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Cf. A117609 (A(n) from Name isin A117609(nname).

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P[n_] := (s = Sum[SquaresR[3, k], {k, 0, n}]) - Round[(4/3)*Pi*n^(3/2)]; record = 0; A000092 = Join[{1}, Reap[For[n = 10, n <= 10^4, n++, If[(p = Abs[P[n]]) > record, record = p; Print[s]; Sow[s]]]][[2, 1]]] (* _]] (* _Jean-François Alcover_, Feb 08 2016, after M. F. Hasler in A000092 *)

A(n) from Name is A117609(n).

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M. F. Hasler: I think it is vastly more natural that a(0) and A92(0) correspond to the radius and index n = 0 with A(0) = A117609(0) = #{(0,0,0)} = 1, and a(1) corresponds to radius 1 and A(1) = A117609(1) = 1+6, and a(2) corresponds to radius 2 and A(2). The problem only arises from the somewhat irrational idea that counting and therefore indices in lists should start at index 1. Indices are just symbols, their numeric value does not have a meaning, so it's not *wrong* to start with 1, but the complications that arise from that choice are to me a clear hint that this is not a good choice. (In an axiomatic approach to natural numbers one has 1 = {0} and therefore indices [e.g., of vectors x \in R^n = R^{0,1,...,n-1}] start with 0.)

Seth A. Troisi: Sorry to keep pressing this point but A in the definition != A000413 but instead is A117609 which is already 0 indexed. I completely agree that A(0) = A117609(0) = #{(0,0,0)} = 1 and that leads to P(0) = | 1 - 0 | = 1. Maybe a good first step to rectifying this is to add the term 0 to A92. I have no open edits so I can't propose such a change but I would appreciate if such a change was proposed (potentially linking to this discussion).

The initial value a(0) = 1 corresponds to an initial A000092(0) = 0 which is the index of a record in the sense that the value P(0) = 0 is larger than all preceding values, because there are none. - M. F. Hasler, May 04 2022

a(n) = A117609(A000092(n)), considering A000092(0) = 0. - M. F. Hasler, May 04 2022

From M. F. Hasler, May 04 2022: (Start)

a(0) = #{(0, 0, 0)} = 1,

a(1) = #{(0, 0, 0), (0, 0, +-1) and cyclic permutations} = 7.

a(2) = #{(0, 0, 0), (0, 0, +-1), +-(0, 1, +-1) & cyclic} = 1 + 6 + 12 = 19. (End)

M. F. Hasler: Yes, one might have started the list of record indices to start with A92(1)=0 instead of A92(1)=1, A92(0)=0, but on one hand we don't change such very old and well established sequences just for an indexing convention; also, I think it is natural to start with index 0 for the initial value equal to 1, since it is a record in a somewhat special sense ("voidly"). All these problems will disappear when people start to accept that the counting = natural numbers start at 0. Unfortunately, mankind seems not yet have reached this state of maturity. Maybe the popularity of Python will help to complete this step within the next generation.

10,2

From M. F. Hasler, May 04 2022: (Start)

a(0) = #{(0, 0, 0)} = 1,

a(1) = #{(0, 0, 0), (0, 0, +-1) and cyclic permutations} = 7.

a(2) = #{(0, 0, 0), (0, 0, +-1), +-(0, 1, +-1) & cyclic} = 1 + 6 + 12 = 19. (End)

Offset changed by Seth A. Troisi, May 03 2022

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M. F. Hasler: Seth: yes, A(n) is clearly defined for n=0, as is A000092 if you look at the definition but not at the offset. Also, there is no ambiguity about the values of a(n) (the sequence here) for 1,2,3...  All that is missing is the value A000092(0) that is forgotten to be listed there: it does not set a *new* record because it sets the *initial* value.... but I suggested in the PROGRAM section already 10 years ago, that there could be an initial a(0)=0.

Seth A. Troisi: I think there's some confusion with having A(n) in the name. Let me try to be over specific. 

"Let *Z*(n) = #{(i,j,k): i^2 + j^2 + k^2 <= n}, V(n) = (4/3)Pi*n^(3/2), P(n) = *Z*(n) - V(n); A000092 gives values of n where |P(n)| sets a new record; *This* sequence gives *Z*(A92(n))."

A(1) = *Z*(A000092(1)) = *Z*(1) = 7
A(2) = *Z*(A000092(2)) = *Z*(2) = 19

If A(1) doesn't equal 1 it's because 0 is omitted from A000092(1).

Seth A. Troisi: IMHO A000092 should have 0 added (but not it's offset changed). A000223 should have 0 added (but not it's offset changed). A000413 (this sequence) should keep 1 as now. A117609(A000092(1)=0) = 1, A117609(A000092(2)=1) =7, A117609(A000092(3)=2) =19, A117609(A000092(4)=5) =57, A117609(A000092(5)=6)=81, A117609(A000092(6)=14)=251

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Michel Marcus: I am not sure; assembling PARI codes from A210639, A117609, A210641, and A000092 I get 7, 19, 57, 81, ... for n>=1

Michel Marcus: I used lista413(nn) = m=0; for(n=1, nn, if(m+0<m=max(abs(A210641(n)), m), print1(A117609(n), ", "))); for last leg

Seth A. Troisi: All the offsets for the 5 linked sequences start at 1. By the definition the 0th term here represents A(A000092(0)) which doesn't exist.

Michel Marcus: I will ask Maximilian to come and see (since I used his codes to check)

Michel Marcus: Also I see Join[{1}, Reap[For[n = 1, n <= 10^4, n++ in Mma

M. F. Hasler: Michel, I think you are right and the change made by Seth is incorrect:
We have A92(1) = 1 and thus a(1) = A(1) = #{(0,0,0), (0,0,+-1), (0,+-1,0), (+-1,0,0)} = 7.

M. F. Hasler: I think the ambiguity would be resolved if the "missing" a(0)=0 would be added to A92 (as I suggest in PARI there, and maybe in history?) Clearly, A(0) = { (0,0,0) } = 1.
Also, A92(2) = 2 and A(2) = # +-{ (0,0,0), (0,0,1), (0,1,+-1) and cyclic permutations } = 1 + 6 + 12 = 19.

Let A(n) = #{(i,j,k): i^2 + j^2 + k^2 <= n}, V(n) = (4/3))*Pi*n^(3/2), P(n) = A(n) - V(n); A000092 gives values of n where |P(n)| sets a new record; sequence gives A(A000092(n)).

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