A265449 Palindromes that are the sums of consecutive fourth powers.

0, 1, 353, 979, 14641, 16561, 998899, 2138312, 104060401, 1004006004001, 10004000600040001, 85045192129154058, 100004000060000400001, 1000004000006000004000001, 10000004000000600000040000001

Offset

1,3

Comments

Subsequence of A217844 and supersequence of A186080.

Links

Table of n, a(n) for n=1..15.

Example

353 = 2^4 + 3^4 + 4^4
979 = 1^4 + 2^4 + 3^4 + 4^4 + 5^4
16561 = 9^4 + 10^4
998899 = 19^4 +...+ 23^4
2138312 = 10^4 +...+ 25^4
85045192129154058 = 5582^4 +...+ 5666^4

Prog

(Python)
import heapq
def ispal(n): s = str(n); return s == s[::-1]
def afind():
  print("0, ") # special case
  N, T = 4, 1  # power, min number of terms
  sigma = sum(i**N for i in range(1, T+1))
  h = [(sigma, 1, T)]
  nextcount = T + 1
  while True:
    (v, s, l) = heapq.heappop(h)
    if ispal(v): print(f"{v}, [= Sum_{{i = {s}..{l}}} i^{N}]")
    if v >= sigma:
      sigma += nextcount**N
      heapq.heappush(h, (sigma, 1, nextcount))
      nextcount += 1
    v -= s**N; s += 1; l += 1; v += l**N
    heapq.heappush(h, (v, s, l))
afind() # Michael S. Branicky, May 16 2021 after Bert Dobbelaere in A344338

Crossrefs

Cf. A186080, A217844.

Sequence in context: A054825 A304385 A142565 * A142785 A259958 A126657

Adjacent sequences: A265446 A265447 A265448 * A265450 A265451 A265452

Keyword

nonn,base,more

Author

Chai Wah Wu, Jan 29 2016

Extensions

a(13)-a(15) from Giovanni Resta, Aug 27 2019

Status

approved