A053610 -id:A053610 - cp's OEIS frontend

A145172 Number of pentagonal numbers needed to represent n with greedy algorithm.

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

Offset: 1

Christina Steffan (christina.steffan(AT)gmx.at), Oct 03 2008

nonn

Sequence is unbounded.

			a(21)=6 since 21 = 12+5+1+1+1+1.

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Cf. A000326 (pentagonal numbers), A053610, A057945, A180447, A192988.

a(n)={my(s=0); forstep(k=(sqrtint(24*n+1)+1)\6, 1, -1, my(t=k*(3*k-1)/2); s+=n\t; n%=t); s} \\ Andrew Howroyd, Apr 21 2021

Terms a(41) and beyond from Andrew Howroyd, Apr 21 2021

A174806 a(n) = n-floor(sqrt(n))^2-floor(sqrt(n-floor(sqrt(n))^2))^2; difference between n and sum of two largest distinct squares <= n.

Original entry on oeis.org

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

Offset: 0

Vladimir Joseph Stephan Orlovsky, Mar 29 2010

nonn

If a(n)=0 then n is a sum of two squares A001481, but not conversely. For the sum of two squares n = 18, 32, 41, ... we have a(n) > 0. - Thomas Ordowski, Jul 11 2014

			24=4^2+8;8-2^2=4, 115=10^2+15;15-3^2=6,..

Michel Marcus, Table of n, a(n) for n = 0..10000

Cf. A001481, A053186, A053610.

a[n_]:=n-Floor[Sqrt[n]]^2-Floor[Sqrt[n-Floor[Sqrt[n]]^2]]^2;
Table[a[n], {n,0,6!}]

a(n) = my(x=sqrtint(n)^2); n - x - sqrtint((n-x))^2; \\ Michel Marcus, Dec 17 2022

a(n) = 0 iff A053610(n) < 3 and 0 < a(n) = m^2 iff A053610(n) = 3. - Thomas Ordowski, Jul 12 2014

A245440 Primes p == 1 (mod 4) such that p - floor(sqrt(p))^2 and 2p - floor(sqrt(2p))^2 are not squares.

Original entry on oeis.org

353, 373, 449, 461, 521, 541, 593, 653, 673, 757, 769, 797, 821, 829, 941, 953, 1009, 1021, 1061, 1069, 1097, 1193, 1217, 1237, 1249, 1277, 1361, 1381, 1481, 1489, 1549, 1597, 1613, 1621, 1657, 1669, 1693, 1709, 1721, 1733, 1777, 1801, 1877, 1889, 1933, 1949

Offset: 1

Thomas Ordowski, Jul 22 2014

nonn

Primes p of the form 4k+1 such that A053610(p) > 2 and A053610(2p) > 2.
Note that p = a^2 + b^2 and 2p = (a+b)^2 + (a-b)^2 is the only way. So according to the definition the greedy algorithm cannot give such the sums of two squares.
Interesting fact: a(n) = A145023(n) for all n < 25. Of course A145023 is a subsequence.
Primes p == 1 (mod 4) such that A245474(p) > 2.

Michael De Vlieger, Table of n, a(n) for n = 1..10461 (first 46 terms from Thomas Ordowski and Colin Barker)

Cf. A002144, A053610, A145016, A145023, A174806, A245474.

[p: p in PrimesUpTo(10000) | p mod 4 eq 1 and not IsSquare(p-Floor(Sqrt(p))^2) and not IsSquare(2*p-Floor(Sqrt(2*p))^2)]; // Vincenzo Librandi, Sep 19 2017

a245440Q[n_Integer] := If[
  And[PrimeQ[n] == True, Mod[n, 4] == 1],
  If[Or[IntegerQ[Sqrt[n - Floor[Sqrt[n]]^2]] == True,
    IntegerQ[Sqrt[2*n - Floor[Sqrt[2*n]]^2]] == True], False, True],
  False]; a245440[n_Integer] :=
Flatten[Position[Thread[a245440Q[Range[n]]],
   True]]; a245440[300000]; (* Michael De Vlieger, Aug 05 2014 *)

s=[]; forprime(p=2, 3000, if(p%4==1 && !issquare(p-floor(sqrt(p))^2) && !issquare(2*p-floor(sqrt(2*p))^2), s=concat(s, p))); s \\ Colin Barker, Jul 22 2014

More terms from Colin Barker, Jul 22 2014

A276585 After a(0)=0, the first differences of A276583.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Sep 07 2016

nonn

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

Cf. A053610, A276583.

a(n) = A053610(A276583(n)).

a(0) = 0; for n >= 1, a(n) = A276583(n) - A276583(n-1).

A263651 Numbers n such that the difference between n and the largest square less than n is a nonzero square.

