OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 112, 770, 801, 1593, 1826, 2320, 2334, 2849, 7561.
(ii) Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Any natural number n can be written as P(x,y) + z^5, where x, y and z are integers with |z^5| <= n, and the polynomial P(x,y) is either of the following ones: T(x)+2*T(y), T(x)+2*pen(y), x^2+pen(y), x^2+y(5y+1)/2, 2*T(x)+pen(y), pen(x)+pen(y), pen(x)+y(3y+j) (j = 1,2), pen(x)+6*T(y), pen(x)+y(7y+j)/2 (j = 1,3,5), pen(x)+y(4y+j) (j = 1,3), pen(x)+y(5y+j) (j = 1,2,3,4), pen(x)+y(13y+7)/2, x(5x+i)/2+y(3y+j) (i = 1,3; j = 1,2), x(5x+j)/2+y(7y+5)/2 (j = 1,3).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.
EXAMPLE
a(1) = 1 since 1 = 1*2/2 + 1^2 + (-1)^5 with |(-1)^5| <= 1.
a(112) = 1 since 112 = 10*11/2 + 5^2 + 2^5.
a(770) = 1 since 770 = 28*29/2 + 11^2 + 3^5.
a(801) = 1 since 801 = 45*46/2 + 3^2 + (-3)^5 with |(-3)^5| < 801.
a(1593) = 1 since 1593 = 49*50/2 + 20^2 + (-2)^5 with |(-2)^5| < 1593.
a(1826) = 1 since 1826 = 55*56/2 + 23^2 + (-3)^5 with |(-3)^5| < 1826.
a(2320) = 1 since 2320 = 5*6/2 + 48^2 + 1^5.
a(2334) = 1 since 2334 = 11*12/2 + 45^2 + 3^5.
a(2849) = 1 since 2849 = 70*71/2 + 11^2 + 3^5.
a(7561) = 1 since 7561 = 97*98/2 + 53^2 + (-1)^5 with |(-1)^5| < 7561.
MATHEMATICA
TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-(-1)^k*x^5-y^2], r=r+1], {k, 0, 1}, {x, 0, n^(1/5)}, {y, 1, Sqrt[n-(-1)^k*x^5]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 25 2016
STATUS
approved