A260734 -id:A260734 - cp's OEIS frontend

A276572 Simple self-inverse permutation of natural numbers: after a(0)=0, list each block of A260734(n) numbers in reverse order, from A260732(n) to A260733(1+n).

0, 1, 3, 2, 5, 4, 9, 8, 7, 6, 13, 12, 11, 10, 18, 17, 16, 15, 14, 23, 22, 21, 20, 19, 30, 29, 28, 27, 26, 25, 24, 37, 36, 35, 34, 33, 32, 31, 44, 43, 42, 41, 40, 39, 38, 52, 51, 50, 49, 48, 47, 46, 45, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 71, 70, 69, 68, 67, 66, 65, 64, 63, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72

Offset: 0

Antti Karttunen, Sep 07 2016

nonn

Maps between A276573 and A276574.

Antti Karttunen, Table of n, a(n) for n = 0..10028
Index entries for sequences that are permutations of the natural numbers

Cf. A260732, A260733, A260734, A276573, A276574.

f[n_] := NestWhileList[# - (If[First@ # > 0, 1, Length[First@ Split@ #] + 1] &@ SquaresR[Range@ 4, #]) &, n^2, # != 0 &]; t = Table[Table[n, {Length[#] - 1 &@ NestWhileList[# - (If[First@ # > 0, 1, Length[First@ Split@ #] + 1] &@ SquaresR[Range@ 4, #]) &, ((n + 1)^2) - 1, # != (n^2) - 1 &]}], {n, 20}] // Flatten ; {0}~Join~Table[Length@ f@ t[[n]] - 1 + Length@ f[t[[n]] + 1] - n - 2, {n, 81}] (* Michael De Vlieger, Sep 08 2016 *)

(define (A276572 n) (if (zero? n) n (+ (- (A260733 (+ 1 (A276571 n))) n) (A260732 (A276571 n)))))

a(0) = 0; for n >= 1, a(n) = (A260733(1+A276571(n))-n)+A260732(A276571(n)).

A276571 After a(0)=0, each n occurs A260734(n) times.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15

Offset: 0

Antti Karttunen, Sep 07 2016

nonn

Auxiliary function for computing A276572 & A276573.

Antti Karttunen, Table of n, a(n) for n = 0..10028

Cf. A260733, A260734, A276572, A276573.

Cf. also A276581.

Table[Table[n, {Length[#] - 1 &@ NestWhileList[# - (If[First@ # > 0, 1, Length[ First@ Split@ #] + 1] &@ SquaresR[Range@ 4, #]) &, ((n + 1)^2) - 1, # != (n^2) - 1 &]}], {n, 15}] // Flatten (* Michael De Vlieger, Sep 08 2016 *)

(define (A276571 n) (let loop ((k 0)) (if (>= (A260733 (+ 1 k)) n) k (loop (+ 1 k)))))

A002828 Least number of squares that add up to n.

Original entry on oeis.org

0, 1, 2, 3, 1, 2, 3, 4, 2, 1, 2, 3, 3, 2, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 3, 1, 2, 3, 4, 2, 3, 4, 2, 3, 2, 3, 1, 2, 3, 4, 2, 2, 3, 3, 3, 2, 3, 4, 3, 1, 2, 3, 2, 2, 3, 4, 3, 3, 2, 3, 4, 2, 3, 4, 1, 2, 3, 3, 2, 3, 3, 4, 2, 2, 2, 3, 3, 3, 3, 4, 2, 1, 2, 3, 3, 2, 3, 4, 3, 2, 2, 3, 4, 3, 3, 4, 3, 2, 2, 3, 1, 2, 3, 4, 2, 3

Offset: 0

N. J. A. Sloane

Lagrange's "Four Squares theorem" states that a(n) <= 4.
It is easy to show that this is also the least number of squares that add up to n^3.
a(n) is the number of iterations in f(...f(f(n))...) to reach 0, where f(n) = A262678(n) = n - A262689(n)^2. Allows computation of this sequence without Lagrange's theorem. - Antti Karttunen, Sep 09 2016
It is also easy to show that a(k^2*n) = a(n) for k > 0: Clearly a(k^2*n) <= a(n) but for all 4 cases of a(n) there is no k which would result in a(k^2*n) < a(n). - Peter Schorn, Sep 06 2021

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..20000 (first 1001 terms from T. D. Noe with corrections from Michel Marcus)
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
N. J. A. Sloane, Transforms.
Eric Weisstein's World of Mathematics, Square Number.

Cf. A000290, A000415, A000419, A002376, A004215, A000378, A001481.

