A053610 -id:A053610 - cp's OEIS frontend

A260740 a(n) = n minus the number of positive squares needed to sum to n using the greedy algorithm: a(n) = n - A053610(n).

0, 0, 0, 0, 3, 3, 3, 3, 6, 8, 8, 8, 8, 11, 11, 11, 15, 15, 15, 15, 18, 18, 18, 18, 21, 24, 24, 24, 24, 27, 27, 27, 27, 30, 32, 32, 35, 35, 35, 35, 38, 38, 38, 38, 41, 43, 43, 43, 43, 48, 48, 48, 48, 51, 51, 51, 51, 54, 56, 56, 56, 56, 59, 59, 63, 63, 63, 63, 66, 66, 66, 66, 69, 71, 71, 71, 71, 74, 74, 74, 78, 80

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

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

Cf. A048760, A053610, A062535, A255131, A261221, A261222, A261223, A261224.

Cf. also A261225.

a(n) = n - A053610(n).
As a recurrence:
a(0) = 0; for n >= 1, a(n) = -1 + A048760(n) + a(n-A048760(n)). [Where A048760(n) gives the largest square <= n.]
Other identities. For all n >= 1:
a(n) = A255131(n) - A062535(n).

A261222 a(n) = number of steps to reach 0 when starting from k = n*n 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

0, 1, 2, 4, 6, 9, 12, 15, 19, 24, 29, 35, 41, 48, 55, 62, 70, 78, 87, 97, 107, 118, 129, 141, 153, 165, 178, 191, 205, 219, 234, 249, 265, 282, 299, 317, 335, 354, 373, 392, 412, 433, 454, 475, 497, 519, 542, 565, 589, 613, 638, 664, 690, 717, 744, 772, 800, 828, 857, 887, 917, 948, 979, 1010, 1042, 1074, 1107, 1140, 1174, 1208, 1243, 1278, 1314, 1351, 1388, 1426, 1464, 1503

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..6000

Essentially one more than A261223.
First differences: A261224.
Cf. A000290, A053610, A260740, A261221.
Cf. also A260732, A261227.

Table[-1 + Length@ NestWhileList[# - Block[{m = #, c = 1}, While[a = (# - Floor[Sqrt@ #]^2) &@ m; a != 0, c++; m = a]; c] &, n^2, # != 0 &], {n, 0, 77}] (* Michael De Vlieger, Sep 08 2016, after Jud McCranie at A053610 *)

a(n) = A261221(n^2).
Other identities. For all n >= 1:
a(n) = 1 + A261223(n).

A261223 a(n) = number of steps to reach 0 when starting from k = (n*n)-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

0, 1, 3, 5, 8, 11, 14, 18, 23, 28, 34, 40, 47, 54, 61, 69, 77, 86, 96, 106, 117, 128, 140, 152, 164, 177, 190, 204, 218, 233, 248, 264, 281, 298, 316, 334, 353, 372, 391, 411, 432, 453, 474, 496, 518, 541, 564, 588, 612, 637, 663, 689, 716, 743, 771, 799, 827, 856, 886, 916, 947, 978, 1009, 1041, 1073, 1106, 1139, 1173, 1207, 1242, 1277, 1313, 1350, 1387, 1425, 1463, 1502, 1541

Offset: 1

Antti Karttunen, Aug 12 2015

nonn

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

One less than A261222.
Cf. A053610, A260740, A261221, A261224.
Cf. also A260733, A261228.

Table[-2 + Length@ NestWhileList[# - Block[{m = #, c = 1}, While[a = (# - Floor[Sqrt@ #]^2) &@ m; a != 0, c++; m = a]; c] &, (n + 1)^2, # != 0 &], {n, 0, 77}] (* Michael De Vlieger, Sep 08 2016, after Jud McCranie at A053610 *)

a(n) = A261221((n^2)-1).

a(n) = A261222(n)-1.

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).

A276583 The infinite trunk of greedy squares beanstalk: The only infinite sequence such that a(n-1) = a(n) - number of squares that sum to a(n) with greedy algorithm (A053610).

