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Revision History for A270559 (Underlined text is an addition; strikethrough text is a deletion.)

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A270559 Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.
(history; published version)
#8 by N. J. A. Sloane at Sat Mar 19 00:07:22 EDT 2016
STATUS

proposed

approved

#7 by Zhi-Wei Sun at Fri Mar 18 22:54:32 EDT 2016
STATUS

editing

proposed

#6 by Zhi-Wei Sun at Fri Mar 18 22:54:08 EDT 2016
COMMENTS

In contrast, the author conjectured in A262813 that any positive integer can be expressed as the sum of a nonnegative cube, a square and a positive triangular number.

#5 by Zhi-Wei Sun at Fri Mar 18 22:49:58 EDT 2016
LINKS

Zhi-Wei Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127(2007), 103-113.

#4 by Zhi-Wei Sun at Fri Mar 18 22:48:36 EDT 2016
LINKS

Zhi-Wei Sun, <a href="/A270559/b270559.txt">Table of n, a(n) for n = 1..10000</a>

EXAMPLE

a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.

#3 by Zhi-Wei Sun at Fri Mar 18 22:47:22 EDT 2016
NAME

Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.

COMMENTS

Conjecture: : (i) a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.

(ii) a(n) = 1 only for n = 1, 2, 8, 20, 62, 97, 296, 1493, 4283, 4346, 5433.

EXAMPLE

a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.

a(2) = 1 since 2 = (-1)^4 + (-1)^3 + 1^2 + 1*2/2.

a(8) = 1 since 8 = 1^4 + 1^3 + 0^2 + 3*4/2.

a(20) = 1 since 20 = (-2)^4 + (-2)^3 + 3^2 + 2*3/2.

a(62) = 1 since 62 = (-2)^4 + (-2)^3 + 3^2 + 9*10/2.

a(97) = 1 since 97 = 1^4 + 1^3 + 2^2 + 13*14/2.

a(296) = 1 since 296 = (-4)^4 + (-4)^3 + 7^2 + 10*11/2.

a(1493) = 1 since 1493 = (-2)^4 + (-2)^3 + 0^2 + 54*55/2.

a(4283) = 1 since 4283 = (-6)^4 + (-6)^3 + 50^2 + 37*38/2.

a(4346) = 1 since 4346 = (-3)^4 + (-3)^3 + 49^2 + 61*62/2.

a(5433) = 1 since 5433 = (-8)^4 + (-8)^3 + 14^2 + 57*58/2.

MATHEMATICA

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]

CROSSREFS

Cf. A000217, A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.

#2 by Zhi-Wei Sun at Fri Mar 18 22:13:39 EDT 2016
NAME

allocated for Zhi-Wei Sun

Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.

DATA

1, 1, 2, 2, 2, 2, 3, 1, 3, 4, 2, 5, 2, 3, 4, 2, 3, 4, 5, 1, 4, 3, 3, 4, 3, 4, 5, 5, 3, 6, 5, 3, 3, 6, 2, 4, 6, 3, 9, 4, 2, 3, 4, 3, 7, 6, 3, 6, 2, 4, 2, 6, 5, 7, 6, 4, 5, 3, 6, 4, 11, 1, 5, 9, 3, 6, 5, 3, 8, 8

OFFSET

1,3

COMMENTS

Conjecture: a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.

MATHEMATICA

TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]

Do[r=0; Do[If[x!=0&&TQ[n-y^2-x^4-x^3], r=r+1], {y, 0, Sqrt[n]}, {x, -1-Floor[(n-y^2)^(1/4)], (n-y^2)^(1/4)}]; Print[n, " ", r]; Continue, {n, 1, 10000}]

CROSSREFS

Cf. A000217, A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533.

KEYWORD

allocated

nonn,new

AUTHOR

Zhi-Wei Sun, Mar 18 2016

STATUS

approved

editing

#1 by Zhi-Wei Sun at Fri Mar 18 22:13:39 EDT 2016
NAME

allocated for Zhi-Wei Sun

KEYWORD

allocated

STATUS

approved

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Last modified August 29 23:09 EDT 2024. Contains 375519 sequences. (Running on oeis4.)