Cf. A010052, A025426, A025427, A053610, A062535, A072401, A229062, A255131, A260728, A260731, A260734, A262678, A262689, A262690, A276573.

a002828 0 = 0  -- confessedly  /= 1, as sum [] == 0
a002828 n | a010052 n == 1 = 1
          | a025426 n > 0 = 2 | a025427 n > 0 = 3 | otherwise = 4
-- Reinhard Zumkeller, Feb 26 2015

with(transforms);
sq:=[seq(n^2, n=1..20)];
LAGRANGE(sq,4,120);
# alternative:
f:= proc(n) local F,x;
   if issqr(n) then return 1 fi;
   if nops(select(t -> t[1] mod 4 = 3 and t[2]::odd, ifactors(n)[2])) = 0 then return 2 fi;
   x:= n/4^floor(padic:-ordp(n,2)/2);
   if x mod 8 = 7 then 4 else 3 fi
end proc:
0, seq(f(n),n=1..200); # Robert Israel, Jun 14 2016
# next Maple program:
b:= proc(n, i) option remember; convert(series(`if`(n=0, 1, `if`(i<1, 0,
      b(n, i-1)+(s-> `if`(s>n, 0, x*b(n-s, i)))(i^2))), x, 5), polynom)
    end:
a:= n-> ldegree(b(n, isqrt(n))):
seq(a(n), n=0..105);  # Alois P. Heinz, Oct 30 2021

SquareCnt[n_] := If[SquaresR[1, n] > 0, 1, If[SquaresR[2, n] > 0, 2, If[SquaresR[3, n] > 0, 3, 4]]]; Table[SquareCnt[n], {n, 150}] (* T. D. Noe, Apr 01 2011 *)
sc[n_]:=Module[{s=SquaresR[Range[4],n]},If[First[s]>0,1,Length[ First[ Split[ s]]]+1]]; Join[{0},Array[sc,110]] (* Harvey P. Dale, May 21 2014 *)

istwo(n:int)=my(f);if(n<3,return(n>=0););f=factor(n>>valuation(n, 2)); for(i=1,#f[,1],if(bitand(f[i,2],1)==1&&bitand(f[i,1],3)==3, return(0)));1
isthree(n:int)=my(tmp=valuation(n,2));bitand(tmp,1)||bitand(n>>tmp,7)!=7
a(n)=if(isthree(n), if(issquare(n), !!n, 3-istwo(n)), 4) \\ Charles R Greathouse IV, Jul 19 2011, revised Mar 17 2022

from sympy import factorint
def A002828(n):
    if n == 0: return 0
    f = factorint(n).items()
    if not any(e&1 for p,e in f): return 1
    if all(p&3<3 or e&1^1 for p,e in f): return 2
    return 3+(((m:=(~n&n-1).bit_length())&1^1)&int((n>>m)&7==7)) # Chai Wah Wu, Aug 01 2023

from sympy.core.power import isqrt
def A002828(n):
    dp = [-1] * (n + 1)
    dp[0] = 0
    for i in range(1, n + 1):
        S = []
        r = isqrt(i)
        for j in range(1, r + 1):
            S.append(1 + dp[i - (j**2)])
        dp[i] = min(S)
    return dp[-1] # Darío Clavijo, Apr 21 2025

;; The first one follows Charles R Greathouse IV's PARI-code above:
(define (A002828 n) (cond ((zero? n) n) ((= 1 (A010052 n)) 1) ((= 1 (A229062 n)) 2) (else (+ 3 (A072401 n)))))
(define (A229062 n) (- 1 (A000035 (A260728 n))))
;; We can also compute this without relying on Lagrange's theorem. The following recursion-formula should be used together with the second Scheme-implementation of A262689 given in the Program section that entry:
(definec (A002828 n) (if (zero? n) n (+ 1 (A002828 (- n (A000290 (A262689 n)))))))
;; Antti Karttunen, Sep 09 2016

From Antti Karttunen, Sep 09 2016: (Start)
a(0) = 0; and for n >= 1, if A010052(n) = 1 [when n is a square], a(n) = 1, otherwise, if A229062(n)=1, then a(n) = 2, otherwise a(n) = 3 + A072401(n). [After Charles R Greathouse IV's PARI program.]
a(0) = 0; for n >= 1, a(n) = 1 + a(n - A262689(n)^2), (see comments).
a(n) = A053610(n) - A062535(n).
(End)

More terms from Arlin Anderson (starship1(AT)gmail.com)

A255131 n minus the least number of squares that add up to n: a(n) = n - A002828(n).

