A159629 Slowest increasing sequence beginning with a(1)=4 such that A002828(a(n)) = A002828(n).

4, 5, 6, 9, 10, 11, 15, 17, 25, 26, 27, 30, 32, 33, 39, 49, 50, 52, 54, 58, 59, 62, 63, 66, 81, 82, 83, 87, 89, 91, 92, 97, 99, 101, 102, 121, 122, 123, 124, 125, 128, 129, 131, 132, 136, 138, 143, 147, 169, 170, 171, 173, 178, 179, 183, 184, 186, 193, 195, 199, 200, 201, 207

Offset: 1

Author

Vladimir Shevelev, Apr 17 2009, May 04 2009

Keywords

• nonn

Comments

Conjecture: For every m>2 there exists a minimum index N(m) such that the minimal increasing recursive sequence S_m(n) beginning with m^2 with the condition A002828(S_m(n)) = A002828(n) coincides with a(n) for all n>N.

Links

• V. Shevelev, Several results on sequences which are similar to the positive integers arXiv:0904.2101 [math.NT], 2009.

Crossrefs

Cf. A002828, A159619, A159615, A007814, A004760, A159559, A159560.

Programs

Mathematica

a2828[n_] := Which[SquaresR[1, n]>0, 1, SquaresR[2, n]>0, 2, SquaresR[3, n] > 0, 3, True, 4];
a[1] = 4; a[n_] := a[n] = For[k = a[n-1]+1, True, k++, If[a2828[k] == a2828[n], Return[k]]];
Array[a, 63] (* Jean-François Alcover, Jul 28 2018 *)

Formula

a(n+1) = min { l > a(n) : A002828(l) = A002828(n+1) }.

Extensions

137 replaced by 136, extended by R. J. Mathar, Sep 17 2009