A138027 -id:A138027 - cp's OEIS frontend

A121561 The number of iterations of "subtract the largest prime less than or equal to the current value" to go from n to the limiting value 0 or 1.

0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2

Offset: 1

Kerry Mitchell, Aug 07 2006

Number of steps to go from n to A121559(n).

The sequence has the form of blocks of numbers; see A121562 for the lengths of those blocks.

			a(9) = 2 because there are 2 steps in going from 9 to 0 in A121559: 9 mod 7 = 2 and 2 mod 2 = 0.

Robert G. Wilson v, Table of n, a(n) for n = 1..100000

Cf. A121559, A064722, a(n)=1: A093515, a(n)=2: A093513, a(n)=3: A138026, a(n)=4: A138027.

LrgstPrm[n_] := Block[{k = n}, While[ !PrimeQ@ k, k-- ]; k]; f[n_] := Block[{c = 0, d = n}, While[d > 1, d = d - LrgstPrm@d; c++ ]; c]; Array[f, 105] (* Robert G. Wilson v, Feb 29 2008 *)

from sympy import prevprime
def a(n): return 0 if n == 0 or n == 1 else 1 + a(n - prevprime(n+1))
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Jul 26 2022

A138026 Numbers k where A121561(k) = 3.

Original entry on oeis.org

122, 123, 148, 190, 208, 209, 220, 221, 250, 292, 302, 303, 326, 327, 346, 418, 430, 476, 477, 518, 519, 532, 533, 538, 539, 556, 586, 628, 629, 640, 670, 671, 700, 718, 782, 783, 796, 806, 807, 820, 838, 848, 849, 872, 873, 896, 897, 902, 903, 928, 962

Offset: 1

Robert G. Wilson v, Feb 27 2008

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A121561, A093515, A093513, A138027.

filter:= proc(n)
  local t;
  t:= n-prevprime(n+1);
  if t <= 1 then return false fi;
  t:= t-prevprime(t+1);
  if t <= 1 then return false fi;
  isprime(t) or isprime(t-1)
end proc:
select(filter,[$2..1000]); # Robert Israel, Jan 15 2019

LrgstPrm[n_] := Block[{k = n}, While[ !PrimeQ@k, k-- ]; k]; f[n_] := Block[{c = 0, d = n}, While[d > 1, d = d - LrgstPrm@ d; c++ ]; c]; lst = {}; Do[ If[f@n == 3, AppendTo[lst,n]], {n, 10^3}]; lst

cp's OEIS Frontend

A121561 The number of iterations of "subtract the largest prime less than or equal to the current value" to go from n to the limiting value 0 or 1.

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