Original entry on oeis.org

0, 0, 0, 0, 3, 3, 3, 3, 6, 8, 8, 8, 9, 11, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 21, 24, 24, 24, 24, 27, 27, 27, 30, 30, 32, 32, 35, 35, 35, 35, 38, 39, 39, 40, 41, 43, 43, 43, 45, 48, 48, 48, 50, 51, 51, 51, 53, 54, 56, 56, 56, 59, 59, 59, 63, 63, 63, 64, 66, 66, 67, 67, 70, 71, 72, 72, 73, 74, 75, 75, 78, 80, 80, 80, 81

Offset: 0

Antti Karttunen, Feb 24 2015

nonn

The associated beanstalk-sequence starts from a(0) as: 0, 3, 6, 8, 11, 15, 16, 18, 21, ... (A276573).

			a(0) = 0, because no squares are needed for an empty sum, and 0 - 0 = 0.
a(3) = 0, because 3 cannot be represented as a sum of less than three squares (1+1+1), and 3 - 3 = 0.
a(4) = 3, because 4 can be represented as a sum of just one square (namely 4 itself), and 4 - 1 = 3.

Robert Israel, Table of n, a(n) for n = 0..10000
Discussion on the SeqFan mailing list

Subsequence: A005563.
Cf. A000290, A002828, A062535, A260731, A260732, A260733, A260734, A262689, A276573.
Cf. also A011371, A236840, A260740.

f:= proc(n) local F, x;
   if issqr(n) then return n-1 fi;
   if nops(select(t -> t[1] mod 4 = 3 and t[2]::odd, ifactors(n)[2])) = 0 then return n-2 fi;
   x:= n/4^floor(padic:-ordp(n, 2)/2);
   if x mod 8 = 7 then n-4 else n-3 fi
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Mar 27 2018

{0}~Join~Table[n - (If[First@ # > 0, 1, Length[First@ Split@ #] + 1] &@ SquaresR[Range@ 4, n]), {n, 84}] (* Michael De Vlieger, Sep 08 2016, after Harvey P. Dale at A002828 *)

a(n) = n - A002828(n).

a(n) = A260740(n) + A062535(n).

A260731 a(n) = Number of steps to reach 0 starting from x=n and using the iterated process: x -> x - A002828(x), where A002828(x) = the least number of squares that add up to x.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 8, 8, 8, 8, 9, 10, 10, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16, 16, 17, 17, 17, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 28, 29, 29, 29, 30, 31, 31, 31, 32, 32, 32, 32, 33, 33, 34, 34, 34, 35, 35, 35, 36, 36, 37, 37, 38

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A002828, A255131, A260732, A260733, A260734.
Left inverse of A276573, A278517 and A278519. A278518(n) gives the number of times n occurs (run lengths).
Cf. also A261221.

A002828[n_] := Which[n == 0, 0, SquaresR[1, n] > 0, 1, SquaresR[2, n] > 0, 2, SquaresR[3, n] > 0, 3, True, 4]; a[0] = 0; a[n_] := a[n] = 1 + a[n - A002828[n]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 14 2016 *)

a(0) = 0; for >= 1, a(n) = 1 + A260731(A255131(n)).
From Antti Karttunen, Nov 28 2016: (Start)
For all n >= 0, a(A278517(n)) = a(A278519(n)) = a(A276573(n)) = n.
(End)

A260733 a(n) = number of steps needed to reach zero when starting from k = (n^2)-1 and repeatedly applying the map that replaces k with k - A002828(k), where A002828(k) = the least number of squares that add up to k.

Original entry on oeis.org

0, 1, 3, 5, 9, 13, 18, 23, 30, 37, 44, 52, 62, 71, 81, 91, 104, 117, 131, 144, 159, 174, 190, 207, 224, 243, 262, 281, 301, 321, 343, 365, 388, 412, 437, 461, 487, 514, 539, 567, 596, 625, 654, 684, 715, 748, 781, 814, 848, 883, 918, 955, 991, 1030, 1067, 1105, 1145, 1187, 1227, 1269, 1311, 1354, 1396, 1441, 1486, 1531, 1579, 1624, 1673, 1723, 1773, 1821

Offset: 1

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..1024

One less than A260732.
Cf. A002828, A255131, A260731, A260734.
Cf. also A261223.

Table[Length[#] - 2 &@ NestWhileList[# - (If[First@ # > 0, 1, Length[ First@ Split@ #] + 1] &@ SquaresR[Range@ 4, #]) &, n^2, # != 0 &], {n, 72}] (* Michael De Vlieger, Sep 08 2016 *)

a(n) = A260731((n^2)-1).

a(n) = A260732(n)-1.

A260732 a(n) = number of steps needed to reach zero when starting from k = n^2 and repeatedly applying the map that replaces k with k - {the least number of squares (A002828) that add up to k}.

