4
$\begingroup$

This problem relates to Prepend 0 to sublists I have the list

t1 = Table[m = n + IntegerReverse[n]; n = m, {n, 3}, {i, 1, 4}]

giving

{{2, 4, 8, 16}, {4, 8, 16, 77}, {6, 12, 33, 66}}

But I want to start with the first value n so I should get

{{1, 2, 4, 8, 16}, {2, 4, 8, 16, 77}, {3, 6, 12, 33, 66}}

It can be done by

Prepend[#, #[[1]]/2] & /@ t1

But this is cheatingly awkward. How can I do it properly or within the Table command? I tried Hold and Prepend inside Table but did not get anywhere.

$\endgroup$
  • $\begingroup$ Flatten /@ Thread[{Range[3], t1}] seems natural enough. $\endgroup$ – LouisB Feb 22 at 5:53
4
$\begingroup$
MapIndexed[Join[#2, #] &, Table[n = n + IntegerReverse[n], {n, 3}, {i, 4}]]

{{1, 2, 4, 8, 16}, {2, 4, 8, 16, 77}, {3, 6, 12, 33, 66}}

To get the same result using only Table:

Table[n = If[i == 0, n, n + IntegerReverse[n]], {n, 3}, {i, 0, 4}]

{{1, 2, 4, 8, 16}, {2, 4, 8, 16, 77}, {3, 6, 12, 33, 66}}

|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.