Original entry on oeis.org

2, 5, 8, 10, 13, 17, 20, 26, 29, 34, 37, 40, 45, 50, 53, 58, 65, 68, 73, 80, 82, 85, 90, 97, 101, 104, 109, 116, 122, 125, 130, 137, 145, 148, 153, 160, 170, 173, 178, 185, 194, 197, 200, 205, 212, 221, 226, 229, 234, 241, 250, 257, 260, 265, 272, 281, 290, 293, 298, 305

Offset: 1

Eli Jaffe, Oct 22 2015

Numbers n such that A053186(n) is a positive square. - Michel Marcus, Oct 23 2015
Numbers of the form a^2 + b^2 where a >= 1 and 1 <= b^2 <= 2a. - Robert Israel, Oct 23 2015
Numbers n such that A053610(n) = 2. - Thomas Ordowski, May 22 2016

			For n=5, the largest square less than 5 is 4, and the difference between 4 and 5 is 1, which is also square.

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

Cf. A053186.

N:= 1000: # to get all terms <= N
sort([seq(seq(a^2 + b^2, b=1..min(floor(sqrt(2*a)),floor(sqrt(N-a^2)))),a=1..floor(sqrt(N-1)))]); # Robert Israel, Oct 23 2015

Select[Range@ 305, And[IntegerQ@ Sqrt[# - Floor[Sqrt@ #]^2], ! IntegerQ@ Sqrt@ #] &] (* Michael De Vlieger, Oct 23 2015 *)

isok(n) = (d = (n - sqrtint(n)^2)) && issquare(d); \\ Michel Marcus, Oct 23 2015

More terms from Michel Marcus, Oct 23 2015

A350178 Take n and subtract the greatest square less than or equal to n. Repeat this process until 0 is reached. a(n) is the sum of all residues after subtractions.

Original entry on oeis.org

0, 0, 1, 3, 0, 1, 3, 6, 4, 0, 1, 3, 6, 4, 6, 9, 0, 1, 3, 6, 4, 6, 9, 13, 12, 0, 1, 3, 6, 4, 6, 9, 13, 12, 9, 11, 0, 1, 3, 6, 4, 6, 9, 13, 12, 9, 11, 14, 18, 0, 1, 3, 6, 4, 6, 9, 13, 12, 9, 11, 14, 18, 17, 20, 0, 1, 3, 6, 4, 6, 9, 13, 12, 9, 11, 14, 18, 17, 20, 24, 16, 0, 1, 3, 6, 4, 6, 9, 13

Offset: 0

Thomas Scheuerle, Dec 18 2021

Let s_1,s_2,s_3,...,s_m be the greedy partition of n into squares (n = s_1+s_2+s_3+...+s_m) such that s_1 >= s_2 >= s_3 >= ... >= s_m then a(n) = 0*s_1 + 1*s_2 + 2*s_3 + ... + (m-1)*s_m.

This sequence contains only numbers which can be written in the form c_1^2 + 2*c_2^2 + ... + m*c_m^2 with c_1 >= c_2 >= c_m. This excludes 2,5,7,8,... .

			a(41): 41 - 6^2 = 5; 5 - 2^2 = 1; 1 - 1^2 = 0 -> 5+1 = 6 = a(41).

Thomas Scheuerle, Table of n, a(n) for n = 0..5000

Cf. A053186, A053610.

Table[Total[Rest[NestWhileList[#-Floor[Sqrt[#]]^2&,n,#>0&]]],{n,0,90}] (* Harvey P. Dale, Apr 27 2025 *)

A350178(n)={my(r=0); while(n-=sqrtint(n)^2, r+=n); r};

a(n) = n - r^2 + a(n - r^2) = a(n - r^2 + (b + r)^2) = a(n + b^2 + 2*b*r), r = floor(sqrt(n)), for any b >= 0. True because a(n) depends only on the distance to the next square <= n.

a(n) = Sum_{k>0} A053186^k(n).

A350698 Consider the positive squares summing to n as obtained by the greedy algorithm; a(n) is the least of these squares.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 1, 4, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 25, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 36, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 1, 49, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 1, 4, 1, 64, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 1, 4, 1, 1, 16, 81, 1, 1, 1

Offset: 1

Rémy Sigrist, Jan 12 2022

nonn

			For n = 13:
- 13 = 3^2 + 2^2,
- so a(13) = 2^2.

Cf. A053610, A350674.

a(n, e=2) = { my (r=0); while (n, r=sqrtnint(n, e); n-=r^e); r^e }

a(n^2) = n^2.

a(n) = A350674(n, A053610(n)).

cp's OEIS Frontend

A145172 Number of pentagonal numbers needed to represent n with greedy algorithm.

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A174806 a(n) = n-floor(sqrt(n))^2-floor(sqrt(n-floor(sqrt(n))^2))^2; difference between n and sum of two largest distinct squares <= n.

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A245440 Primes p == 1 (mod 4) such that p - floor(sqrt(p))^2 and 2p - floor(sqrt(2p))^2 are not squares.

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Magma

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A276585 After a(0)=0, the first differences of A276583.

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A263651 Numbers n such that the difference between n and the largest square less than n is a nonzero square.

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Maple

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A350178 Take n and subtract the greatest square less than or equal to n. Repeat this process until 0 is reached. a(n) is the sum of all residues after subtractions.

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A350698 Consider the positive squares summing to n as obtained by the greedy algorithm; a(n) is the least of these squares.

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