Original entry on oeis.org

0, 1, 2, 4, 6, 10, 14, 19, 24, 31, 38, 45, 53, 63, 72, 82, 92, 105, 118, 132, 145, 160, 175, 191, 208, 225, 244, 263, 282, 302, 322, 344, 366, 389, 413, 438, 462, 488, 515, 540, 568, 597, 626, 655, 685, 716, 749, 782, 815, 849, 884, 919, 956, 992, 1031, 1068, 1106, 1146, 1188, 1228, 1270, 1312, 1355, 1397, 1442, 1487, 1532, 1580, 1625

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..1024

Partial sums of A260734.
Essentially one more than A260733.
Cf. A002828, A260731.
Cf. also A261222.

Table[Length[#] - 1 &@ NestWhileList[# - (If[First@ # > 0, 1, Length[ First@ Split@ #] + 1] &@ SquaresR[Range@ 4, #]) &, n^2, # != 0 &], {n, 0, 68}] (* Michael De Vlieger, Sep 08 2016, after Harvey P. Dale at A002828 *)

a(n) = A260731(n^2).

For all n >= 1: a(n) = 1 + A260733(n).

A261224 a(n) = number of steps needed to reach (n^2)-1 when starting from k = ((n+1)^2)-1 and repeatedly applying the map that replaces k with k - A053610(k), where A053610(k) = the number of positive squares that sum to k using the greedy algorithm.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 17, 18, 18, 19, 19, 19, 20, 21, 21, 21, 22, 22, 23, 23, 24, 24, 25, 26, 26, 27, 27, 28, 28, 28, 29, 30, 30, 31, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 37, 37, 38, 38, 39, 39, 39, 40, 41, 41, 42, 42, 42, 43, 43, 44, 44, 45, 45, 46

Offset: 1

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..6000

First differences of both A261222 and A261223.
Cf. A000290, A053610, A260740, A261221.
Cf. also A260734, A261229.

Table[-1 + Length@ NestWhileList[# - Block[{m = #, c = 1}, While[a = (# - Floor[Sqrt@ #]^2) &@ m; a != 0, c++; m = a]; c] &, ((n + 1)^2) - 1, # != n^2 - 1 &], {n, 91}] (* Michael De Vlieger, Sep 08 2016, after Jud McCranie at A053610 *)

a(n) = A261221(((n+1)^2)-1) - A261221((n^2)-1). [The definition.]
Equally, for all n >= 1:
a(n) = A261221((n+1)^2) - A261221(n^2).
a(n) = A261222(n+1) - A261222(n).
a(n) = A261223(n+1) - A261223(n).

A276574 The infinite trunk of least squares beanstalk with reversed subsections.

Original entry on oeis.org

0, 3, 8, 6, 15, 11, 24, 21, 18, 16, 35, 32, 30, 27, 48, 45, 43, 40, 38, 63, 59, 56, 53, 51, 80, 78, 75, 72, 70, 67, 64, 99, 96, 93, 90, 88, 85, 83, 120, 117, 115, 112, 108, 105, 102, 143, 139, 136, 134, 131, 128, 126, 123, 168, 165, 162, 160, 158, 155, 152, 149, 147, 144, 195, 192, 189, 186, 183, 179, 176, 173, 171

Offset: 0

Antti Karttunen, Sep 07 2016

			This can be viewed as an irregular table, where after 0, each row has A260734(n) = 1, 2, 2, 4, 4, 5, 5, 7, ... terms:
0;
3;
8, 6;
15, 11;
24, 21, 18, 16;
35, 32, 30, 27;
48, 45, 43, 40, 38;
63, 59, 56, 53, 51;
80, 78, 75, 72, 70, 67, 64;
99, 96, 93, 90, 88, 85, 83;
120, 117, 115, 112, 108, 105, 102;
...
Each row begins with (n^2)-1 (see A005563), and each successive term is obtained by subtracting A002828(k) from the previous term k, until ((n-1)^2)-1 would be encountered, which is not listed second time (as it already occurs as the first term of the previous row), but instead, the current row is finished and the next row is started with the term ((n+1)^2)-1.

Antti Karttunen, Table of n, a(n) for n = 0..10028

Cf. A005563 (left edge), A277023 (right edge).
Cf. A000196, A000290, A002828, A010052, A255131, A260734, A262689, A276572.
Used to construct A276573.
Cf. A277015 (tells which rows end with squares, listed in A277016).

