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A271608 Number of ordered ways to write n as pen(u) + pen(v) + pen(x) + pen(y) + pen(z) with u,v,x,y,z nonnegative integers such that u + 2*v + 4*x + 5*y + 6*z is a pentagonal number, where pen(k) denotes the pentagonal number k*(3k-1)/2. 40
1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 3, 3, 3, 3, 3, 3, 4, 2, 6, 4, 2, 1, 1, 8, 4, 5, 2, 2, 7, 10, 9, 2, 3, 4, 5, 6, 6, 5, 2, 7, 11, 11, 4, 1, 5, 8, 13, 8, 6, 5, 3, 8, 8, 12, 7, 3, 8, 18, 16, 12, 2, 7, 10, 15, 11, 10, 4, 4, 11, 15, 22 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 2, 4, 5, 7, 9, 21, 22, 43. Also, every n = 0,1,2,... can be written as pen(u) + pen(v) + pen(x) + pen(y) + pen(z) with u,v,x,y,z nonnegative integers such that 3*u + 5*v + 11*x + 16*y + 19*z is also a pentagonal number.
(ii) Any integer n > 43 can be written as the sum of five pentagonal numbers u, v, x, y and z such that u + 2*v + 5*x + 7*y + 10*z is also a pentagonal number. Also, each integer n > 10 can be written as the sum of five pentagonal numbers u, v, x, y and z such that u + 2*v + 5*x + 7*y + 10*z is a square.
(iii) Any natural number n can be written as u^2 + v^2 + x^2 + y^2 + z^2 with u^2 + 2*v^2 + 3*x^2 + 4*y^2 + 5*z^2 a square, where u, v, x, y and z are integers.
As conjectured by Fermat and proved by Cauchy, each natural number can be written as the sum of five pentagonal numbers.
See also A271510, A271513, A271518 and A271644 for some similar conjectures refining Lagrange's four-square theorem.
LINKS
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
EXAMPLE
a(7) = 1 since 7 = 5 + 0 + 1 + 0 + 1 = pen(2) + pen(0) + pen(1) + pen(0) + pen(1) with 2 + 2*0 + 4*1 + 5*0 + 6*1 = 12 = pen(3).
a(9) = 1 since 9 = 1 + 1 + 5 + 1 + 1 = pen(1) + pen(1) + pen(2) + pen(1) + pen(1) with 1 + 2*1 + 4*2 + 5*1 + 6*1 = 22 = pen(4).
a(22) = 1 since 22 = 0 + 0 + 5 + 12 + 5 = pen(0) + pen(0) + pen(2) + pen(3) + pen(2) with 0 + 2*0 + 4*2 + 5*3 + 6*2 = 35 = pen(5).
a(43) = 1 since 43 = 5 + 1 + 35 + 1 + 1 = pen(2) + pen(1) + pen(5) + pen(1) + pen(1) with 2 + 2*1 + 4*5 + 5*1 + 6*1 = 35 = pen(5).
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
pen[x_]:=pen[x]=x*(3x-1)/2
pQ[n_]:=pQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1, 6]==0)
Do[r=0; Do[If[pQ[n-pen[x]-pen[y]-pen[z]-pen[w]]&&pQ[x+2y+4z+5w+6*Floor[(Sqrt[24(n-pen[x]-pen[y]-pen[z]-pen[w])+1]+1)/6]], r=r+1], {x, 0, (Sqrt[24n+1]+1)/6}, {y, 0, (Sqrt[24(n-pen[x])+1]+1)/6}, {z, 0, (Sqrt[24(n-pen[x]-pen[y])+1]+1)/6}, {w, 0, (Sqrt[24(n-pen[x]-pen[y]-pen[z])+1]+1)/6}]; Print[n, " ", r]; Label[aa]; Continue, {n, 0, 70}]
CROSSREFS
Sequence in context: A133083 A083921 A119672 * A087740 A029439 A225743
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 10 2016
STATUS
approved

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Last modified April 19 08:39 EDT 2024. Contains 371782 sequences. (Running on oeis4.)