A271063 a(1) = 2, a(2) = 3; thereafter a(n) = a(n-1) + a(|n-a(T)|), where a(T) is the largest term in the sequence before a(n) such that 0 < |n-a(T)| < n.
2, 3, 5, 7, 10, 17, 22, 25, 50, 72, 89, 161, 322, 411, 483, 533, 558, 580, 597, 607, 614, 619, 622, 624, 629, 1253, 1875, 2494, 3108, 3715, 4312, 4892, 5450, 5983, 6466, 6877, 13343, 19326, 24776, 29668, 33980, 37695, 40803, 43297, 86594, 127397, 165092, 199072, 228740, 253516
Offset: 1
Keywords
Examples
a(11) = 89 a(T) = a(7) = 22 |12-22| = 10 a(10) = 72 89 + 72 = 161 a(12) = 161
Links
- Cody M. Haderlie, Table of n, a(n) for n = 1..1982
Programs
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Mathematica
a = {2, 3}; Do[AppendTo[a, a[[n - 1]] + Part[a, #] &@ SelectFirst[ Reverse@ Abs[a - n], 0 < # < n &]], {n, 3, 50}]; a (* Michael De Vlieger, Apr 08 2016 *) -
PARI
lista(nn) = {va = vector(nn); print1(va[1] = 2, ", "); print1(va[2] = 3, ", "); for (n=3, nn, T = 0; forstep(k = n-1, 1, -1, vabs = abs(n - va[k]); if ((vabs < n) && (vabs > 0), T = k; break);); va[n] = va[n-1] + va[abs(n-va[T])]; print1(va[n], ", "););} \\ Michel Marcus, Apr 08 2016
Comments