A130052 -id:A130052 - cp's OEIS frontend

A298467 Smallest positive integer that can be expressed as the sum of consecutive positive squares in exactly n ways.

1, 25, 20449, 554503705

Offset: 1

Seiichi Manyama, Jan 19 2018

a(5) > 10^9. - Rémy Sigrist, Jan 19 2018

a(5) > 10^15. - Michael S. Branicky, Feb 18 2023

			a(2) = 25 because 3^2 + 4^2 = 5^2 = 25,
a(3) = 20449 because 7^2 + 8^2 + ... + 39^2 = 38^2 + 39^2 + ... + 48^2 = 143^2 = 20449.
a(4) = 554503705 because 480^2 + 481^2 + ... + 1210^2 = 3570^2 + 3571^2 + ... + 3612^2 = 3613^2 + 3614^2 + ... + 3654^2 = 7442^2 + 7443^2 ... + 7451^2 = 554503705. - _Rémy Sigrist_, Jan 19 2018

Cf. A038547, A111044, A130052, A234304, A234311.

a(n) <= A234311(n).

a(4) from Rémy Sigrist, Jan 19 2018

A296338 a(n) = number of partitions of n into consecutive positive squares.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1

Offset: 1

Seiichi Manyama, Jan 14 2018

nonn

			   1 = 1^2,                   so  a(1) = 1.
   4 = 2^2,                   so  a(4) = 1.
   5 = 1^2 + 2^2,             so  a(5) = 1.
   9 = 3^2,                   so  a(9) = 1.
  13 = 2^2 + 3^2,             so a(13) = 1.
  14 = 1^2 + 2^2 + 3^2,       so a(14) = 1.
  16 = 4^2,                   so a(16) = 1.
  25 = 3^2 + 4^2 = 5^2,       so a(25) = 2.
  29 = 2^2 + 3^2 + 4^2,       so a(29) = 1.
  30 = 1^2 + 2^2 + 3^2 + 4^2, so a(30) = 1.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A000290, A001227, A034705, A130052, A234304, A297199, A298467, A299173.

nMax = 100; t = {0}; Do[k = n; s = 0; While[s = s + k^2; s <= nMax, AppendTo[t, s]; k++], {n, 1, nMax}]; tt = Tally[t]; a[] = 0; Do[a[tt[[i, 1]]] = tt[[i, 2]], {i, 1, Length[tt]}]; Table[a[n], {n, 1, nMax}] (* _Jean-François Alcover, Feb 04 2018, using T. D. Noe's program for A034705 *)

def A296338(n)
  m = Math.sqrt(n).to_i
  ary = Array.new(n + 1, 0)
  (1..m).each{|i|
    sum = i * i
    ary[sum] += 1
    i += 1
    sum += i * i
    while sum <= n
      ary[sum] += 1
      i += 1
      sum += i * i
    end
  }
  ary[1..-1]
end
p A296338(100)

a(A034705(n)) >= 1 for n > 1.

G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k^2). - Ilya Gutkovskiy, Apr 18 2019

A234304 Numbers with at least 3 representations as a sum of consecutive square numbers.

Original entry on oeis.org

20449, 147441, 910805, 1026745, 2403800, 2513434, 3198550, 8116801, 11739805, 15053585, 18646301, 33313175, 93812510, 102939515, 134910295, 136448235, 151443110, 163998695, 195435485, 197780465, 213872920, 267043455, 461498779, 482204660, 554503705, 559990541

Offset: 1

David W. Wilson, Dec 22 2013

nonn

			20449 = 7^2 + .. + 39^2 = 38^2 + .. + 48^2 = 143^2.

David W. Wilson, Table of n, a(n) for n = 1..761

Cf. A130052 (numbers with at least 2 representations).

A309783 Numbers that are sums of one or more consecutive positive triangular numbers in more than one way.

