OFFSET
1,1
COMMENTS
Shiraishi's solutions to a^3 + b^3 + c^3 = d^3 are a = 3n^2; b = 6n^2 - 3n + 1 or 6n^2 + 3n + 1; c = 9n^3 - 6n^2 + 3n - 1 or 9n^3 + 6n^2 + 3n, respectively, for n > 0; and d = c+1. See Smith and Mikami for a derivation.
Shiraishi's formulas show that the sequence is infinite. Hence the sequences A023042 (solutions to x^3 + y^3 + z^3 = w^3), A225908 (solutions to a^3 + b^3 = c^3 - d^3), A225909 (solutions to a^3 + b^3 = (c+1)^3 - c^3) and A226902 (numbers c in A225909) are also infinite.
Shiraishi's solution b = 6n^2 +/- 3n + 1 is the centered triangular numbers A005448 except 1.
REFERENCES
Shiraishi Chochu (aka Shiraishi Nagatada), Shamei Sampu (Sacred Mathematics), 1826.
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..10000
David Eugene Smith and Yoshio Mikami, A History of Japanese Mathematics, Open Court, Chicago, 1914; Dover reprint, 2004; pp. 233-235.
Wikipedia (French), Shiraishi Nagatada
Wikipedia (German), Shiraishi Nagatada
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).
FORMULA
a(2*n-1) = 9*n^3 - 6*n^2 + 3*n - 1.
a(2*n) = 9*n^3 + 6*n^2 + 3*n.
G.f.: x*(5 + 13*x + 20*x^2 + 10*x^3 + 5*x^4 + x^5) / ((1 + x)^3*(1 - x)^4). - Bruno Berselli, Jun 22 2013
a(n) = (18*n^3 + 27*n^2 + 27*n + 1 - (3*n^2 + 3*n + 1)*(-1)^n)/16. - Bruno Berselli, Jun 22 2013
a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n > 7. - Chai Wah Wu, Aug 05 2025
EXAMPLE
The first two terms are a(1) = 9 - 6 + 3 - 1 = 5 and a(2) = 9 + 6 + 3 = 18. Then Shiraishi's formulas give 3^3 + 4^3 + 5^3 = 6^3 and 3^3 + 10^3 + 18^3 = 19^3.
MATHEMATICA
LinearRecurrence[{1, 3, -3, -3, 3, 1, -1}, {5, 18, 53, 102, 197, 306, 491}, 50] (* Paolo Xausa, May 09 2026 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jonathan Sondow, Jun 22 2013
STATUS
approved
