cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

References

  • 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).

Links

Crossrefs

Cf. A000290, A000415, A000419, A002376, A004215, A000378, A001481.
Cf. A010052, A025426, A025427, A053610, A062535, A072401, A229062, A255131, A260728, A260731, A260734, A262678, A262689, A262690, A276573.

Programs

  • Haskell
    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
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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
  • Python
    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
  • Python
    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
  • Scheme
    ;; 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

Formula

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)

Extensions

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

A053610 Number of positive squares needed to sum to n using the greedy algorithm.

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 4, 2, 1, 2, 3, 4, 2, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 3, 4, 5, 2, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 3, 4, 5, 2, 3, 4, 1, 2, 3, 4
Offset: 1

Views

Author

Jud McCranie, Mar 19 2000

Keywords

Comments

Define f(n) = n - x^2 where (x+1)^2 > n >= x^2. a(n) = number of iterations in f(...f(f(n))...) to reach 0.
a(n) = 1 iff n is a perfect square.
Also sum of digits when writing n in base where place values are squares, cf. A007961. - Reinhard Zumkeller, May 08 2011
The sequence could have started with a(0)=0. - Thomas Ordowski, Jul 12 2014
The sequence is not bounded, see A006892. - Thomas Ordowski, Jul 13 2014

Examples

			7=4+1+1+1, so 7 requires 4 squares using the greedy algorithm, so a(7)=4.

Links

Crossrefs

Cf. A006892 (positions of records), A055401, A007961.
Cf. A000196, A000290, A057945 (summing triangular numbers).

Programs

  • Haskell
    a053610 n = s n $ reverse $ takeWhile (<= n) $ tail a000290_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
  • Maple
    A053610 := proc(n)
    local a,x;
    a := 0 ;
    x := n ;
    while x > 0 do
        x := x-A048760(x) ;
        a := a+1 ;
    end do:
    a ;
end proc: # R. J. Mathar, May 13 2016
  • Mathematica
    f[n_] := (n - Floor[Sqrt[n]]^2); g[n_] := (m = n; c = 1; While[a = f[m]; a != 0, c++; m = a]; c); Table[ g[n], {n, 1, 105}]
  • PARI
    A053610(n,c=1)=while(n-=sqrtint(n)^2,c++);c \\ M. F. Hasler, Dec 04 2008
  • Python
    from math import isqrt
def A053610(n):
    c = 0
    while n:
        n -= isqrt(n)**2
        c += 1
    return c # Chai Wah Wu, Aug 01 2023

Formula

a(n) = A007953(A007961(n)). - Henry Bottomley, Jun 01 2000
a(n) = a(n - floor(sqrt(n))^2) + 1 = a(A053186(n)) + 1 [with a(0) = 0]. - Henry Bottomley, May 16 2000
A053610 = A002828 + A062535. - M. F. Hasler, Dec 04 2008

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

Views

Author

Antti Karttunen, Feb 24 2015

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • Maple
    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
  • Mathematica
    {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 *)

Formula

a(n) = n - A002828(n).
a(n) = A260740(n) + A062535(n).

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

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 12 2015

Keywords

Links

Crossrefs

Cf. A048760, A053610, A062535, A255131, A261221, A261222, A261223, A261224.
Cf. also A261225.

Formula

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).
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.