Original entry on oeis.org

0, 3, 6, 8, 11, 15, 18, 21, 24, 27, 32, 35, 38, 43, 48, 51, 56, 59, 63, 66, 71, 74, 78, 80, 83, 88, 91, 95, 99, 102, 107, 110, 114, 117, 120, 123, 128, 131, 135, 138, 143, 146, 151, 154, 158, 161, 164, 168, 171, 176, 179, 183, 186, 192, 195, 198, 203, 206, 210, 213, 219, 224, 227, 232, 235, 239, 242, 248, 251, 255

Offset: 0

Antti Karttunen, Sep 07 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..10095

Cf. A053610, A260740, A276582, A276584, A276585 (first differences).

Cf. also A179016, A259934, A276573, A276613, A276623 for similar constructions.

(define (A276583 n) (A276584 (A276582 n)))

a(n) = A276584(A276582(n)).

A261221 a(n) = number of steps to reach 0 when starting from k = n and repeatedly applying the map that replaces k with k - A053610(k), where A053610(k) = the number of positive squares needed to sum to k using the greedy algorithm.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

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

Cf. A260740, A261222, A261223, A261224.

Cf. also A260731, A261226.

a(0) = 0; for n >= 1, a(n) = 1 + a(A260740(n)).

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)

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 5, 4, 5, 4, 5, 6, 5, 6, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2

Offset: 0

Antti Karttunen, Dec 17 2015

Sum of digits in "Jacobsthal greedy base", A265747.
It would be nice to know for sure whether this sequence gives also the least number of Jacobsthal numbers that add to n, i.e., that there cannot be even better nongreedy solutions.
The integer 63=21+21+21 has 3 for its 'non-greedy' solution, and a(63) = 5 for its greedy solution 63=43+11+5+3+1. - Yuriko Suwa, Jul 11 2021
Positions where a(n) is different from A372555(n) are n=63, 84, 148, 169, 191, 212, 234, 255, etc. See A372557. - Antti Karttunen, May 07 2024

			a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
For n=1 we need just A001045(2) = 1, thus a(1) = 1.
For n=2 we need A001045(2) + A001045(2) = 1 + 1, thus a(2) = 2.
For n=4 we need A001045(3) + A001045(2) = 3 + 1, thus a(4) = 2.
For n=6 we form the greedy sum as A001045(4) + A001045(2) = 5 + 1, thus a(6) = 2. Alternatively, we could form the sum as A001045(3) + A001045(3) = 3 + 3, but the number of summands in that case is no less.
For n=7 we need A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = 3.
For n=8 we need A001045(4) + A001045(3) = 5 + 3, thus a(8) = 2.
For n=10 we need A001045(4) + A001045(4) = 5 + 5, thus a(10) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..10923

Cf. A001045, A130249, A197911, A003158, A265746, A265747, A364377, A364379, A364380, A372288, A372555, A372557.
Cf. A054111 (apparently the positions of the first occurrence of each n > 0).
Cf. also A007895, A014420, A053610, A265404, A265743, A265744.

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265745[n_] := A265745[n] = 1 + A265745[n - jacob[maxInd[n]]]; A265745[0] = 0; Array[A265745, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

A130249(n) = floor(log(3*n + 1)/log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = {if(n == 0, 0, my(d = n - A001045(A130249(n))); if(d == 0, 1, 1 + A265745(d)));} \\ Amiram Eldar, Jul 21 2023

def greedyJ(n): n1 = (3*n+1).bit_length() - 1; return (2**n1 - (-1)**n1)//3
def a(n): return 0 if n == 0 else 1 + a(n - greedyJ(n))
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 11 2021

a(0) = 0; for n >= 1, a(n) = 1 + a(n - A001045(A130249(n))). [This formula uses a simple greedy algorithm.]

A055401 Number of positive cubes needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7

Offset: 0

Henry Bottomley, May 16 2000

Define f(n) = n - k^3 where (k+1)^3 > n >= k^3; a(n) = number of steps such that f(f(...f(n)))= 0.

Also sum of digits when writing n in base where place values are positive cubes, cf. A000433. [Reinhard Zumkeller, May 08 2011]

			a(32)=6 because 32=27+1+1+1+1+1 (not 32=8+8+8+8).
a(33)=7 because 33=27+1+1+1+1+1+1 (not 33=8+8+8+8+1).

