OFFSET
1,1
COMMENTS
Begin with the left (most significant) k digits and sequentially remove the first j leading digits until a palindrome is found; continue.
a(33) is actually '00', a(34) is actually '030' (which should be obvious), a(38) is actually '00'.
If e is normal number then all multidigit palindromes should appear.
LINKS
EXAMPLE
e = 2.7182818284590452353602874713526624977572470936999595749669676...
a(1) = 828, as the first nontrivial palindrome in E is '828', which appears in the digits 4 through 6.
a(2) = 18281, as the second nontrivial palindrome in E is '18281', which appears in the digits 3 through 7.
Please note that a(1) runs from digits 4-6 whereas a(2) runs from 3-7. This is why a(1) appears before a(2).
MATHEMATICA
e = RealDigits[E, 10, 250][[1]]; palQ[n_] := n == Reverse[n]; k = 2; lst = {}; While[k < 251, While[j < k, If[ palQ[ Take[e, {j, k}]], p = FromDigits[Take[e, {j, k}]]; AppendTo[lst, p]; Print[p]]; j++]; k++; j = 1]
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Reinhard Zumkeller and Robert G. Wilson v, Jun 09 2013
STATUS
approved