A097962 Slowest increasing sequence where the digits, taken one by one, show the pattern even/odd/even/odd/even...

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 23, 25, 27, 29, 41, 43, 45, 47, 49, 61, 63, 65, 67, 69, 81, 83, 85, 87, 89, 210, 301, 410, 501, 610, 701, 810, 901, 2101, 2103, 2105, 2107, 2109, 2121, 2123, 2125, 2127, 2129, 2141, 2143, 2145, 2147, 2149, 2161, 2163, 2165, 2167, 2169

Offset: 0

Author

Eric Angelini, Sep 06 2004

Keywords

• nonn
• base

Comments

Distinct from A098951, which is not required to be increasing. The first 31 terms are identical, but here a(30) = 210 must be followed by a(31) = 301, while there 210 is followed by 10. - M. F. Hasler, Mar 23 2019

Links

• M. F. Hasler, Table of n, a(n) for n = 0..10000

Programs

Mathematica

nn = 57; c[_] := False; a[0] = j = 0; p = 1; c[0] = True;
Do[k = j;
  While[(Set[q, Mod[#[[-1]], 2]];
    Nand[! c[k], Mod[#[[1]], 2] == p,
    Union[Length /@ SplitBy[#, EvenQ]] == {1}]) &[IntegerDigits[k]],
    k++]; Set[{a[n], j, p, c[k]}, {k, k, 1 - q, True}], {n, nn}];
Array[a, nn + 1, 0] (* Michael De Vlieger, Dec 09 2024 *)

PARI

nxt(n,d=digits(n))={if(!bittest(#d,0), forstep(i=#d,1,-1, 10>(d[i]+=2)&& return(fromdigits(d)); d[i]-=10);  d||return(1); d[#d]=if(d[1]%=2,10,21); fromdigits(Vecrev(d)), 10>d[1]+=1, d[1]=d[1]*10+d[#d]; fromdigits(d)\10, d[1]=21; fromdigits(d))}
vector(50,i,t=if(i>1,nxt(t),0)) \\ M. F. Hasler, Mar 23 2019