Original entry on oeis.org

10, 36, 55, 64, 100, 120, 136, 164, 210, 276, 361, 435, 460, 514, 560, 596, 676, 760, 1176, 1225, 1320, 1326, 1460, 1484, 1485, 1505, 1540, 1684, 1736, 1770, 1891, 1936, 2014, 2080, 2145, 2180, 2314, 2485, 2596, 2890, 3156, 3244, 3275, 3364, 3486, 3570, 3710, 3916

Offset: 1

Ilya Gutkovskiy, Aug 17 2019

nonn

The first number that is the sum in three ways is 2180. The first that is the sum in four ways is 10053736. - Robert Israel, Aug 20 2019

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (first 2759 terms from Robert Israel)

Cf. A000217, A034706, A130052, A307666.

N:= 10000: # for terms <= N
V:= Vector(N):
for i from 1 while i*(i+1)/2 <= N do
  s:= i*(i+1)*(i+2)/6;
  for j from i-1 to 0 by -1 do
    t:= j*(j+1)*(j+2)/6;
    if s-t > N then break fi;
    V[s-t]:= V[s-t]+1
  od;
od:
select(t -> V[t]>1, [$1..N]); # Robert Israel, Aug 20 2019

A307666(a(n)) > 1.

A299173 a(n) is the maximum number of squared consecutive positive integers into which the integer n can be partitioned.

Original entry on oeis.org

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

Offset: 1

Jean-François Alcover, Feb 04 2018

a(k^2)>=1, the inequality being strict if k is in A097812.

			25 = 5^2 = 3^2 + 4^2 and no such partition is longer, so a(25) = 2.
30 = 1^2 + 2^2 + 3^2 + 4^2 and no such partition is longer, so a(30) = 4.
2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2 and no such partition is longer, so a(2018) = 12. (This special example is due to _Seiichi Manyama_.) - _Jean-François Alcover_, Feb 05 2018

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

Cf. A034705, A097812, A130052, A111044, A234304, A234311, A296338, A298467.

N:= 200: # to get a(1)..a(N)
A:= Vector(N):
S:= n -> n*(n+1)*(2*n+1)/6:
M:= floor(sqrt(N)):
for d from 1 to M do
  for b from d to M do
    s:= S(b) - S(b-d);
    if s > N then break fi;
    A[s]:= d
od od:
convert(A,list); # Robert Israel, Feb 04 2018

terms = 100; jmax = Ceiling[Sqrt[terms]]; kmax = Ceiling[(3*terms)^(1/3)]; Clear[a]; a[_] = 0; Do[r = Range[j, j + k - 1]; n = r . r; If[k > a[n], a[n] = k], {j, jmax}, {k, kmax}]; Array[a, terms]

A322135 Table of truncated square pyramid numbers, read by antidiagonals.

Original entry on oeis.org

1, 4, 5, 9, 13, 14, 16, 25, 29, 30, 25, 41, 50, 54, 55, 36, 61, 77, 86, 90, 91, 49, 85, 110, 126, 135, 139, 140, 64, 113, 149, 174, 190, 199, 203, 204, 81, 145, 194, 230, 255, 271, 280, 284, 285, 100, 181, 245, 294, 330, 355, 371, 380, 384, 385, 121, 221, 302

Offset: 1

Allan C. Wechsler, Nov 27 2018

The n-th row contains n numbers: n^2, n^2 + (n-1)^2, ..., n^2 + (n-1)^2 + ... + 1^2.
All numbers that appear in the table are listed in ascending order at A034705.
All numbers that appear twice or more are listed at A130052.
The left column is A000290 (the squares).
The top row is A000330 (the square pyramidal numbers).
The columns are A000290, A099776 (or a tail of A001844), a tail of A005918 or A120328, a tail of A027575, a tail of A027578, a tail of A027865, ...
The first two rows are A000330 and a tail of A168599, but subsequent rows are not currently in the OEIS, and are all tails of A000330 minus various constants.
The main diagonal is A050410.

			The 17th term is entry 2 on antidiagonal 6, so we sum two terms: 6^2 + 5^2 = 61.
Table begins:
   1   5  14  30  55  91 140 204 ...
   4  13  29  54  90 139 203 ...
   9  25  50  86 135 199 ...
  16  41  77 126 190 ...
  25  61 110 174 ...
  36  85 149 ...
  49 113 ...
  64 ...
  ...

See comments; also cf. A000330, A059255.

