This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A337566 #40 Dec 13 2021 16:13:11 %S A337566 0,0,1,1,1,3,1,2,3,3,1,5,1,3,5,3,1,6,1,5,5,3,1,7,3,3,5,5,1,9,1,4,5,3, %T A337566 5,9,1,3,5,7,1,9,1,5,9,3,1,9,3,6,5,5,1,9,5,7,5,3,1,13,1,3,9,5,5,9,1,5, %U A337566 5,9,1,12,1,3,9,5,5,9,1,9,7,3,1,13,5,3,5,7,1,15,5,5 %N A337566 a(n) is the number of possible decompositions of the polynomial n * (x + x^2 + ... + x^q), where q > 1, into a sum of k polynomials, not necessarily all different; each of these polynomials is to be of the form b_1 * x + b_2 * x^2 + ... + b_q * x^q where each b_i is one of the numbers 1, 2, 3, ..., q and no two b_i are equal. %C A337566 Inspired by the 6th problem of the 13th British Mathematical Olympiad in 1977 (see the link BMO) where the problem asked to find for n = 26 all the values of q for which this decomposition is possible (see 2nd example). %C A337566 As mentioned by Tony Gardiner in his book (see reference), "the wording" of this problem "is very strange". Letter n in Olympiad exercise becomes q in the Name. %C A337566 If a solution is the sum of k polynomials of degree q, then, the relation between (n,k,q) is: k*(q+1) = 2*n with q > 1 (as in the problem) and q < n (because one proves there is no solution when q >= n); then, a(n) is the number of pairs (k,q) that are solutions of this last relation. %D A337566 A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 6 pp. 212-213 (1977). %H A337566 Antti Karttunen, <a href="/A337566/b337566.txt">Table of n, a(n) for n = 1..20000</a> %H A337566 British Mathematical Olympiad 1977, <a href="https://bmos.ukmt.org.uk/home/bmo-1977.pdf">Problem 6</a>. %H A337566 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>. %F A337566 a(1) = 0 then, for n >= 2, a(n) = tau(2n) - 3 = A099777(n) - 3. %F A337566 a(n) = 1 iff n = 4 or n = p odd prime (A065091). %F A337566 a(n) = p-3, p odd prime > 3 iff n = 2^(p-2). %e A337566 For n = 3, the only solution, that corresponds to q = 2 and k = 2, is: %e A337566 3 * (x + x^2) = (x + 2x^2) + (2x + x^2). %e A337566 For n = 26 as in the British Olympiad problem, a(26) = 3, and these three possible decompositions are: %e A337566 for k = 2, q = 25: %e A337566 26 * (x + x^2 + x^3 + ... + x^24 + x^25) = %e A337566 (x + 2x^2 + 3x^3 + ... + 24x^24 + 25x^25) + %e A337566 (25x + 24x^2 + 23x^3 + ... + 2x^24 + x^25); %e A337566 for k = 4, q = 12: %e A337566 26 * (x + x^2 + x^3 + ... + x^11 + x^12) = %e A337566 (x + 2x^2 + 3x^3 + ... + 11x^11 + 12x^12) + %e A337566 (12x + 11x^2 + 10x^3 + ... + 2x^11 + x^12) + %e A337566 (x + 2x^2 + 3x^3 + ... + 11x^11 + 12x^12) + %e A337566 (12x + 11x^2 + 10x^3 + ... + 2x^11 + x^12); %e A337566 for k = 13, q = 3: %e A337566 26 * (x + x^2 + x^3) = %e A337566 4 * (x + 2x^2 + 3x^3) + %e A337566 4 * (2x + 3x^2 + x^3) + %e A337566 3 * (3x + x^2 + 2x^3) + %e A337566 (2x + x^2 + 3x^3) + %e A337566 (3x + 2x^2 + x^3). %p A337566 with(numtheory): %p A337566 Data:= 0, seq(tau(2*n)-3, n=2..150); %t A337566 MapAt[# + 1 &, Array[DivisorSigma[0, 2 #] - 3 &, 92], 1] (* _Michael De Vlieger_, Dec 12 2021 *) %o A337566 (PARI) a(n) = if (n==1, 0, numdiv(2*n)-3); \\ _Michel Marcus_, Sep 06 2020 %Y A337566 Cf. A065091, A099777. %K A337566 nonn %O A337566 1,6 %A A337566 _Bernard Schott_, Sep 01 2020