(definec (A276574 n) (cond ((zero? n) n) ((= 1 n) 3) (else (let ((maybe_next (A255131 (A276574 (- n 1))))) (if (zero? (A010052 (+ 1 maybe_next))) maybe_next (+ -1 (A000290 (+ 2 (A000196 (+ 1 maybe_next))))))))))

a(0) = 0; a(1) = 3; for n > 1, let k = A255131(a(n-1)). If k+1 is not a square, then a(n) = k, otherwise a(n) = A000290(2+A000196(k+1)) - 1.

Example section added and the formula rewritten to a simpler form (which is now correct) - Antti Karttunen, Oct 16 2016

A277890 Number of even numbers encountered before (n^2)-1 is reached when starting from k = ((n+1)^2)-1 and iterating map k -> k - A002828(k).

Original entry on oeis.org

0, 2, 0, 3, 2, 3, 1, 5, 3, 4, 4, 6, 3, 5, 3, 7, 8, 8, 6, 8, 9, 10, 6, 8, 10, 10, 7, 11, 10, 13, 11, 12, 12, 14, 10, 13, 12, 13, 14, 15, 13, 15, 15, 18, 18, 16, 15, 17, 21, 18, 18, 18, 19, 20, 16, 21, 20, 20, 22, 20, 23, 20, 22, 23, 21, 23, 23, 27, 25, 24, 22, 28, 22, 27, 24, 26, 25, 25, 29, 29, 28, 26, 30, 31, 28, 28, 31, 30, 32, 33, 27, 32, 34, 34, 30, 33, 33

Offset: 1

Antti Karttunen, Nov 08 2016

nonn

The starting point ((n+1)^2)-1 of the iteration is included if it is even, but the ending point (n^2)-1 is never included in the count.
a(n) = number of even numbers on row n of A276574, after the initial zero-row.
See also comments in A277891.

			For n=6, we start iterating from k = ((6+1)^2)-1 = 48, and then 48 - A002828(48) = 45, 45 - A002828(45) = 43, 43 - A002828(43) = 40, 40 - A002828(40) = 38, and 38 - A002828(38) = 35 (which is 6^2 - 1), and three of these numbers are even, thus a(6) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000035, A000290, A002828, A010052, A260734, A276573, A276574, A277486, A277488, A277889, A277891.

istwo(n:int)=my(f); if(n<3, return(n>=0); ); f=factor(n>>valuation(n, 2)); for(i=1, #f[, 1], if(bitand(f[i, 2], 1)==1&&bitand(f[i, 1], 3)==3, return(0))); 1
isthree(n:int)=my(tmp=valuation(n, 2)); bitand(tmp, 1)||bitand(n>>tmp, 7)!=7
A002828(n)=if(issquare(n), !!n, if(istwo(n), 2, 4-isthree(n))) \\ From Charles R Greathouse IV, Jul 19 2011
A277890(n) = { my(orgk = ((n+1)^2)-1); my(k = orgk, s = 0); while(((k == orgk) || !issquare(1+k)), s = s + (1-(k%2)); k = k - A002828(k)); s; };
for(n=1, 10000, write("b277890.txt", n, " ", A277890(n)));

(define (A277890 n) (let ((org_k (- (A000290 (+ 1 n)) 1))) (let loop ((k org_k) (s 0)) (if (and (< k org_k) (= 1 (A010052 (+ 1 k)))) s (loop (- k (A002828 k)) (+ s (- 1 (A000035 k))))))))

a(n) + A277891(n) = A260734(n).
For n >= 2, a(n) >= A277486(n).
a(n) >= A277488(n).

cp's OEIS Frontend

A276572 Simple self-inverse permutation of natural numbers: after a(0)=0, list each block of A260734(n) numbers in reverse order, from A260732(n) to A260733(1+n).

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A276571 After a(0)=0, each n occurs A260734(n) times.

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A002828 Least number of squares that add up to n.

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A255131 n minus the least number of squares that add up to n: a(n) = n - A002828(n).

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A260731 a(n) = Number of steps to reach 0 starting from x=n and using the iterated process: x -> x - A002828(x), where A002828(x) = the least number of squares that add up to x.

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A260733 a(n) = number of steps needed to reach zero when starting from k = (n^2)-1 and repeatedly applying the map that replaces k with k - A002828(k), where A002828(k) = the least number of squares that add up to k.

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A260732 a(n) = number of steps needed to reach zero when starting from k = n^2 and repeatedly applying the map that replaces k with k - {the least number of squares (A002828) that add up to k}.

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A261224 a(n) = number of steps needed to reach (n^2)-1 when starting from k = ((n+1)^2)-1 and repeatedly applying the map that replaces k with k - A053610(k), where A053610(k) = the number of positive squares that sum to k using the greedy algorithm.