A364143 a(n) is the minimal number of consecutive squares needed to sum to A216446(n).

2, 5, 3, 2, 2, 3, 10, 2, 7, 9, 12, 11, 6, 11, 14, 3, 11, 29, 14, 7, 23, 4, 49, 8, 24, 5, 17, 12, 38, 46, 27, 34, 6, 14, 22, 66, 11, 66, 14, 11, 6, 77, 36, 63, 96, 11, 50, 3, 19, 96, 52, 41, 66, 33, 11, 3, 14, 121, 66, 89, 34, 127, 51, 2, 86, 54, 181, 48, 8

Offset

1,1

Links

Table of n, a(n) for n=1..69.

Project Euler, Problem 125: Palindromic Sums.

Example

a(8) = 7 is because 7 consecutive squares are needed to sum to A216446(8) = 595 = 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.

Prog

(Python)
is_palindrome = lambda n: str(n) == str(n)[::-1]
def g(L):
  L2, squares, D = L*L, [x*x for x in range(0, L + 1)], {}
  for i in range(1, L + 1):
    for j in range(i + 1, L + 1):
      candidate = sum(squares[i:j+1])
      if candidate < L2 and is_palindrome(candidate):
        if candidate in D:
          D[candidate]= min(D[candidate], j-i-1)
        else:
          D[candidate] = j-i+1
  return [D[k] for k in sorted(D.keys())]
print(g(1000))

Crossrefs

Cf. A216446, A034705, A180436, A267600.

Sequence in context: A019709 A076840 A328823 * A346265 A188923 A078375

Adjacent sequences: A364140 A364141 A364142 * A364144 A364145 A364146

Keyword

nonn,base

Author

Darío Clavijo, Jul 10 2023

Status

approved