A339951 Similar to A100787, but alternate with addition and multiplication instead. See Comments section for more information.

1, 2, 4, 8, 64, 70, 280, 287, 0, 2, 16, 16, 32, 40, 280, 280, 560, 561, 3366, 3367, 20202, 20205, 40410, 40414, 0, 2, 16, 16, 32, 40, 0, 5, 30, 30, 150, 156, 156, 159, 477, 483, 2898, 2901, 8703, 8709, 60963, 60965, 0, 2, 0, 2, 4, 4, 8, 8, 40, 44, 0, 4, 4, 4

Offset

0,2

Comments

We start with a(0) as 1, placing an addition sign below the digit. As such, we add 1 to a(0) to receive the next term which is 2. This time, place a multiplication sign below the new digit and double a(1) to receive a(2) which is 4. Place add under the next unused digit and add this to a(2), followed by multiply under the next a(3) and repeat.

Unlike A100787, terms within this sequence are not strictly increasing, as the 8th digit in this sequence is a 0 where we apply multiplication. Also, the differences do not reflect the digits of this sequence, due to alternation with multiplication. The records are 1, 2, 4, 8, 64, 70, 280, 287, 280, 560, 561, 3366 ...

Links

Table of n, a(n) for n=0..59.

Example

1, 2, 4, 8, 64, 70, 280, 287, 0, 2, 16, ...
+  *  +  *  +*  +*  +*+  *+*  +  *  +*
To calculate a(5), we add 6 to 64, as the 5th digit is a 6 with the add operator underneath. Thus, a(5) is 70.
To calculate a(8), we multiply 287 by 0, as the 8th digit is a 0 with the multiply operator underneath. Thus, a(8) is 0.

Crossrefs

Cf. A100787, A107974, A107975, A107976, A107977.

Sequence in context: A036447 A371966 A263451 * A069027 A362343 A065549

Adjacent sequences: A339948 A339949 A339950 * A339952 A339953 A339954

Keyword

base,nonn

Author

Jamie Robert Creasey, Dec 24 2020

Status

approved