cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A306225 Number of ways to write n as w + x^5 + pen(y) + pen(z), where w is 0 or 1, and x,y,z are integers with x >= w and pen(y) < pen(z), and where pen(m) denotes the pentagonal number m*(3*m-1)/2.

Original entry on oeis.org

1, 2, 3, 2, 2, 2, 4, 4, 4, 2, 1, 2, 3, 4, 3, 3, 4, 3, 3, 2, 2, 3, 3, 4, 2, 2, 4, 5, 5, 2, 2, 1, 3, 4, 5, 4, 5, 6, 6, 6, 6, 7, 4, 4, 4, 3, 5, 5, 6, 4, 3, 6, 5, 5, 5, 4, 5, 6, 9, 7, 4, 4, 5, 5, 3, 5, 4, 4, 4, 5, 4, 6, 8, 7, 5, 2, 6, 5, 8, 6, 3, 3, 5, 7, 6, 4, 3, 3, 4, 5, 5, 6, 7, 9, 5, 4, 4, 5, 6, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 30 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer can be written as the sum of two fifth powers of nonnegative integers one of which is 0 or 1, and two distinct generalized pentagonal numbers.
We have verified a(n) > 0 for all n = 1..2*10^6. The conjecture implies that the set A = {x^5 + pen(y): x = 0,1,2,... and y is an integer} is an additive basis of order two (i.e., the sumset A + A coincides with {0,1,2,...}).
See also A306227 for a similar conjecture.

Examples

			a(11) = 1 with 11 = 1 + 1^5 + pen(-1) + pen(-2).
a(1000) = 1 with 1000 = 0 + 2^5 + pen(8) + pen(-24).
a(5104) = 1 with 5104 = 1 + 3^5 + pen(-3) + pen(57).
a(8196) = 1 with 8196 = 0 + 2^5 + pen(48) + pen(-56).
a(9537) = 1 with 9537 = 1 + 6^5 + pen(17) + pen(30).
a(15049) = 1 with 15049 = 0 + 6^5 + pen(-44) + pen(54).
a(16775) = 1 with 16775 = 1 + 5^5 + pen(-17) + pen(94).

Links

Crossrefs

Cf. A000584, A001318, A270920, A306227.

Programs

  • Mathematica
    PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-x-y^5-z(3z-1)/2],r=r+1],{x,0,Min[1,(n-1)/2]},{y,x,(n-1-x)^(1/5)},{z,-Floor[(Sqrt[12(n-1-x-y^5)+1]-1)/6],(Sqrt[12(n-1-x-y^5)+1]+1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]

A306239 Number of ways to write n as x^3 + y^3 + pen(z) + pen(w), where x, y, z, w are nonnegative integers with x <= y and z <= w, and pen(k) denotes the pentagonal number k*(3*k-1)/2.

Original entry on oeis.org

1, 2, 3, 2, 1, 1, 2, 2, 2, 2, 3, 2, 2, 3, 4, 2, 1, 2, 3, 2, 1, 3, 3, 2, 3, 3, 3, 2, 4, 4, 2, 2, 3, 4, 2, 4, 5, 4, 3, 2, 5, 3, 2, 3, 4, 4, 1, 2, 3, 3, 3, 4, 6, 3, 3, 3, 4, 3, 3, 4, 4, 2, 3, 4, 5, 5, 4, 4, 2, 2, 5, 8, 7, 4, 4, 5, 3, 5, 6, 7, 2, 3, 5, 3, 5, 2, 5, 5, 4, 4, 3, 6, 5, 4, 6, 3, 2, 4, 8, 5, 5
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 31 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 4, 5, 16, 20, 46. Also, any nonnegative integer not equal to 16 can be written as x^6 + y^3 + pen(z) + pen(w) with x, y, z, w nonnegative integers.
We have verified a(n) > 0 for all n = 0..5*10^6.

Examples

			a(4) = 1 with 4 = 1^3 + 1^3 + pen(1) + pen(1).
a(5) = 1 with 5 = 0^3 + 0^3 + pen(0) + pen(2).
a(16) = 1 with 16 = 2^3 + 2^3 + pen(0) + pen(0).
a(20) = 1 with 20 = 0^3 + 2^3 + pen(0) + pen(3).
a(46) = 1 with 46 = 1^3 + 1^3 + pen(4) + pen(4).