T[n_,k_] = Sum[(n+i)^2, {i,0,k-1}]; Table[T[n-k+1, k], {n,1,10},  {k,1,n}] // Flatten (* Amiram Eldar, Nov 28 2018 *)
f[n_] := Table[SeriesCoefficient[-((y (y (1 + y) + x (1 - 2 y - 3 y^2) + x^2 (1 - 3 y + 4 y^2)))/((-1 + x)^3 (-1 + y)^4)) , {x, 0,
i + 1 - j}, {y, 0, j}], {i, n, n}, {j, 1, n}]; Flatten[Array[f, 10]] (* Stefano Spezia, Nov 28 2018 *)

T(n,k) = n^2 + (n+1)^2 + ... + (n+k-1)^2 = A000330(n + k - 1) - A000330(n - 1) = T(n, k) = k*n^2 + (k^2 - k)*n + (1/3*k^3 - 1/2*k^2 + 1/6*k)

G.f.: -y*(y*(1 + y) + x*(1 - 2*y - 3*y^2) + x^2*(1 - 3*y + 4*y^2))/((- 1 + x)^3*(- 1 + y)^4). - Stefano Spezia, Nov 28 2018

A346163 Numbers k such that there exist equal sums of k and 2k consecutive positive squares.

Original entry on oeis.org

1, 17, 23, 25, 49, 55, 71, 73, 79, 89, 95, 103, 113, 127, 143, 161, 167, 175, 185, 191, 193, 199, 215, 217, 233, 239, 241, 265, 271, 287, 289, 305, 361, 377, 391, 409, 415, 431, 433, 457, 473, 481, 505, 511, 521, 535, 545, 553, 569, 593, 599, 617, 631, 647

Offset: 1

Johan Westin, Jul 08 2021

nonn

a(n) is congruent to 1 or 5 (mod 6).
a(n) is not congruent to 3, 4 or 5 (mod 8) or to 7, 11, 16 or 20 (mod 27), see Alder and Alfred.
k is in the sequence if the quadratic Diophantine equation 6*(k*x^2 - 2*k*y^2 + k*(k-1)*x + 2*k*(1-2*k)*y) - 14*k^3 + 9*k^2 - k = 0 has solutions x, y in the positive integers.

			a(1): 5^2 = 3^2 + 4^2. Here the left-hand side has k = 1 term, and the right-hand side has 2k = 2 terms. Hence k = 1 is in the sequence.
a(2): 29^2 + 30^2 + ... + 44^2 + 45^2 = 8^2 + 9^2 + ... + 40^2 + 41^2 = 23681. Here the left and right sums have k = 17 and 2k = 34 terms, respectively. Hence k = 17 is in the sequence.

H. L. Alder and U. Alfred, n and n+k Consecutive Integers with Equal Sums of Squares, The American Mathematical Monthly, 71:7 (1964), 749-754.
Index entries for sequences related to sums of squares

Cf. A000330, A001032, A130052.

import sympy # Version 1.8
xx, yy = sympy.symbols("x y")
def pyramidal(n):
    return n*(n+1)*(2*n+1)/6 # A000330(n)
def expanded_diophantine(k,n):
    left_hand_side =  pyramidal(xx+n-1) - pyramidal(xx-1)
    right_hand_side =  pyramidal(yy+n+k-1) - pyramidal(yy-1)
    return sympy.expand(right_hand_side-left_hand_side)
def has_solutions(k,n):
    return len(sympy.solvers.diophantine(expanded_diophantine(k,n))) != 0
def k_in_a346163(k):
    return has_solutions(k,k)

cp's OEIS Frontend

A298467 Smallest positive integer that can be expressed as the sum of consecutive positive squares in exactly n ways.

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A296338 a(n) = number of partitions of n into consecutive positive squares.

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A234304 Numbers with at least 3 representations as a sum of consecutive square numbers.

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A309783 Numbers that are sums of one or more consecutive positive triangular numbers in more than one way.

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Maple

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A299173 a(n) is the maximum number of squared consecutive positive integers into which the integer n can be partitioned.

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Maple

Mathematica

A322135 Table of truncated square pyramid numbers, read by antidiagonals.

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Mathematica

Formula

A346163 Numbers k such that there exist equal sums of k and 2k consecutive positive squares.

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Python