

A306606


Number of ways to write n as w^2 + T(x)^2 + Pen(y)^2 + 2*Pen(z)^2, where w,x,y,z are integers with w > 0 and x >= 0, and T(m) = m*(m+1)/2 and Pen(m) = m*(3m1)/2.


2



1, 2, 2, 3, 4, 3, 3, 3, 3, 6, 5, 4, 7, 5, 2, 5, 5, 6, 6, 3, 4, 5, 1, 2, 5, 6, 5, 5, 6, 3, 3, 2, 2, 6, 5, 3, 9, 7, 3, 6, 4, 5, 8, 3, 6, 7, 2, 4, 7, 6, 8, 9, 9, 6, 6, 5, 3, 9, 9, 6, 11, 7, 4, 10, 5, 6, 13, 5, 5, 9, 2, 5, 7, 7, 7, 10, 7, 3, 5, 3, 3, 6, 6, 5, 11, 6, 6, 9, 4, 6, 11, 3, 5, 9, 2, 5, 4, 4, 7, 11
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OFFSET

1,2


COMMENTS

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as s^2 + t^2 + u^2 + 2*v^2, where s is a positive integer, t is a triangular number, u and v are generalized pentagonal numbers.
I have verified a(n) > 0 for all n = 1..5*10^6.
It is well known that any positive odd integer can be written as x^2 + y^2 + 2*z^2 with x,y,z integers.
See also A306614 for similar conjectures.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.


EXAMPLE

a(1) = 1 with 1 = 1^2 + T(0)^2 + Pen(0)^2 + 2*Pen(0)^2.
a(23) = 1 with 23 = 4^2 + T(1)^2 + Pen(1)^2 + 2*Pen(1)^2.
a(335) = 1 with 335 = 18^2 + T(2)^2 + Pen(0)^2 + 2*Pen(1)^2.
a(3695) = 1 with 3695 = 53^2 + T(7)^2 + Pen(1)^2 + 2*Pen(2)^2.


MATHEMATICA

t[x_]:=t[x]=x(x+1)/2; p[x_]:=p[x]=x(3x1)/2;
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[nt[x]^2p[y]^22*p[z]^2], r=r+1], {x, 0, (Sqrt[8*Sqrt[n1]+1]1)/2}, {y, Floor[(Sqrt[24*Sqrt[n1t[x]^2]+1]1)/6], (Sqrt[24*Sqrt[n1t[x]^2]+1]+1)/6},
{z, Floor[(Sqrt[24*Sqrt[(n1t[x]^2p[y]^2)/2]+1]1)/6], (Sqrt[24*Sqrt[(n1t[x]^2p[y]^2)/2]+1]+1)/6}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]


CROSSREFS

Cf. A000217, A000290, A001318, A275344, A306614.
Sequence in context: A307981 A262824 A242899 * A229989 A126688 A054703
Adjacent sequences: A306603 A306604 A306605 * A306607 A306608 A306609


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 28 2019


STATUS

approved