Links

Crossrefs

Cf. A000326, A000578, A306225, A306227.

Programs

  • Mathematica
    PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0);
tab={};Do[r=0;Do[If[PenQ[n-x^3-y^3-z(3z-1)/2],r=r+1],{x,0,(n/2)^(1/3)},{y,x,(n-x^3)^(1/3)},{z,0,(Sqrt[12(n-x^3-y^3)+1]+1)/6}];tab=Append[tab,r],{n,0,100}];Print[tab]

A334086 Positive numbers not of the form 2*x^4 + y*(y+1)/2 + z*(z+1)/2 with x,y,z nonnegative integers.

Original entry on oeis.org

19, 82, 109, 118, 145, 149, 271, 280, 296, 349, 350, 371, 392, 454, 491, 643, 670, 692, 754, 755, 923, 937, 986, 989, 1021, 1031, 1150, 1189, 1210, 1294, 1346, 1372, 1610, 1682, 1699, 1720, 1819, 1913, 2050, 2065, 2141, 2227, 2479, 2524, 2753, 2996, 3184, 3451, 3590, 3805, 3968, 4129, 4139, 4199, 4261, 4706
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 14 2020

Keywords

Comments

Conjecture: The sequence has totally 216 terms as listed in the b-file.
As none of the 216 terms in the b-file is divisible by 3, the conjecture implies that for each nonnegative integer n we can write 3*n as 2*x^4 + y*(y+1)/2 + z*(z+1)/2 and hence 12*n+1 = 8*x^4 + (y+z+1)^2 + (y-z)^2, where x,y,z are integers.
Our computation indicates that after the 216-th term 4592329 there are no other terms below 10^8.
It is known that each n = 0,1,2,... can be written as the sum of two triangular numbers and twice a square.
a(217) > 10^9, if it exists. - Giovanni Resta, Apr 14 2020

Examples

			a(1) = 19 since 19 is the first nonnegative integer which cannot be written as the sum of two triangular numbers and twice a fourth power.

Links

Crossrefs

Cf. A000217, A000583, A115160, A290491, A306227, A334113.

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[Do[If[TQ[n-2x^4-y(y+1)/2],Goto[aa]],{x,0,(n/2)^(1/4)},{y,0,(Sqrt[4(n-2x^4)+1]-1)/2}];tab=Append[tab,n];Label[aa],{n,0,5000}];Print[tab]

A334113 Positive numbers not of the form 4*x^4 + y*(y+1)/2 + z*(z+1)/2, where x,y,z are nonnegative integers.

Original entry on oeis.org

23, 44, 54, 63, 117, 138, 149, 162, 180, 188, 243, 251, 261, 270, 287, 294, 398, 401, 458, 512, 611, 657, 684, 693, 734, 797, 842, 863, 914, 932, 936, 945, 987, 1029, 1047, 1098, 1323, 1401, 1449, 1472, 1484, 1494, 1574, 1608, 1637, 1769, 1792, 1799, 1823, 1839, 1902, 1995
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 14 2020

Keywords

Comments

Conjecture: The sequence only has 602 terms as listed in the b-file.
Our computation indicates that after the 602-th term 31737789 there are no other terms below 10^8.
It is known that each n = 0,1,2,... can be written as the sum of an even square and two triangular numbers.

Links

Crossrefs

Cf. A000217, A000583, A115160, A306227, A334086.

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[Do[If[TQ[n-4x^4-y(y+1)/2],Goto[aa]],{x,0,(n/4)^(1/4)},{y,0,(Sqrt[4(n-4x^4)+1]-1)/2}];tab=Append[tab,n];Label[aa],{n,0,2000}];Print[tab]