Antti Karttunen & Reinhard Zumkeller (terms 1-10000), Table of n, a(n) for n = 0..10000

Cf. A018888, A055400.
Cf. A002376 (least number of positive cubes needed to represent n; differs from this sequence for the first time at n=32, where a(32)=6, while A002376(32)=4).
Cf. A053610, A048766, A000578, A000433.
Cf. also A261225, A261226, A261227, A261228, A261229.

a055401 n = s n $ reverse $ takeWhile (<= n) $ tail a000578_list where
  s _ []                 = 0
  s m (x:xs) | x > m     = s m xs
             | otherwise = m' + s r xs where (m',r) = divMod m x
-- Reinhard Zumkeller, May 08 2011

f:= proc(n,k) local m, j;
if n = 0 then return 0 fi;
for j from k by -1 while j^3 > n do od:
m:= floor(n/j^3);
m + procname(n-m*j^3, j-1);
end proc:
seq(f(n,floor(n^(1/3))),n=0..100); # Robert Israel, Aug 17 2015

a[0] = 0; a[n_] := {n} //. {b___, c_ /; !IntegerQ[c^(1/3)], d___} :> {b, f = Floor[c^(1/3)]^3, c - f, d} // Length; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 17 2015 *)

F=vector(30,n,n^3); /* modify to get other sequences of "greedy representations" */
last_leq(v,F)={local(j=1); while ( F[j]<=v, j+=1 ); F[j-1]} /* Return last element <=v in sorted array F[] */
greedy(n,F)={local(v=n,ct=0); while ( v, v-=last_leq(v,F); ct+=1; ); ct}
vector(min(100,F[#F-1]),n,greedy(n,F)) /* show terms */
/* Joerg Arndt, Apr 08 2011 */

;; With memoization-macro definec.
(definec (A055401 n) (if (zero? n) n (+ 1 (A055401 (A055400 n)))))
;; Antti Karttunen, Aug 16 2015

a(0) = 0; for n >= 1, a(n) = a(n-floor(n^(1/3))^3)+1 = a(A055400(n))+1 = a(n-A048762(n))+1.

a(0) = 0 prepended by Antti Karttunen, Aug 16 2015

A007961 n written in base where place values are positive squares.

Original entry on oeis.org

1, 2, 3, 10, 11, 12, 13, 20, 100, 101, 102, 103, 110, 111, 112, 1000, 1001, 1002, 1003, 1010, 1011, 1012, 1013, 1020, 10000, 10001, 10002, 10003, 10010, 10011, 10012, 10013, 10020, 10100, 10101, 100000, 100001, 100002, 100003, 100010, 100011

Offset: 1

R. Muller

For n>1: A000196(n) = number of digits of a(n); A190321(n) = number of nonzero digits of a(n); A053610(n) = sum of digits of a(n). [Reinhard Zumkeller, May 08 2011]

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
F. Smarandache, Only Problems, Not Solutions!.

Cf. A000290, A000433.

import Data.Char (intToDigit)
a007961 :: Integer -> Integer
a007961 n = read $ map intToDigit $
  t n $ reverse $ takeWhile (<= n) $ tail a000290_list where
    t _ []          = []
    t m (x:xs)
        | x > m     = 0 : t m xs
        | otherwise = (fromInteger m') : t r xs
        where (m',r) = divMod m x
-- Reinhard Zumkeller, May 08 2011

A007961 := proc(n)
    local k,nrem,L,b,d;
    k := floor(sqrt(n)) ;
    nrem := n ;
    L := [] ;
    for b from k to 1 by -1 do
        d := floor(nrem/b^2) ;
        L := [d,op(L)] ;
        nrem := nrem -d*b^2 ;
    end do:
    add( op(i,L)*10^(i-1),i=1..nops(L)) ;
end proc: # R. J. Mathar, Jul 25 2015

cp's OEIS Frontend

A260740 a(n) = n minus the number of positive squares needed to sum to n using the greedy algorithm: a(n) = n - A053610(n).

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A261222 a(n) = number of steps to reach 0 when starting from k = n*n 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.

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A261223 a(n) = number of steps to reach 0 when starting from k = (n*n)-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.

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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.

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Mathematica

Formula

A276583 The infinite trunk of greedy squares beanstalk: The only infinite sequence such that a(n-1) = a(n) - number of squares that sum to a(n) with greedy algorithm (A053610).

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Formula

A261221 a(n) = number of steps to reach 0 when starting from k = n and repeatedly applying the map that replaces k with k - A053610(k), where A053610(k) = the number of positive squares needed to sum to k using the greedy algorithm.

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Formula

A002828 Least number of squares that add up to n.

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Haskell

Maple

Mathematica

PARI

Python

Python

Scheme

Formula

Extensions

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

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Mathematica

PARI