

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*(3k1)/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 foursquare theorem.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..1500
Z.W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 13671396.


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*(3x1)/2
pQ[n_]:=pQ[n]=SQ[24n+1]&&(n==0Mod[Sqrt[24n+1]+1, 6]==0)
Do[r=0; Do[If[pQ[npen[x]pen[y]pen[z]pen[w]]&&pQ[x+2y+4z+5w+6*Floor[(Sqrt[24(npen[x]pen[y]pen[z]pen[w])+1]+1)/6]], r=r+1], {x, 0, (Sqrt[24n+1]+1)/6}, {y, 0, (Sqrt[24(npen[x])+1]+1)/6}, {z, 0, (Sqrt[24(npen[x]pen[y])+1]+1)/6}, {w, 0, (Sqrt[24(npen[x]pen[y]pen[z])+1]+1)/6}]; Print[n, " ", r]; Label[aa]; Continue, {n, 0, 70}]


CROSSREFS

Cf. A000290, A000326, A271510, A271513, A271518, A271644.
Sequence in context: A133083 A083921 A119672 * A087740 A029439 A225743
Adjacent sequences: A271605 A271606 A271607 * A271609 A271610 A271611


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 10 2016


STATUS

approved