A306240 Number of ways to write n as x^9 + y^3 + z*(z+1) + w*(w+1), where x,y,z,w are nonnegative integers with x <= 2 and z <= w.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 1, 3, 5, 4, 3, 2, 1, 1, 2, 4, 4, 3, 3, 2, 2, 3, 4, 3, 2, 3, 4, 5, 6, 4, 2, 2, 2, 2, 3, 5, 5, 4, 4, 4, 4, 2, 1, 3, 4, 5, 5, 3, 2, 2, 2, 3, 4, 4, 5, 4, 2, 4, 6, 5, 2, 2, 3, 4, 6, 6, 4, 4, 5, 3, 3, 6, 6, 4, 3, 3, 3, 3, 3, 5, 7, 6, 5, 3, 3, 4, 3, 5, 6, 4, 3, 4, 4, 3, 5, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 31 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 11, 17, 18, 47, 108, 109, 234, 359. Also, any nonnegative integer can be written as x^6 + y^3 + z*(z+1) + w*(w+1), where x,y,z,w are nonnegative integers with x <= 2.
We have verified a(n) > 0 for all n = 0..2*10^7.

Examples

			a(11) = 1 with 11 = 1^9 + 2^3 + 0*1 + 1*2.
a(18) = 1 with 18 = 0^9 + 0^3 + 2*3 + 3*4.
a(109) = 1 with 109 = 1^9 + 4^3 + 1*2 + 6*7.
a(234) = 1 with 234 = 0^9 + 6^3 + 2*3 + 3*4.
a(359) = 1 with 359 = 1^9 + 2^3 + 10*11 + 15*16.
a(1978) = 3 with 1978 = 2^9 + 2^3 + 26*27 + 27*28 = 2^9 + 6^3 + 19*20 + 29*30 = 2^9 + 6^3 + 24*25 + 25*26.

Links

Crossrefs

Cf. A000578, A001014, A001017, A002378, A262813, A262815, A270488, A306225, A306227, A306239.

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[4n+1]];
tab={};Do[r=0;Do[If[TQ[n-x^9-y^3-z(z+1)],r=r+1],{x,0,Min[2,n^(1/9)]},{y,0,(n-x^9)^(1/3)},{z,0,(Sqrt[2(n-x^9-y^3)+1]-1)/2}];tab=Append[tab,r],{n,0,100}];Print[tab]

A306249 Number of ways to write n as x*(2x-1) + y*(3y-1) + z*(4z-1) + w*(5w-1), where x,y,z are nonnegative integers and w is 0 or 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 1, 2, 3, 4, 2, 3, 3, 3, 4, 2, 2, 1, 4, 3, 1, 1, 5, 4, 3, 3, 3, 4, 4, 3, 1, 3, 3, 5, 1, 2, 4, 5, 4, 4, 2, 3, 7, 3, 3, 2, 5, 3, 3, 2, 2, 3, 4, 5, 1, 4, 6, 6, 2, 3, 5, 3, 3, 3, 5, 4, 5, 5, 3, 6, 6, 4, 3, 4, 5, 2, 3, 4, 4, 5, 2, 2, 5, 6, 6, 1, 5, 3, 6, 2, 4, 3, 4, 4, 2
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 31 2019

Keywords

Comments

Conjecture: a(n) > 0 for any nonnegative integer n.
This has been verified for n up to 10^6. By Theorem 1.2 of the linked 2017 paper of the author, any nonnegative integer can be written as x*(2x-1) + y*(3y-1) + z*(4z-1) with x,y,z integers.
We have some other similar conjectures. For example, we conjecture that each n = 0,1,2,... can be written as x*(3x-1)/2 + y*(5y-1)/2 + z*(7z-1)/2 + w*(9w-1)/2) (or x*(x-1) + y*(2y-1) + z*(3z-1) + w*(4w-1)) with x,y,z,w nonnegative integers.

Examples

			a(1) = 1 with 1 = 1*(2*1-1) + 0*(3*0-1) + 0*(4*0-1) + 0*(5*0-1).
a(2) = 1 with 2 = 0*(2*0-1) + 1*(3*1-1) + 0*(4*0-1) + 0*(5*0-1).
a(12) = 1 with 12 = 2*(2*2-1) + 1*(3*1-1) + 0*(4*0-1) + 1*(5*1-1).
a(26) = 1 with 26 = 2*(2*2-1) + 1*(3*1-1) + 2*(4*2-1) + 1*(5*1-1).
a(220) = 1 with 220 = 6*(2*6-1) + 7*(3*7-1) + 2*(4*2-1) + 0*(5*0-1).
a(561) = 1 with 561 = 17*(2*17-1) + 0*(3*0-1) + 0*(4*0-1) + 0*(5*0-1).
a(1356) = 1 with 1356 = 23*(2*23-1) + 1*(3*1-1) + 9*(4*9-1) + 1*(5*1-1).

