments

The sequence can be generated from strings of 2^n terms starting (1, 1,... then the next string of 2^(n+1) terms is obtained by appending a "reverse and increment" substring to the previous substring.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..4096

Formula

a(n) = A000110(A005811(n-1)). - Andrew Howroyd, Sep 24 2025

Example

Given terms in the Bell sequence, A000110; A165195 begins (1, 1, 2, 1,... then to obtain the first 2^3 terms, the first 2^2 terms = (1, 1, 2, 1,... then append to the latter the reversal of (1, 1, 2, 1) = (1, 2, 1, 1) but incremented with the next higher Bell number = (2, 5, 2, 1). The first 2^3 terms are thus (1, 1, 2, 1, 2, 5, 2, 1). Repeat with analogous operations to obtain 2^4 terms, and so on..

Prog

(PARI) seq(n)={my(v=vector(n, i, i--; hammingweight(bitxor(i, i>>1))), b=Vec(serlaplace(exp(exp(x+O(x*x^vecmax(v))) - 1)))); vector(#v, i, b[1+v[i]])} \\ Andrew Howroyd, Sep 24 2025

Crossrefs

Cf. A000110, A005811, A165194, A165196.

Sequence in context: A351517 A337224 A090079 * A121487 A057031 A230219

Adjacent sequences: A165192 A165193 A165194 * A165196 A165197 A165198

Keyword

nonn

Author

Gary W. Adamson, Sep 06 2009

Extensions

a(33) onwards from Andrew Howroyd, Sep 24 2025

Status

approved