Links

Crossrefs

Cf. A000326, A000384, A002378, A033991, A049450, A255350, A306225, A306227, A306239, A306242.

Programs

  • Mathematica
    HexQ[n_]:=HexQ[n]=IntegerQ[Sqrt[8n+1]]&&(n==0||Mod[Sqrt[8n+1]+1,4]==0);
tab={};Do[r=0;Do[If[HexQ[n-x(5x-1)-y(4y-1)-z(3z-1)],r=r+1],{x,0,Min[1,(Sqrt[20n+1]+1)/10]},{y,0,(Sqrt[16(n-x(5x-1))+1]+1)/8},{z,0,(Sqrt[12(n-x(5x-1)-y(4y-1))+1]+1)/6}];tab=Append[tab,r],{n,0,100}];Print[tab]

A306260 Number of ways to write n as w*(4w+1) + x*(4x-1) + y*(4y-2) + z*(4z-3) with w,x,y,z nonnegative integers.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 4, 2, 3, 3, 2, 4, 4, 3, 1, 2, 1, 2, 3, 1, 2, 5, 5, 4, 5, 5, 4, 3, 1, 2, 4, 4, 4, 4, 5, 5, 7, 2, 2, 5, 3, 4, 5, 5, 3, 7, 4, 2, 5, 2, 4, 7, 6, 6, 6, 5, 6, 5, 3, 5, 6, 5, 8, 9, 8, 4, 7, 2, 4, 9, 2, 6, 5, 8, 6, 7, 7, 2, 6, 4, 4, 12, 6, 5, 5, 7, 9, 8, 5, 6, 9, 8
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 01 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 1, 2, 4, 7, 9, 11, 14, 23, 25, 28, 37.
Conjecture 2: Each n = 0,1,2,... can be written as w*(4w+2) + x*(4x-1) + y*(4y-2) + z*(4z-3) with w,x,y,z nonnegative integers.
Conjecture 3: Each n = 0,1,2,... can be written as 4*w^2 + x*(4x+1) + y*(4y-2) + z*(4z-3) with w,x,y,z nonnegative integers.
We have verified that a(n) > 0 for all n = 0..2*10^6. By Theorem 1.3 in the linked 2017 paper of the author, any nonnegative integer can be written as x*(4x-1) + y*(4y-2) + z*(4z-3) with x,y,z integers.

Examples

			a(11) = 1 with 11 = 1*(4*1+1) + 1*(4*1-1) + 1*(4*1-2) + 1*(4*1-3).
a(23) = 1 with 23 = 2*(4*2+1) + 1*(4*1-1) + 1*(4*1-2) + 0*(4*0-3).
a(25) = 1 with 25 = 0*(4*0+1) + 1*(4*1-1) + 2*(4*2-2) + 2*(4*2-3).
a(28) = 1 with 28 = 2*(4*2+1) + 0*(4*0-1) + 0*(4*0-2) + 2*(4*2-3).
a(37) = 1 with 37 = 1*(4*1+1) + 1*(4*1-1) + 1*(4*1-2) + 3*(4*3-3).

Links

Crossrefs

Cf. A001107, A002939, A007742, A033991, A255350, A306225, A306227, A306239, A306240, A306249, A306250.

Programs

  • Mathematica
    QQ[n_]:=QQ[n]=IntegerQ[Sqrt[16n+1]]&&Mod[Sqrt[16n+1],8]==1;
tab={};Do[r=0;Do[If[QQ[n-x(4x-1)-y(4y-2)-z(4z-3)],r=r+1],{x,0,(Sqrt[16n+1]+1)/8},{y,0,(Sqrt[4(n-x(4x-1))+1]+1)/4},{z,0,(Sqrt[16(n-x(4x-1)-y(4y-2))+9]+3)/8}];tab=Append[tab,r],{n,0,100}];Print[tab]
Showing 1-7 of 7 results.

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